If a star emitting light of wavelength 500 nm is moving towards earth with a velocity of $1.5{\rm{ }} \times {\rm{ }}{10^6}m/s$ then the shift in the wavelength due to doppler’s effect will be:

$25\,\,\mathop {\rm{A}}\limits^ \circ $

$250\,\,\mathop {\rm{A}}\limits^ \circ $

$2.5\,\,\mathop {\rm{A}}\limits^ \circ $

WAVE OPTICS MCQs for NEET UG PHYSICS — Medium, Page 2

31 Medium

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Understanding +4 | -1

To increase the resolving power of a telescope

View answer and explanation

Correct option: C

Both (1) & (2)

Option C is correct. The resolving power of a telescope increases when the wavelength of light is decreased and the aperture of the objective lens is increased.

Solution approach

Correct Answer + Core Reason: Both (1) & (2). Resolving power increases with decreasing wavelength and increasing aperture.

Step-by-Step Solution:

Step 1: Recall the formula for the resolving power (RP) of a telescope: $RP = \frac{d}{1.22 \lambda}$ , where $d$ is the diameter (aperture) of the objective lens and $\lambda$ is the wavelength of light.

Step 2: Analyze the formula. The resolving power is directly proportional to the aperture of the objective lens ( $d$ ) and inversely proportional to the wavelength of light ( $\lambda$ ).

Step 3: Determine the conditions for increasing resolving power. To increase the resolving power, the wavelength of light has to be decreased and the aperture of the objective lens has to be increased.

Why other options are wrong:
- Option A: Only decreasing the wavelength is not the complete answer.
- Option B: Only increasing the aperture is not the complete answer.
- Option D: Decreasing both focal length and aperture would decrease resolving power (aperture) and not directly relate to wavelength in the resolving power formula.

Key Concepts:
- Resolving power of a telescope
- Wavelength of light
- Aperture of objective lens

Memory Aids:
- Remember that shorter wavelengths and larger apertures allow for better resolution.

Quick Tips:
- Consider both wavelength and aperture when trying to improve the resolving power of a telescope.

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Application +4 | -1

Find the thickness of a plate which will produce a change in optical path equal to half the wavelength

\lambda

of the light passing through it normally. The refractive index of the plate

\mu

is

View answer and explanation

Correct option: D

\frac{\lambda }{{2\left( {\mu - 1} \right)}}

Option D (

\frac{\lambda }{{2\left( {\mu - 1} \right)}}

) is correct because it represents the thickness required to introduce an optical path difference of half the wavelength.

Solution approach

Correct Answer + Core Reason: $\frac{\lambda }{{2\left( {\mu - 1} \right)}}$ . This represents the thickness required to introduce an optical path difference of half the wavelength.

Step-by-Step Solution:

Step 1: Identify the given information:\nOptical path difference, $\Delta x = \frac{\lambda}{2}$ \nRefractive index of the plate, $\mu$ \nThickness of the plate, $t = ?$

Step 2: Key Formula: Optical path difference due to a plate of thickness $t$ and refractive index $\mu$ is given by:
$\Delta x = (\mu - 1)t$

Step 3: Apply the formula and solve for $t$ :
$\frac{\lambda}{2} = (\mu - 1)t$
$t = \frac{\lambda}{{2(\mu - 1)}}$

Why other options are wrong:
- Option A: $\frac{\lambda }{{4\left( {\mu - 1} \right)}}$ : This is incorrect because it would produce a change in optical path equal to one-fourth the wavelength.
- Option B: $\frac{{2\lambda }}{{4\left( {\mu - 1} \right)}}$ : This simplifies to $\frac{\lambda }{{2\left( {\mu - 1} \right)}}$ , which is the correct answer. However, the form is not the simplest.
- Option C: $\frac{\lambda }{{\left( {\mu - 1} \right)}}$ : This is incorrect because it would produce a change in optical path equal to the full wavelength.

Key Concepts:
- Optical path difference
- Refractive index

Memory Aids:
- $\Delta x = (\mu - 1)t$

Quick Tips:
- Remember the formula for optical path difference.

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Application +4 | -1

A beam of ordinary unpolarised light passes through a tourmaline crystal

{C_1}

and then it passes through another tourmaline crystal

{C_2}

, which is oriented such that its principal plane is parallel to that of

{C_2}

. The intensity if emergent light is

{I_0}

. Now

{C_2}

is rotated by

60^\circ

about the ray. The emergent ray will have an intensity

View answer and explanation

Correct option: C

{I_0}/4

The correct answer is C (

{I_0}/4

) because the intensity is reduced by a factor of cos²(θ) after passing through the second polarizer, and the initial intensity is already halved by the first polarizer.

Solution approach

Correct Answer + Core Reason: ${I_0}/4$ . The intensity of the emergent light is calculated using Malus's law.

Step-by-Step Solution:

Step 1: When unpolarized light passes through the first tourmaline crystal $C_1$ , the intensity is reduced to half, i.e., $I_1 = \frac{I_0}{2}$ .

Step 2: The second tourmaline crystal $C_2$ is initially oriented such that its principal plane is parallel to that of $C_1$ . So, the intensity remains the same, $I_2 = I_1 = \frac{I_0}{2}$ .

Step 3: When $C_2$ is rotated by $60^\circ$ , the angle between the principal planes of $C_1$ and $C_2$ is $\theta = 60^\circ$ .

Step 4: According to Malus's law, the intensity of the transmitted light is $I = I_1 \cos^2 \theta$ .

Step 5: $I = \frac{I_0}{2} \cos^2 (60^\circ) = \frac{I_0}{2} \left( \frac{1}{2} \right)^2 = \frac{I_0}{2} \times \frac{1}{4} = \frac{I_0}{8}$ . However, the initial intensity after the first crystal is I0/2. After rotating the second crystal by 60 degrees, the intensity becomes (I0/2)cos²(60) = (I0/2)(1/4) = I0/8. The question states that the initial intensity is I0, so the intensity after the first crystal is I0/2. When the second crystal is rotated by 60 degrees, the intensity becomes I = (I0/2)cos²(60) = (I0/2)(1/4) = I0/8. The question is asking for the intensity *after* the rotation, given the initial intensity *after* the first crystal is I0. So, the answer should be I0/8. There seems to be an error in the options. Let's re-evaluate. The initial intensity is I0. After the first crystal, it's I0/2. The second crystal is rotated by 60 degrees, so the intensity becomes (I0/2)cos²(60) = (I0/2)(1/4) = I0/8. But the closest answer is I0/4, which would be the answer if the rotation was 45 degrees. Let's assume the question meant the *first* crystal reduces the intensity to I0/2, and we are looking for the intensity *relative* to that I0/2. Then, the answer is (I0/2) * cos²(60) = (I0/2) * (1/4) = I0/8. However, none of the options match this. Let's assume the initial I0 is *after* the first polarizer. Then the intensity after the second polarizer is I = I0 cos²(60) = I0 * (1/4) = I0/4.

Why other options are wrong:
- Option A: This is incorrect as it is not the correct intensity.
- Option B: This is incorrect as it is not the correct intensity.
- Option D: This is incorrect as it is not the correct intensity.

Key Concepts:
- Polarization
- Malus's law

Memory Aids:
- Remember Malus's law: $I = I_0 \cos^2 \theta$ .

Quick Tips:
- Pay attention to the angle between the polarizers.

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Understanding +4 | -1

In Young’s double slit experiment, the distance between the two slits is made half, then the fringe width will become

View answer and explanation

Correct option: B

Double

Option B (Double) is correct because fringe width is inversely proportional to the distance between the slits. Halving the distance doubles the fringe width.

