WAVE OPTICS MCQs for NEET UG PHYSICS — Easy

1 Easy

Application +4 | -1

Two waves of amplitude

5 \mathrm{~mm}

and

7 \mathrm{~mm}

reach a point in opposite phase. The resultant amplitude is

View answer and explanation

Correct option: C

2 \mathrm{~mm}

Option C (2 mm) is correct because when waves are in opposite phase, the resultant amplitude is the difference between the individual amplitudes.

Solution approach

Correct Answer + Core Reason: 2 mm. The resultant amplitude is the difference between the individual amplitudes when they are in opposite phase.

Step-by-Step Solution:

Step 1: Identify the given information:\nAmplitude 1, $A_1 = 5 \text{{ mm}}$ \nAmplitude 2, $A_2 = 7 \text{{ mm}}$ \nWaves are in opposite phase.

Step 2: Key Concept: When two waves are in opposite phase, the resultant amplitude is the difference between their individual amplitudes.
$A_{resultant} = |A_1 - A_2|$

Step 3: Apply the formula:
$A_{resultant} = |5 - 7| = |-2| = 2 \text{{ mm}}$

Why other options are wrong:
- Option A: 24 mm: This would be the result if the amplitudes were added incorrectly or squared and added.
- Option B: 12 mm: This is an incorrect calculation.
- Option D: 1.4 mm: This is an incorrect calculation, likely from misunderstanding the phase relationship.

Key Concepts:
- Superposition of waves
- Interference
- Amplitude

Memory Aids:
- Opposite phase = Subtract amplitudes.

Quick Tips:
- Remember to subtract amplitudes when waves are in opposite phase.

2 Easy

2 of 23Question permalink

Analyzing +4 | -1

Four independent light waves are expressed as

\mathrm{y}_1=\mathrm{a}_1 \sin \omega \mathrm{t}, \mathrm{y}_2=\mathrm{a}_2 \sin \omega \mathrm{t}, \mathrm{y}_3=\mathrm{a}_3 \cos \omega \mathrm{t}

and

\mathrm{y}_4=\mathrm{a}_4 \sin (\omega \mathrm{t}+\pi / 3)

. The interference is possible between

View answer and explanation

Correct option: D

Not possible at all

Option D (Not possible at all) is correct because for interference to occur, the waves must have the same frequency and a constant phase relationship. Waves 1 and 2 are in phase, wave 3 is in quadrature, and wave 4 has a constant phase shift, but none of these pairs can produce sustained interference.

Solution approach

Correct Answer + Core Reason: Not possible at all. For interference to occur, the waves must have the same frequency and a constant phase relationship.

Step-by-Step Solution:

Step 1: Analyze the given waves:
- $y_1 = a_1 \sin \omega t$
- $y_2 = a_2 \sin \omega t$
- $y_3 = a_3 \cos \omega t = a_3 \sin (\omega t + \frac{\pi}{2})$
- $y_4 = a_4 \sin (\omega t + \frac{\pi}{3})$

Step 2: Check for possible interference between pairs of waves:
- Waves 1 and 2: They have the same frequency and a constant phase difference of 0. However, the question states that the waves are independent, implying they are not derived from the same source and therefore not coherent. Thus, no sustained interference.
- Waves 1 and 3: They have the same frequency, but a constant phase difference of $\frac{\pi}{2}$ . However, the question states that the waves are independent, implying they are not derived from the same source and therefore not coherent. Thus, no sustained interference.
- Waves 1 and 4: They have the same frequency, but a constant phase difference of $\frac{\pi}{3}$ . However, the question states that the waves are independent, implying they are not derived from the same source and therefore not coherent. Thus, no sustained interference.
- Waves 3 and 4: They have the same frequency, but a constant phase difference. However, the question states that the waves are independent, implying they are not derived from the same source and therefore not coherent. Thus, no sustained interference.

Step 3: Since the waves are independent, they are not coherent, and no sustained interference is possible.

Why other options are wrong:
- Option A (1 and 3): Although they have the same frequency, the waves are independent and therefore not coherent, so no sustained interference is possible.
- Option B (1 and 4): Although they have the same frequency, the waves are independent and therefore not coherent, so no sustained interference is possible.
- Option C (3 and 4): Although they have the same frequency, the waves are independent and therefore not coherent, so no sustained interference is possible.

Key Concepts:
- Interference of Light Waves
- Coherence
- Phase Difference
- Superposition Principle

Memory Aids:
- Remember that for sustained interference, waves must be coherent (same frequency and constant phase relationship).

Quick Tips:
- Always check for coherence before assuming interference is possible.

3 Easy

3 of 23Question permalink

Application +4 | -1

A beam of monochromatic light of wavelength

5.82 \times 10^{-7} \mathrm{~m}

falls normally on a glass wedge with the wedge angle of 20 second of an arc. If the refractive index of glass is

1.5,

find the number of dark interference fringes per

\mathrm{cm}

of the wedge length:

View answer and explanation

Correct option: B

5 per cm

Option B (5 per cm) is correct because the fringe width is calculated using the wedge angle and refractive index, then the number of fringes per cm is found.