Solution approach

Correct Answer + Core Reason: Double. Fringe width is inversely proportional to the slit separation.

Step-by-Step Solution:

Step 1: Problem Identification: The problem describes Young's double-slit experiment and asks how the fringe width changes when the slit separation is halved.

Step 2: Key Formula: Fringe width, $\beta = \frac{{\lambda D}}{d}$ , where $\lambda$ is the wavelength, $D$ is the distance to the screen, and $d$ is the distance between the slits.

Step 3: Application: If the distance between the slits ( $d$ ) is halved (i.e., $d' = \frac{d}{2}$ ), the new fringe width $\beta'$ becomes: $\beta' = \frac{{\lambda D}}{{\frac{d}{2}}} = 2\frac{{\lambda D}}{d} = 2\beta$

Step 4: Conclusion: The fringe width doubles when the distance between the slits is halved.

Why other options are wrong:
- Option A: Half: This is incorrect because the fringe width is inversely proportional to the slit separation, not directly proportional.
- Option C: One fourth: This is incorrect because the fringe width doubles, not becomes one-fourth.
- Option D: Unchanged: This is incorrect because changing the slit separation directly affects the fringe width.

Key Concepts:
- Young's double-slit experiment
- Fringe width formula

Memory Aids:
- $\beta = \frac{{\lambda D}}{d}$ (Fringe width formula)

Quick Tips:
- Remember the inverse relationship between fringe width and slit separation.

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Application +4 | -1

The wavelength of light observed on the earth from a moving star is found to decrease by

0.05\,\,\%

. the star is

View answer and explanation

Correct option: C

Coming closer with a velocity of

1.5 \times {10^5}\,\,\mathrm{m}{\mathrm{s}^{ - 1}}

Option C (Coming closer with a velocity of

1.5 \times {10^5}\,\mathrm{m}{\mathrm{s}^{ - 1}}

) is correct because a decrease in wavelength indicates a blueshift, meaning the star is approaching, and the velocity can be calculated using the Doppler shift formula.

Solution approach

Correct Answer + Core Reason: Coming closer with a velocity of $1.5 \times {10^5}\,\mathrm{m}{\mathrm{s}^{ - 1}}$ . The decrease in wavelength indicates the star is approaching, and the velocity is calculated using the Doppler effect.

Step-by-Step Solution:

Step 1: Identify the given values: Fractional change in wavelength, $\frac{\Delta \lambda}{\lambda} = -0.05\% = -0.0005$ . The negative sign indicates a decrease in wavelength.

Step 2: Use the Doppler shift formula for small velocities: $\frac{\Delta \lambda}{\lambda} = -\frac{v}{c}$ , where $v$ is the velocity of the star and $c$ is the speed of light.

Step 3: Solve for the velocity: $v = -c \frac{\Delta \lambda}{\lambda} = - (3 \times 10^8 \, \mathrm{m/s}) (-0.0005) = 1.5 \times 10^5 \, \mathrm{m/s}$ .

Step 4: Since the wavelength is decreasing, the star is coming closer.

Why other options are wrong:
- **Option A (Coming closer with a velocity of $1.5\, \times \,{10^4}\,\mathrm{m}{\mathrm{s}^{ - 1}}$ ): This is incorrect because the velocity is calculated incorrectly. The percentage change in wavelength is not properly converted to velocity.
- Option B (Moving away with a velocity of $1.5 \times {10^4}\,\mathrm{m}{\mathrm{s}^{ - 1}}$ ): This is incorrect because the star is moving closer, not away, as indicated by the decrease in wavelength.
- Option D (Moving away with a velocity of $1.5\, \times {10^{ - 1}}$ ): This is incorrect because the velocity is calculated incorrectly and the star is moving closer, not away.

Key Concepts:
- Doppler effect
- Redshift and blueshift

Memory Aids:
- Remember that a decrease in wavelength (blueshift) indicates the source is approaching, and an increase in wavelength (redshift) indicates the source is moving away.

Quick Tips:**
- Pay attention to the sign of the change in wavelength to determine whether the source is approaching or moving away.

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Analyzing +4 | -1

Statement A: In the interference pattern the intensity is same at all points in a bright band

Statement B: In Young's double slit experiment, as we move away from the central maximum, the third maximum always comes before the third minimum.

View answer and explanation

Correct option: A

A and B are correct

Option A (A and B are correct) is correct because both statements about interference patterns and Young's double-slit experiment are true.

Solution approach

Correct Answer + Core Reason: A and B are correct. Both statements are factually accurate regarding interference patterns and YDSE.

Step-by-Step Solution:

Step 1: Analyze Statement A: In an ideal interference pattern, the intensity within a bright fringe is uniform.

Step 2: Analyze Statement B: In YDSE, the maxima and minima alternate, and the third maximum will indeed appear before the third minimum as we move away from the central maximum.

Step 3: Combine the analyses: Both statements hold true under standard conditions for interference and YDSE.

Why other options are wrong:
- Option B: This is incorrect as both statements are true.
- Option C: This is incorrect as statement A is true.
- Option D: This is incorrect as statement B is true.

Key Concepts:
- Interference of light
- Young's Double Slit Experiment (YDSE)
- Maxima and Minima in Interference

Memory Aids:
- Remember the alternating pattern of bright and dark fringes in YDSE.

Quick Tips:
- Visualize the interference pattern to confirm the order of maxima and minima.

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Understanding +4 | -1

The phenomenon of polarization of light indicates that

View answer and explanation

Correct option: B

Light is a transverse wave

The phenomenon of polarization of light indicates that light is a transverse wave. Only transverse waves can be polarized, as their oscillations are perpendicular to the direction of propagation.

Solution approach

Correct Answer + Core Reason: Light is a transverse wave. Polarization is a property unique to transverse waves.

Step-by-Step Solution:

Step 1: Define Polarization: Polarization is the phenomenon in which the oscillations of a wave are confined to a single plane.

Step 2: Relate Polarization to Wave Type: Only transverse waves can be polarized because their oscillations are perpendicular to the direction of propagation. Longitudinal waves cannot be polarized because their oscillations are parallel to the direction of propagation.

Why other options are wrong:
- Option A: Longitudinal waves cannot be polarized.
- Option C: Light *is* a wave, as demonstrated by phenomena like interference and diffraction.
- Option D: The speed of light is a constant, but it doesn't directly explain polarization.

Key Concepts:
- Polarization
- Transverse Waves
- Longitudinal Waves

Memory Aids:
- Remember that only transverse waves can be polarized.

Quick Tips:
- Polarization is a key characteristic that distinguishes transverse waves from longitudinal waves.

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Analyzing +4 | -1

Statement A: In Fraunhofer diffraction central observation point is always bright where as in Fresnel diffraction observation point may bright or dark.

Statement B: In Fresnel light waves approaches at observation point are always in same phase, whereas in Fraunhofer light waves approaches at central observation point may be in same phase or in opposite phase.

View answer and explanation

Correct option: C

A is correct and B is wrong

Option C (A is correct and B is wrong) is correct. In Fraunhofer diffraction, the central observation point is always bright. In Fresnel diffraction, it can be bright or dark. In Fresnel diffraction, the light waves arriving at the observation point are not always in the same phase.

Solution approach

Correct Answer + Core Reason: A is correct and B is wrong. Statement A is correct about Fraunhofer and Fresnel diffraction, while Statement B incorrectly describes the phase of light waves in Fresnel diffraction.

Step-by-Step Solution:

Step 1: Analyze Statement A: In Fraunhofer diffraction, the source and the screen are at infinite distances from the diffracting object (or effectively at infinity using lenses). This results in plane wavefronts incident on the diffracting object, and the central observation point is always bright due to constructive interference. In Fresnel diffraction, the source or the screen (or both) are at finite distances from the diffracting object. The wavefronts are not planar, and the central observation point can be either bright or dark depending on the path difference. Therefore, Statement A is correct.

Step 2: Analyze Statement B: In Fresnel diffraction, the light waves arriving at the central observation point are not always in the same phase. The phase depends on the path difference between the waves. In Fraunhofer diffraction, the light waves arriving at the central observation point are in phase. Therefore, Statement B is wrong.

Why other options are wrong:
- Option A: A and B are correct. This is incorrect because Statement B is wrong.
- Option B: A and B are wrong. This is incorrect because Statement A is correct.
- Option C: A is correct and B is wrong. This is the correct answer.
- Option D: A is wrong and B is correct. This is incorrect because Statement A is correct and Statement B is wrong.

Key Concepts:
- Fraunhofer diffraction
- Fresnel diffraction
- Wavefronts
- Phase difference
- Interference

Memory Aids:
- Fraunhofer: Source and screen at infinity, plane wavefronts, central point always bright.
- Fresnel: Source or screen at finite distance, non-planar wavefronts, central point can be bright or dark.

Quick Tips:
- Remember the key differences between Fraunhofer and Fresnel diffraction based on the distances of the source and screen.

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Understanding +4 | -1

Interference may be seen using two independent

View answer and explanation

Correct option: C

Lasers

Option C (Lasers) is correct because lasers are highly coherent light sources, making them suitable for observing interference.

Solution approach

Correct Answer + Core Reason: Lasers. Lasers provide coherent light necessary for sustained interference.

Step-by-Step Solution:

Step 1: Problem Identification: The question asks which light sources can be used to see interference.

Step 2: Key Concept: Interference requires coherent light sources. Coherent sources emit light waves with a constant phase difference and the same frequency.

Step 3: Explanation: Lasers produce highly coherent light due to stimulated emission. Other options like sodium lamps, fluorescent tubes, and mercury vapor lamps produce incoherent light.

Step 4: Conclusion: Lasers are the only suitable option for observing interference among the given choices.

Why other options are wrong:
- Option A: Sodium lamps: Sodium lamps produce nearly monochromatic light, but it is not highly coherent.
- Option B: Fluorescet tubes: Fluorescent tubes produce incoherent light with a broad spectrum.
- Option D: Mercury vapour lamps: Mercury vapor lamps produce incoherent light with a discrete spectrum.

Key Concepts:
- Coherence of light
- Interference of light
- Laser light

Memory Aids:
- Lasers = Coherent light.

Quick Tips:
- Remember that only coherent light sources can produce sustained interference patterns.

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Remembering +4 | -1

Brewster’s angle in terms of refractive index (n) of the medium

View answer and explanation

Correct option: D

{\tan ^{ - 1}}\left[ n \right]

Brewster's angle is given by the formula

\theta_B = \tan^{-1}(n)

, where n is the refractive index of the medium. Therefore, option D is correct.

Solution approach

Correct Answer + Core Reason: ${\tan ^{ - 1}}\left[ n \right]$ . Brewster's angle is defined as the angle of incidence for which light with a particular polarization is perfectly transmitted through a transparent dielectric surface, with no reflection.

Step-by-Step Solution:

Step 1: Recall Brewster's Law: Brewster's law states that the tangent of the polarizing angle ( $\theta_B$ ) is equal to the refractive index (n) of the medium.

Step 2: Write the Formula: Mathematically, this is expressed as $\tan(\theta_B) = n$ .

Step 3: Solve for Brewster's Angle: Therefore, Brewster's angle is given by $\theta_B = \tan^{-1}(n)$ .

Why other options are wrong:
- Option A: This is incorrect; the formula involves the inverse tangent, not the inverse tangent of the square root of n.
- Option B: This is incorrect; the formula involves the inverse tangent, not the inverse sine.
- Option C: This is incorrect; the formula involves the inverse tangent, not the inverse sine of the square root of n.

Key Concepts:
- Brewster's Law
- Polarization by Reflection
- Refractive Index

Memory Aids:
- Remember the formula: $\tan(\theta_B) = n$ .

Quick Tips:
- When dealing with Brewster's angle problems, always remember the relationship between the angle of incidence and the refractive index.

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Understanding +4 | -1

{V_O}\;and\;{V_E}

represent the velocities,

{\mu _O}\;and\;{\mu _E}

the refractive indices of ordinary and extraordinary rays for a doubly refracting crystal. Then

View answer and explanation

Correct option: C

{V_o} \le {V_E},{\mu _o} \ge {\mu _E}

if the crystal is calcite

For a calcite crystal, the refractive index for the ordinary ray is greater than that for the extraordinary ray (

\mu_o > \mu_E

), which implies that the velocity of the ordinary ray is less than that of the extraordinary ray (

V_o < V_E

).

Solution approach

Correct Answer + Core Reason: ${V_o} \le {V_E},{\mu _o} \ge {\mu _E}$ if the crystal is calcite. In calcite, the ordinary ray experiences a higher refractive index and thus a lower velocity compared to the extraordinary ray.

Step-by-Step Solution:

Step 1: Understand Double Refraction: Double refraction (birefringence) is the phenomenon where a ray of light splits into two rays (ordinary and extraordinary) when passing through certain anisotropic materials.

Step 2: Define Ordinary and Extraordinary Rays: The ordinary ray (O-ray) obeys Snell's law, while the extraordinary ray (E-ray) does not.

Step 3: Relate Refractive Index and Velocity: The refractive index ( $\mu$ ) is inversely proportional to the velocity (V) of light in the medium: $\mu = \frac{c}{V}$ , where c is the speed of light in vacuum.

Step 4: Consider Calcite Crystal: For calcite, the refractive index of the ordinary ray is greater than that of the extraordinary ray ( $\mu_o > \mu_E$ ). Therefore, the velocity of the ordinary ray is less than that of the extraordinary ray ( $V_o < V_E$ ).

Why other options are wrong:
- Option A: This is incorrect for calcite; the relationships are reversed.
- Option B: This is incorrect for quartz; the relationships are different for quartz.
- Option D: This is incorrect for quartz; the relationships are different for quartz.

Key Concepts:
- Double Refraction (Birefringence)
- Ordinary Ray
- Extraordinary Ray
- Refractive Index
- Crystal Optics

Memory Aids:
- Remember that in calcite, the ordinary ray has a higher refractive index and lower velocity.

Quick Tips:
- The relationship between refractive index and velocity is inverse: higher refractive index means lower velocity.

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Analyzing +4 | -1

Assertion: If Fraunhoffer diffraction through single slit is observed under monochromatic light, then central maxima cannot be identified from other maxima.

Reason: In Fraunhoffer single slit diffraction under monochromatic light, all maxima has equal width.

View answer and explanation

Correct option: D

If both Assertion and Reason are false statements.

Option D (If both Assertion and Reason are false statements) is correct. The central maxima *can* be identified, and the maxima do *not* have equal width.

Solution approach

Correct Answer + Core Reason: If both Assertion and Reason are false statements. The central maxima can be identified, and the maxima do not have equal width.

Step-by-Step Solution:

Step 1: Analyze the Assertion: In Fraunhofer diffraction through a single slit, the central maxima is the brightest and widest maxima. It can be easily identified from other maxima because its intensity is much higher and its width is twice that of the other secondary maxima. Therefore, the assertion is false.

Step 2: Analyze the Reason: In Fraunhofer single-slit diffraction under monochromatic light, the central maxima has twice the width of the other maxima, and the intensity of the maxima decreases as we move away from the center. Therefore, all maxima do not have equal width, and the reason is false.

Step 3: Since both the assertion and the reason are false, the correct option is D.

Why other options are wrong:
- Option A: If both Assertion & Reason are true and the reason is the correct explanation of the assertion. This is incorrect because both the assertion and reason are false.
- Option B: If both Assertion & Reason are true but the reason is not the correct explanation of the assertion. This is incorrect because both the assertion and reason are false.
- Option C: If Assertion is true statement but Reason is false. This is incorrect because the assertion is also false.
- Option D: If both Assertion and Reason are false statements. This is the correct answer.

Key Concepts:
- Fraunhofer diffraction
- Single-slit diffraction
- Central maxima
- Secondary maxima

Memory Aids:
- The central maxima in single-slit diffraction is the brightest and widest.

Quick Tips:
- Remember the characteristics of the central maxima in single-slit diffraction: it is the brightest and widest.

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Application +4 | -1

In young’s two slit experiment the distance between the two coherent sources is

2\;mm

and the screen is at a distance of

1\;m

. If the fringe width is found to be

0.03\;cm

, then the wavelength of the light used is

View answer and explanation

Correct option: D

6000\mathop \mathrm{A}\limits^ \circ

Option D (

6000\mathop \mathrm{A}\limits^ \circ

) is correct because it is calculated using the formula for fringe width in Young's double-slit experiment.

Solution approach

Correct Answer + Core Reason: $6000\mathop \mathrm{A}\limits^ \circ$ . The wavelength is calculated using the fringe width formula.

Step-by-Step Solution:

Step 1: Identify the given information: Distance between the slits $d = 2\,mm = 2 \times 10^{-3}\,m$ , distance to the screen $D = 1\,m$ , fringe width $\beta = 0.03\,cm = 0.03 \times 10^{-2}\,m = 3 \times 10^{-4}\,m$ .

Step 2: Recall the formula for fringe width: $\beta = \frac{\lambda D}{d}$

Step 3: Solve for the wavelength $\lambda$ : $\lambda = \frac{\beta d}{D}$

Step 4: Substitute the given values: $\lambda = \frac{3 \times 10^{-4} \times 2 \times 10^{-3}}{1}$

Step 5: Calculate the wavelength: $\lambda = 6 \times 10^{-7}\,m = 6000 \times 10^{-10}\,m = 6000\mathop \mathrm{A}\limits^ \circ$

Step 6: Final Answer: $6000\mathop \mathrm{A}\limits^ \circ$

Why other options are wrong:
- **Option A ( $4000\mathop \mathrm{A}\limits^ \circ$ ): This is incorrect because it is a result of incorrect calculation or using the wrong formula.
- Option B ( $5000\mathop \mathrm{A}\limits^ \circ$ ): This is incorrect because it is a result of incorrect calculation or using the wrong formula.
- Option C ( $5890\mathop \mathrm{A}\limits^ \circ$ ): This is incorrect because it is a result of incorrect calculation or using the wrong formula.

Key Concepts:
- Young's Double Slit Experiment
- Fringe Width
- Wavelength of Light

Memory Aids:**
- Remember the formula for fringe width: $\beta = \frac{\lambda D}{d}$

Quick Tips:
- Always convert all units to the same system before calculations.

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Application +4 | -1

Two coherent sources are 0.3 mm apart. They are 1 m away from the screen. The second dark fringe is at a distance of 0.3 cm from the centre. Distance of fourth bright fringe from the centre is

View answer and explanation

Correct option: B

0.8 cm

Option B (0.8 cm) is correct because it is calculated using the formula for the position of bright fringes in a double-slit experiment.

Solution approach

Correct Answer + Core Reason: 0.8 cm. The distance of the fourth bright fringe is calculated using the formula for bright fringes in Young's double-slit experiment.

Step-by-Step Solution:

Step 1: The position of the $n$ -th dark fringe is given by $y_n = (n - \frac{1}{2})\frac{\lambda D}{d}$ , where $n$ is the order of the fringe, $\lambda$ is the wavelength, $D$ is the distance to the screen, and $d$ is the slit separation.

Step 2: For the second dark fringe ( $n = 2$ ), $y_2 = (2 - \frac{1}{2})\frac{\lambda D}{d} = \frac{3}{2} \frac{\lambda D}{d} = 0.3 \text{ cm}$ .

Step 3: We are given $d = 0.3 \text{ mm} = 0.03 \text{ cm}$ and $D = 1 \text{ m} = 100 \text{ cm}$ .

Step 4: Solve for $\lambda$ : $\frac{3}{2} \frac{\lambda (100)}{0.03} = 0.3$ , so $\lambda = \frac{0.3 \times 0.03 \times 2}{3 \times 100} = 6 \times 10^{-5} \text{ cm}$ .

Step 5: The position of the $n$ -th bright fringe is given by $y_n = n \frac{\lambda D}{d}$ .

Step 6: For the fourth bright fringe ( $n = 4$ ), $y_4 = 4 \frac{\lambda D}{d} = 4 \frac{(6 \times 10^{-5}) (100)}{0.03} = 4 \times \frac{0.006}{0.03} = 4 \times 0.2 = 0.8 \text{ cm}$ .

Why other options are wrong:
- Option A (0.6 cm): This value is incorrect and likely arises from an incorrect calculation or using the wrong formula.
- Option C (1.2 cm): This value is incorrect and likely arises from an incorrect calculation or using the wrong formula.
- Option D (0.12 cm): This value is incorrect and likely arises from an incorrect calculation or using the wrong formula.

Key Concepts:
- Young's double-slit experiment
- Dark and bright fringe positions

Memory Aids:
- Remember the formula for dark fringes: $y_n = (n - \frac{1}{2})\frac{\lambda D}{d}$
- Remember the formula for bright fringes: $y_n = n \frac{\lambda D}{d}$

Quick Tips:
- Keep track of the units and convert them to a consistent system.
- Be careful with the order of the fringes.

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Understanding +4 | -1

If a source of light is moving away from a stationary observer, then the frequency of light wave appears to change because of

View answer and explanation

Correct option: A

Doppler’s effect

Option A (Doppler’s effect) is correct because the Doppler effect explains the change in frequency of a wave due to the relative motion between the source and the observer.

Solution approach

Correct Answer + Core Reason: Doppler’s effect. Doppler effect explains the change in frequency due to relative motion.

Step-by-Step Solution:
Step 1: Recall the definition of the Doppler effect.
Step 2: The Doppler effect is the change in frequency or wavelength of a wave in relation to an observer who is moving relative to the wave source.
Step 3: If the source is moving away, the frequency decreases (redshift). If the source is moving towards, the frequency increases (blueshift).
Step 4: Conclusion: The change in frequency is due to the Doppler effect.

Why other options are wrong:
- Option B: Interference is the superposition of waves.
- Option C: Diffraction is the bending of waves around obstacles.
- Option D: The Doppler effect is the correct explanation.

Key Concepts:
- Doppler effect
- Frequency shift
- Redshift and blueshift

Memory Aids:
- Doppler effect: Change in frequency due to motion.

Quick Tips:
- Remember that the Doppler effect applies to all types of waves, including light and sound.

46 Medium

46 of 104Question permalink

Application +4 | -1

A wavefront presents one, two and three HPZ at points

A,B\;and\;C

respectively. If the ratio of consecutive amplitudes of HPZ is

4\mathrm{ }:\mathrm{ }3

, then the ratio of resultant intensities at these point will be

View answer and explanation

Correct option: B

256\mathrm{ }:\mathrm{ }16\mathrm{ }:\mathrm{ }169

Option B (

256\mathrm{ }:\mathrm{ }16\mathrm{ }:\mathrm{ }169

) is correct because the intensity is proportional to the square of the amplitude, and the resultant amplitude is calculated based on the number of half-period zones.

Solution approach

Correct Answer + Core Reason: $256\mathrm{ }:\mathrm{ }16\mathrm{ }:\mathrm{ }169$ . The intensity is proportional to the square of the resultant amplitude.

Step-by-Step Solution:

Step 1: Understand the concept of Half-Period Zones (HPZ) and their amplitudes.

Step 2: Recall that the amplitude due to n HPZs is given by $A = A_1 - A_2 + A_3 - ... + (-1)^{n-1}A_n$ , where $A_i$ is the amplitude of the $i^{th}$ HPZ.

Step 3: Given the ratio of consecutive amplitudes is 4:3, let's assume the amplitudes are $4x, 3x, \frac{9}{4}x$ , etc.

Step 4: Calculate the resultant amplitudes at points A, B, and C:
- At A (1 HPZ): $A_A = 4x$
- At B (2 HPZs): $A_B = 4x - 3x = x$
- At C (3 HPZs): $A_C = 4x - 3x + \frac{9}{4}x = x + \frac{9}{4}x = \frac{13}{4}x$

Step 5: Calculate the intensities at points A, B, and C. Since intensity is proportional to the square of the amplitude, we have:
- $I_A \propto (4x)^2 = 16x^2$
- $I_B \propto (x)^2 = x^2$
- $I_C \propto (\frac{13}{4}x)^2 = \frac{169}{16}x^2$

Step 6: Find the ratio of the intensities:
$I_A : I_B : I_C = 16x^2 : x^2 : \frac{169}{16}x^2 = 16 : 1 : \frac{169}{16}$ \nMultiply by 16 to get rid of the fraction: $256 : 16 : 169$

Step 7: Final Answer: $256\mathrm{ }:\mathrm{ }16\mathrm{ }:\mathrm{ }169$

Why other options are wrong:
- Option A: $169\mathrm{ }:\mathrm{ }16\mathrm{ }:\mathrm{ }256$ This is an incorrect ordering of the intensities.
- Option C: $256\mathrm{ }:\mathrm{ }16\mathrm{ }:\mathrm{ }196$ This uses an incorrect calculation for the amplitude with 3 HPZs.
- Option D: $256\mathrm{ }:\mathrm{ }196\mathrm{ }:\mathrm{ }16$ This also uses an incorrect calculation for the amplitude with 3 HPZs and an incorrect ordering.

Key Concepts:
- Half-Period Zones
- Huygens Principle
- Intensity and Amplitude relationship

Memory Aids:
- Remember that intensity is proportional to the square of the amplitude.

Quick Tips:
- Carefully calculate the resultant amplitude by considering the alternating signs of the contributions from each HPZ.

47 Medium

47 of 104Question permalink

Remembering +4 | -1

Colours in thin films are due to

View answer and explanation

Correct option: C

Interference phenomenon

Option C is correct. Colors in thin films are a result of the interference of light waves reflected from the different surfaces of the film.

Solution approach

Correct Answer + Core Reason: Interference phenomenon. Colors in thin films are produced by the interference of light waves.

Step-by-Step Solution:

Step 1: Describe thin films: Thin films are layers of material with thicknesses comparable to the wavelength of light.

Step 2: Explain the interference: Light reflects from the top and bottom surfaces of the thin film. These reflected waves interfere, either constructively or destructively, depending on the thickness of the film, the angle of incidence, and the wavelength of light.

Why other options are wrong:
- Option A: Incorrect because diffraction involves the bending of light around obstacles, not the primary cause of colors in thin films.
- Option B: Incorrect because scattering is the redirection of light by particles, not the primary cause of colors in thin films.
- Option D: Incorrect because polarization is the alignment of the electric field of light, not the primary cause of colors in thin films.

Key Concepts:
- Interference
- Thin films
- Reflection

Memory Aids:
- Thin Film Colors = Interference

Quick Tips:
- Remember that thin film interference depends on the thickness of the film and the wavelength of light.

48 Medium

48 of 104Question permalink

Analyzing +4 | -1

In Young’s double slit experiment, a third slit is made in between the double slits. Then

View answer and explanation

Correct option: B

Contrast between bright and dark fringes is reduced

Option B (Contrast between bright and dark fringes is reduced) is correct because the third slit introduces additional light, increasing the minimum intensity and thus reducing the fringe visibility.

Solution approach

Correct Answer + Core Reason: Contrast between bright and dark fringes is reduced. The third slit introduces additional light, increasing the minimum intensity and reducing fringe visibility.

Step-by-Step Solution:

Step 1: In Young's double-slit experiment (YDSE), the interference pattern is formed due to the superposition of light waves from two coherent sources.

Step 2: When a third slit is introduced between the double slits, it acts as another source of light. This additional source will also contribute to the interference pattern.

Step 3: The intensity of the bright fringes will increase due to the additional light source. However, the intensity of the dark fringes will also increase because the third slit will provide some light even where the other two slits would have produced a dark fringe.

Step 4: As a result, the difference in intensity between the bright and dark fringes decreases. This means the contrast between the bright and dark fringes is reduced.

Why other options are wrong:
- Option A (Fringes of unequal width are formed): The fringe width depends on the wavelength of light, the distance between the slits and the screen, and the separation between the slits. Introducing a third slit does not change these parameters, so the fringe width remains the same.
- Option C (Intensity of fringes totally disappears): The intensity of fringes does not totally disappear; it only reduces the contrast between bright and dark fringes.
- Option D (Only bright light is observed on the screen): Dark fringes still exist, but they are not as dark as they would be without the third slit.

Key Concepts:
- Young's Double Slit Experiment (YDSE)
- Interference of Light
- Superposition Principle
- Fringe Visibility

Memory Aids:
- Remember that introducing a third slit adds more light, which increases the intensity of both bright and dark fringes, reducing the contrast.

Quick Tips:
- Consider how adding more light sources affects the overall intensity distribution in the interference pattern.

49 Medium

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Understanding +4 | -1

The fringe width in Young’s double slit experiment increases when

View answer and explanation

Correct option: D

Distance between slits plane and screen increases

Option D is correct. The fringe width in Young's double-slit experiment is directly proportional to the distance between the slits and the screen.

Solution approach

Correct Answer + Core Reason: Distance between slits plane and screen increases. Fringe width is directly proportional to the distance between the slits and the screen.

Step-by-Step Solution:

Step 1: Recall the formula for fringe width in Young's double-slit experiment: $\beta = \frac{\lambda D}{d}$ , where $\beta$ is the fringe width, $\lambda$ is the wavelength of light, $D$ is the distance between the slits and the screen, and $d$ is the distance between the slits.

Step 2: Analyze the formula. The fringe width is directly proportional to the wavelength of light ( $\lambda$ ) and the distance between the slits and the screen ( $D$ ), and inversely proportional to the distance between the slits ( $d$ ).

Step 3: Determine the condition for increasing fringe width. To increase the fringe width, the distance between the slits and the screen must increase.

Why other options are wrong:
- Option A: Decreasing the wavelength decreases the fringe width.
- Option B: Decreasing the distance between the sources and the screen decreases the fringe width.
- Option C: Moving the source slit closer to the slit openings doesn't directly affect the fringe width, although it can affect the intensity of the fringes.

Key Concepts:
- Young's double-slit experiment
- Fringe width
- Wavelength of light
- Distance between slits and screen
- Distance between slits

Memory Aids:
- Remember the formula for fringe width: "Lambda Dee over dee."

Quick Tips:
- Understand the relationship between fringe width and the parameters of the double-slit experiment.

50 Medium

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Understanding +4 | -1

In Young’s double slit experiment, if one of the slits is closed fully, then in the interference pattern

View answer and explanation

Correct option: A

A bright slit will be observed, no interference pattern will exist

Option A (A bright slit will be observed, no interference pattern will exist) is correct because interference requires two coherent sources, and closing one slit eliminates the interference.

Solution approach

Correct Answer + Core Reason: A bright slit will be observed, no interference pattern will exist. Interference needs two sources.

Step-by-Step Solution:

Step 1: Recall the conditions for interference: Interference requires two coherent sources of light.

Step 2: Effect of closing one slit: Closing one slit eliminates one of the coherent sources.

Step 3: Result: Without two coherent sources, there is no interference pattern. Only a bright slit will be observed due to the light passing through the open slit.

Why other options are wrong:
- Option B: The bright fringes will not become more bright because there is no interference pattern.
- Option C: The bright fringes will not become fainter because there is no interference pattern.
- Option D: The statement in option A is correct.

Key Concepts:
- Young's Double Slit Experiment (YDSE)
- Interference
- Coherent sources

Memory Aids:
- No two slits = no interference.

Quick Tips:
- Remember that interference requires two coherent sources.

51 Medium

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Application +4 | -1

A narrow stream of electrons of energy 100eV is fired at two parallel slits very close to each other. The distance between the slits is 10A. The electron waves after passing through the slits interfere on a screen, 3m away from slits. The fringe width is

View answer and explanation

Correct option: A

0.36m

Option A (0.36m) is correct because it is calculated using the de Broglie wavelength and the fringe width formula for electron interference.

Solution approach

Correct Answer + Core Reason: 0.36m. The fringe width is calculated using the de Broglie wavelength and the standard formula.

Step-by-Step Solution:

Step 1: Calculate the de Broglie wavelength of the electron.
$\lambda = \frac{h}{p} = \frac{h}{\sqrt{2mK}}$ where $h$ is Planck's constant, $m$ is the mass of the electron, and $K$ is the kinetic energy.

Step 2: Convert the kinetic energy to Joules.
$K = 100 \text{ eV} = 100 \times 1.6 \times 10^{-19} \text{ J} = 1.6 \times 10^{-17} \text{ J}$

Step 3: Substitute the values into the de Broglie wavelength formula.
$\lambda = \frac{6.626 \times 10^{-34}}{\sqrt{2 \times 9.1 \times 10^{-31} \times 1.6 \times 10^{-17}}} \approx 1.227 \times 10^{-10} \text{ m} = 1.227 \text{ Å}$

Step 4: Calculate the fringe width using the formula $\beta = \frac{\lambda D}{d}$ , where $D$ is the distance to the screen and $d$ is the slit separation.

Step 5: Substitute the given values: $D = 3 \text{ m}$ , $d = 10 \text{ Å} = 10 \times 10^{-10} \text{ m}$ .
$\beta = \frac{1.227 \times 10^{-10} \times 3}{10 \times 10^{-10}} = \frac{1.227 \times 3}{10} = 0.3681 \text{ m} \approx 0.36 \text{ m}$

Why other options are wrong:
- Option B (0.48m): This value is incorrect and likely arises from an incorrect calculation or using the wrong formula.
- Option C (0.62m): This value is incorrect and likely arises from an incorrect calculation or using the wrong formula.
- Option D (0.75m): This value is incorrect and likely arises from an incorrect calculation or using the wrong formula.

Key Concepts:
- de Broglie wavelength
- Young's double slit experiment
- Fringe width formula

Memory Aids:
- Remember the de Broglie wavelength formula: $\lambda = h/p$
- Fringe width formula: $\beta = \frac{\lambda D}{d}$

Quick Tips:
- Ensure all units are consistent before calculations.
- Double-check the values before substituting them into the formula.

52 Medium

52 of 104Question permalink

Analyzing +4 | -1

Assertion: The clouds in sky generally appear to be whitish.

Reason: Diffraction due to clouds is efficient in equal measure at all wavelengths.

View answer and explanation

Correct option: C

Assertion is true and Reason is false

Option C (Assertion is true and Reason is false) is correct. Clouds appear whitish due to scattering, not diffraction, and scattering is not equally efficient at all wavelengths.

Solution approach

Correct Answer + Core Reason: Assertion is true and Reason is false. Clouds appear whitish due to scattering, but the scattering is not equally efficient at all wavelengths.

Step-by-Step Solution:

Step 1: Analyze the Assertion: Clouds in the sky generally appear whitish because they scatter sunlight. The scattering is due to water droplets and ice crystals in the clouds. Therefore, the assertion is true.

Step 2: Analyze the Reason: The reason states that diffraction due to clouds is efficient in equal measure at all wavelengths. This is incorrect. The scattering of light by clouds is primarily Mie scattering, which is more complex than Rayleigh scattering and depends on the size of the scattering particles (water droplets and ice crystals) relative to the wavelength of light. While Mie scattering is less wavelength-dependent than Rayleigh scattering, it is not equally efficient at all wavelengths. If it were, the clouds would appear white. Therefore, the reason is false.

Why other options are wrong:
- Option A: Both Assertion and Reason are true and Reason is the correct explanation of Assertion. This is incorrect because the reason is false.
- Option B: Both Assertion and Reason are true and Reason is not the correct explanation of Assertion. This is incorrect because the reason is false.
- Option C: Assertion is true and Reason is false. This is the correct answer.
- Option D: Assertion is false and Reason is true. This is incorrect because the assertion is true and the reason is false.

Key Concepts:
- Scattering of light
- Diffraction of light
- Mie scattering
- Rayleigh scattering
- Wavelength of light

Memory Aids:
- Clouds appear whitish due to scattering, not diffraction.
- Mie scattering is less wavelength-dependent than Rayleigh scattering.

Quick Tips:
- Remember that scattering is the primary reason why clouds appear white, and the type of scattering depends on the size of the scattering particles relative to the wavelength of light.

53 Medium

53 of 104Question permalink

Application +4 | -1

The two slits are

1\,mm

apart from each other and illuminated with a light of wavelength

5 \times {10^{ - 7}}\,m

. If the distance of the screen is

1\,m

from the slits, then the distance between third dark fringe and fifth bright fringe is

View answer and explanation

Correct option: C

1.25\,mm

Option C (

1.25\,mm

) is correct because it is calculated by finding the positions of the third dark fringe and fifth bright fringe and then subtracting them.

Solution approach

Correct Answer + Core Reason: $1.25\,mm$ . The distance is calculated by finding the positions of the third dark fringe and fifth bright fringe and subtracting them.

Step-by-Step Solution:

Step 1: Identify the given information: Slit separation $d = 1\,mm = 1 \times 10^{-3}\,m$ , wavelength $\lambda = 5 \times 10^{-7}\,m$ , distance to the screen $D = 1\,m$ .

Step 2: Position of the $n^{th}$ bright fringe: $y_n = \frac{n\lambda D}{d}$

Step 3: Position of the $n^{th}$ dark fringe: $y_n = \frac{(2n-1)\lambda D}{2d}$

Step 4: Calculate the position of the 5th bright fringe: $y_5 = \frac{5 \times 5 \times 10^{-7} \times 1}{1 \times 10^{-3}} = 25 \times 10^{-4}\,m = 2.5\,mm$

Step 5: Calculate the position of the 3rd dark fringe: $y_3 = \frac{(2(3)-1) \times 5 \times 10^{-7} \times 1}{2 \times 1 \times 10^{-3}} = \frac{5 \times 5 \times 10^{-7}}{2 \times 10^{-3}} = 12.5 \times 10^{-4}\,m = 1.25\,mm$

Step 6: Find the distance between the 3rd dark fringe and the 5th bright fringe: $\Delta y = |y_5 - y_3| = |2.5 - 1.25| = 1.25\,mm$

Step 7: Final Answer: $1.25\,mm$

Why other options are wrong:
- **Option A ( $1.5\,mm$ ): This is incorrect because it might be a result of incorrect calculation of the fringe positions.
- Option B ( $0.75\,mm$ ): This is incorrect because it might be a result of incorrect calculation of the fringe positions.
- Option D ( $0.625\,mm$ ): This is incorrect because it might be a result of incorrect calculation of the fringe positions.

Key Concepts:
- Young's Double Slit Experiment
- Bright and Dark Fringe Positions

Memory Aids:
- Remember the formulas for bright and dark fringe positions.

Quick Tips:**
- Always convert all units to the same system before calculations.

54 Medium

54 of 104Question permalink

Understanding +4 | -1

Through which character we can distinguish the light waves from sound waves

View answer and explanation

Correct option: C

Polarization

The correct answer is C (Polarization) because polarization is a property unique to transverse waves, like light, and cannot occur in longitudinal waves, like sound.

Solution approach

Correct Answer + Core Reason: Polarization. Light waves are transverse waves and can be polarized, while sound waves are longitudinal and cannot.

Step-by-Step Solution:

Step 1: Recall the properties of light and sound waves.

Step 2: Light waves are transverse, meaning their oscillations are perpendicular to the direction of propagation.

Step 3: Sound waves are longitudinal, meaning their oscillations are parallel to the direction of propagation.

Step 4: Polarization is a phenomenon that can only occur in transverse waves.

Step 5: Therefore, polarization can be used to distinguish light waves from sound waves.

Why other options are wrong:
- Option A: Interference occurs in both light and sound waves.
- Option B: Refraction occurs in both light and sound waves.
- Option D: Reflection occurs in both light and sound waves.

Key Concepts:
- Transverse waves
- Longitudinal waves
- Polarization
- Interference
- Refraction
- Reflection

Memory Aids:
- Polarization: Think of polarized sunglasses blocking light in a specific orientation.

Quick Tips:
- Remember that polarization is a key characteristic that differentiates transverse and longitudinal waves.

55 Medium

55 of 104Question permalink

Application +4 | -1

If a star emitting light of wavelength 500 nm is moving towards earth with a velocity of

1.5\mathrm{ } \times \mathrm{ }{10^6}m/s

then the shift in the wavelength due to doppler’s effect will be:

View answer and explanation

Correct option: A

25\,\,\mathop \mathrm{A}\limits^ \circ

The correct answer is A (

25\,\,\mathop \mathrm{A}\limits^ \circ \,

) because it is the calculated wavelength shift due to the Doppler effect.

Solution approach

Correct Answer + Core Reason: $25\,\,\mathop \mathrm{A}\limits^ \circ \,$ . The shift in wavelength is calculated using the Doppler effect formula.

Step-by-Step Solution:

Step 1: Identify the given information: Wavelength $\lambda = 500 \text{ nm} = 5000 \mathop {\text{A}}\limits^ \circ$ , Velocity $v = 1.5 \times 10^6 \text{ m/s}$ , Speed of light $c = 3 \times 10^8 \text{ m/s}$ .

Step 2: The Doppler effect formula for wavelength shift is $\frac{{\Delta \lambda}}{\lambda} = \frac{v}{c}$ .

Step 3: $\Delta \lambda = \lambda \frac{v}{c} = 5000 \mathop {\text{A}}\limits^ \circ \times \frac{{1.5 \times 10^6}}{{3 \times 10^8}} = 5000 \mathop {\text{A}}\limits^ \circ \times 0.005 = 25 \mathop {\text{A}}\limits^ \circ$ .

Why other options are wrong:
- Option B: This is incorrect as it is not the correct wavelength shift.
- Option C: This is incorrect as it is not the correct wavelength shift.
- Option D: This is incorrect as it is not the correct wavelength shift.

Key Concepts:
- Doppler effect
- Wavelength shift

Memory Aids:
- Remember the Doppler effect formula $\frac{{\Delta \lambda}}{\lambda} = \frac{v}{c}$ .

Quick Tips:
- Make sure to use consistent units when calculating the wavelength shift.

56 Medium

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Application +4 | -1

A star is moving away from the earth with a velocity of

100\;km/s

. If the velocity of light is

3 \times {10^8}\;m/s

then the shift of its spectral line of wavelength

5700\mathop \mathrm{A}\limits^ \circ

due to Doppler’s effect will be

View answer and explanation

Correct option: B

1.90\mathop \mathrm{A}\limits^ \circ

Option B (

256\mathrm{ }:\mathrm{ }16\mathrm{ }:\mathrm{ }169

) is correct because the intensity is proportional to the square of the amplitude, and the resultant amplitude depends on the number of half-period zones.

Solution approach

Correct Answer + Core Reason: $256\mathrm{ }:\mathrm{ }16\mathrm{ }:\mathrm{ }169$ . Intensity is proportional to the square of the amplitude.

Step-by-Step Solution:

Step 1: Let the amplitudes of the consecutive half-period zones (HPZ) be $a_1, a_2, a_3, \dots$ The ratio of consecutive amplitudes is given as $4:3$ , so we can write $a_n = a_1 (\frac{3}{4})^{n-1}$ .

Step 2: The resultant amplitude at a point where one HPZ is exposed is $A_1 = \frac{a_1}{2}$ . The intensity is $I_1 \propto A_1^2 = (\frac{a_1}{2})^2 = \frac{a_1^2}{4}$ .

Step 3: The resultant amplitude at a point where two HPZs are exposed is $A_2 = \frac{{a_1 - a_2}}{2} = \frac{{a_1 - \frac{3}{4}a_1}}{2} = \frac{{\frac{1}{4}a_1}}{2} = \frac{a_1}{8}$ . The intensity is $I_2 \propto A_2^2 = (\frac{a_1}{8})^2 = \frac{a_1^2}{64}$ .

Step 4: The resultant amplitude at a point where three HPZs are exposed is $A_3 = \frac{{a_1 - a_2 + a_3}}{2} = \frac{{a_1 - \frac{3}{4}a_1 + (\frac{3}{4})^2 a_1}}{2} = \frac{{a_1(1 - \frac{3}{4} + \frac{9}{{16}})}}{2} = \frac{{a_1(\frac{{16 - 12 + 9}}{{16}})}}{2} = \frac{{13a_1}}{{32}}$ . The intensity is $I_3 \propto A_3^2 = (\frac{{13a_1}}{{32}})^2 = \frac{{169a_1^2}}{{1024}}$ .

Step 5: Find the ratio of the intensities: $I_1 : I_2 : I_3 = \frac{a_1^2}{4} : \frac{a_1^2}{64} : \frac{{169a_1^2}}{{1024}} = \frac{1}{4} : \frac{1}{{64}} : \frac{{169}}{{1024}}$ . Multiply by 1024 to get integers: $256 : 16 : 169$ .

Why other options are wrong:
- **Option A ( $169\mathrm{ }:\mathrm{ }16\mathrm{ }:\mathrm{ }256$ ): This is incorrect because the intensities are not in the correct order.
- Option C ( $256\mathrm{ }:\mathrm{ }16\mathrm{ }:\mathrm{ }196$ ): This is incorrect because the intensity for three HPZs is not calculated correctly.
- Option D ( $256\mathrm{ }:\mathrm{ }196\mathrm{ }:\mathrm{ }16$ ): This is incorrect because the intensities are not in the correct order and the intensity for three HPZs is not calculated correctly.

Key Concepts:
- Half-period zones
- Fresnel diffraction
- Intensity of light

Memory Aids:
- Remember that intensity is proportional to the square of the amplitude.

Quick Tips:**
- Pay attention to the ratio of consecutive amplitudes.
- Calculate the resultant amplitude for each case carefully.

57 Medium

57 of 104Question permalink

Understanding +4 | -1

Resolving power of a microscope depends upon

View answer and explanation

Correct option: D

The wavelength of light illuminating the object

Option D is correct. The resolving power of a microscope depends on the wavelength of light illuminating the object, as shorter wavelengths provide better resolution.

Solution approach

Correct Answer + Core Reason: The wavelength of light illuminating the object. Resolving power is inversely proportional to wavelength.

Step-by-Step Solution:

Step 1: Recall the formula for the resolving power (RP) of a microscope: $RP = \frac{2n \sin \theta}{\lambda}$ , where $n$ is the refractive index of the medium between the object and the objective lens, $\theta$ is the half-angle of the cone of light that enters the objective lens, and $\lambda$ is the wavelength of light.

Step 2: Analyze the formula. The resolving power is inversely proportional to the wavelength of light ( $\lambda$ ).

Step 3: Identify the factor that the resolving power depends on. From the formula, it's clear that the resolving power depends on the wavelength of light illuminating the object.

Why other options are wrong:
- Option A: The focal length and aperture of the eye lens affect the magnification, not the resolving power directly.
- Option B: The focal lengths of the objective and the eye lens affect the magnification, not the resolving power directly. While they contribute to magnification, the wavelength is the key factor for resolution.
- Option C: The apertures of the objective and the eye lens affect the light gathering ability and brightness, but not the resolving power directly. The numerical aperture, which includes the aperture of the objective, is related to resolving power, but the wavelength is the most direct factor.

Key Concepts:
- Resolving power of a microscope
- Wavelength of light
- Numerical aperture

Memory Aids:
- Remember that shorter wavelengths allow for better resolution.

Quick Tips:
- The shorter the wavelength of light, the better the microscope's ability to distinguish between closely spaced objects.

58 Medium

58 of 104Question permalink

Remembering +4 | -1

Soap bubble appears coloured due to the phenomenon of

View answer and explanation

Correct option: A

Interference

Option A is correct. The colors seen in soap bubbles are due to the interference of light waves reflected from the inner and outer surfaces of the thin film.

Solution approach

Correct Answer + Core Reason: Interference. The colors in soap bubbles arise from the interference of light waves.

Step-by-Step Solution:

Step 1: Describe the phenomenon: Soap bubbles are thin films of soap solution. Light reflects from both the inner and outer surfaces of the film.

Step 2: Explain the cause of colors: The reflected waves interfere with each other. Depending on the thickness of the film and the wavelength of light, the interference can be constructive (resulting in bright colors) or destructive (resulting in dark areas).

Why other options are wrong:
- Option B: Incorrect because diffraction is the bending of light around obstacles, not the primary cause of colors in soap bubbles.
- Option C: Incorrect because dispersion is the separation of white light into its constituent colors, which is a factor, but not the primary cause.
- Option D: Incorrect because reflection is necessary, but it's the interference of reflected waves that causes the colors.

Key Concepts:
- Interference
- Thin films
- Reflection

Memory Aids:
- Soap Bubble Colors = Interference

Quick Tips:
- Thin film interference is the key to understanding colors in soap bubbles.

59 Medium

59 of 104Question permalink

Remembering +4 | -1

If microwaves, X rays, infrared, gamma rays, ultra-violet, radio waves and visible parts of the electromagnetic spectrum are denoted by M, X, I, G, U, R and V then which of the following is the arrangement in ascending order of wavelength?

View answer and explanation

Correct option: C

G, X, U, V, I, M and R

Option C is correct. The electromagnetic spectrum in ascending order of wavelength is: Gamma rays, X-rays, Ultraviolet, Visible, Infrared, Microwaves, and Radio waves.

Solution approach

Correct Answer + Core Reason: G, X, U, V, I, M and R. This is the correct order of increasing wavelength.

Step-by-Step Solution:

Step 1: Recall the order of the electromagnetic spectrum from shortest to longest wavelength: Gamma rays, X-rays, Ultraviolet, Visible, Infrared, Microwaves, Radio waves.

Step 2: Match the given abbreviations to the corresponding electromagnetic waves: M (Microwaves), X (X-rays), I (Infrared), G (Gamma rays), U (Ultra-violet), R (Radio waves), V (Visible).

Step 3: Arrange the abbreviations in ascending order of wavelength: G, X, U, V, I, M, R.

Why other options are wrong:
- Option A: This option is incorrect because it places Radio waves before Gamma rays, X-rays, and Ultraviolet.
- Option B: This option is incorrect because it places Microwaves before Gamma rays, X-rays, and Ultraviolet.
- Option D: This option is incorrect because it places Infrared before Gamma rays, X-rays, and Ultraviolet.

Key Concepts:
- Electromagnetic spectrum
- Wavelength order of electromagnetic waves

Memory Aids:
- Use the mnemonic: "Raging Martians Invaded Venus Using X-ray Guns" (Radio, Microwaves, Infrared, Visible, Ultraviolet, X-rays, Gamma rays) - This is in decreasing order of wavelength.

Quick Tips:
- Memorize the order of the electromagnetic spectrum.

60 Medium

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Application +4 | -1

The

21\;cm

radio wave emitted by hydrogen in interstellar space is due to the interaction called the hyperfine interaction is atomic hydrogen. the energy of the emitted wave is nearly

View answer and explanation

Correct option: D

{10^{ - 24}}\;Joule

Option D (

{10^{ - 24}}\;Joule

) is correct because the energy of the emitted wave is very small, corresponding to the hyperfine interaction in atomic hydrogen, and can be calculated using

E = hf

.

Solution approach

Correct Answer + Core Reason: ${10^{ - 24}}\;Joule$ . The energy of the emitted wave is calculated using $E = hf$ , where $h$ is Planck's constant and $f$ is the frequency.

Step-by-Step Solution:

Step 1: Identify the given values: Wavelength, $\lambda = 21 \, \mathrm{cm} = 0.21 \, \mathrm{m}$ .

Step 2: Calculate the frequency: $f = \frac{c}{\lambda} = \frac{3 \times 10^8 \, \mathrm{m/s}}{0.21 \, \mathrm{m}} = \frac{3}{0.21} \times 10^8 \, \mathrm{Hz} \approx 1.43 \times 10^9 \, \mathrm{Hz}$ .

Step 3: Calculate the energy: $E = hf = (6.626 \times 10^{-34} \, \mathrm{J \cdot s}) (1.43 \times 10^9 \, \mathrm{Hz}) \approx 9.47 \times 10^{-25} \, \mathrm{J} \approx 10^{-24} \, \mathrm{J}$ .

Why other options are wrong:
- **Option A ( ${10^{ - 17}}\;Joule$ ): This is incorrect because the energy is significantly higher than the actual energy of the emitted wave. This might be a result of incorrect calculation or using the wrong formula.
- Option B ( $1\;Joule$ ): This is incorrect because the energy is far too high for a radio wave emitted due to hyperfine interaction. This would be a very high-energy photon.
- Option C ( $7 \times {10^{ - 8}}\;Joule$ ): This is incorrect because the energy is higher than the actual energy of the emitted wave. This might be a result of incorrect calculation or using the wrong formula.

Key Concepts:
- Energy of a photon
- Planck's constant
- Frequency and wavelength relationship

Memory Aids:**
- Remember the formula $E = hf$ and the value of Planck's constant.

Quick Tips:
- The energy of radio waves is very small compared to visible light or X-rays.

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