Solution approach

Correct Answer + Core Reason: 5 per cm. The fringe width is calculated using the wedge angle and refractive index.

Step-by-Step Solution:

Step 1: Identify the given information:\nWavelength, $\lambda = 5.82 \times 10^{-7} \text{{ m}}$ \nWedge angle, $\theta = 20 \text{{ seconds of an arc}} = 20 \times \frac{1}{{3600}} \text{{ degrees}} = 20 \times \frac{1}{{3600}} \times \frac{\pi }{{180}} \text{{ radians}}$ \nRefractive index, $n = 1.5$

Step 2: Calculate the fringe width, $\beta$ :
$\beta = \frac{{\lambda }}{{2n\theta}}$
$\beta = \frac{{5.82 \times 10^{-7}}}{{2 \times 1.5 \times 20 \times \frac{\pi }{{3600 \times 180}}}} = \frac{{5.82 \times 10^{-7} \times 3600 \times 180}}{{2 \times 1.5 \times 20 \times \pi }} = \frac{{5.82 \times 10^{-7} \times 3600 \times 180}}{{60\pi }}$
$\beta = \frac{{5.82 \times 10^{-7} \times 60 \times 180}}{{ \pi }} = \frac{{5.82 \times 10^{-7} \times 10800}}{{ \pi }} \approx \frac{{6.2856 \times 10^{-3}}}{{\pi }} \approx 2 \times 10^{-3} \text{{ m}} = 0.2 \text{{ cm}}$

Step 3: Calculate the number of fringes per cm:\nNumber of fringes per cm $= \frac{1}{\beta} = \frac{1}{{0.2}} = 5 \text{{ fringes/cm}}$

Why other options are wrong:
- Option A: 2 per cm: Incorrect calculation of fringe width.
- Option C: 7 per cm: Incorrect calculation of fringe width.
- Option D: 6.5 per cm: Incorrect calculation of fringe width.

Key Concepts:
- Interference
- Wedge film
- Fringe width

Memory Aids:
- $\beta = \frac{{\lambda }}{{2n\theta}}$

Quick Tips:
- Convert all units to be consistent before calculating.

4 Easy

4 of 23Question permalink

Application +4 | -1

In a Young's double slit experiment the slit separation is

0.5 \mathrm{~mm}

and the screen is

0.5 \mathrm{~m}

from the slit. For a monochromatic light of wavelength

500 \mathrm{~nm}

the distance of

3^{\text {rd}}

maxima from the

2^{\text {nd}}

minima on the other side is:

View answer and explanation

Correct option: D

2.25 mm

The correct answer is D (2.25 mm) because it is the calculated distance between the 3rd maxima and the 2nd minima on the opposite side in Young's double slit experiment.

Solution approach

Correct Answer + Core Reason: 2.25 mm. The distance is calculated by finding the positions of the 3rd maxima and 2nd minima and then summing their distances from the central maxima.

Step-by-Step Solution:

Step 1: Identify the given information: Slit separation $d = 0.5 \text{ mm} = 0.5 \times 10^{-3} \text{ m}$ , Screen distance $D = 0.5 \text{ m}$ , Wavelength $\lambda = 500 \text{ nm} = 500 \times 10^{-9} \text{ m}$ .

Step 2: The position of the $m$ -th maxima is given by $y_m = \frac{{m\lambda D}}{d}$ .

Step 3: The position of the $n$ -th minima is given by $y_n = \frac{{(2n-1)\lambda D}}{{2d}}$ .

Step 4: The position of the 3rd maxima is $y_3 = \frac{{3\lambda D}}{d} = \frac{{3 \times 500 \times 10^{-9} \times 0.5}}{{0.5 \times 10^{-3}}} = 1.5 \times 10^{-3} \text{ m} = 1.5 \text{ mm}$ .

Step 5: The position of the 2nd minima is $y_2 = \frac{{(2(2)-1)\lambda D}}{{2d}} = \frac{{3\lambda D}}{{2d}} = \frac{{3 \times 500 \times 10^{-9} \times 0.5}}{{2 \times 0.5 \times 10^{-3}}} = 0.75 \times 10^{-3} \text{ m} = 0.75 \text{ mm}$ .

Step 6: The distance between the 3rd maxima and the 2nd minima on the other side is $y_3 + y_2 = 1.5 \text{ mm} + 0.75 \text{ mm} = 2.25 \text{ mm}$ .

Why other options are wrong:
- Option A: This is incorrect as it is not the correct distance between the 3rd maxima and the 2nd minima.
- Option B: This is incorrect as it is not the correct distance between the 3rd maxima and the 2nd minima.
- Option C: This is incorrect as it is not the correct distance between the 3rd maxima and the 2nd minima.

Key Concepts:
- Young's double-slit experiment
- Maxima and minima positions

Memory Aids:
- Remember the formulas for the positions of maxima and minima.

Quick Tips:
- Pay attention to whether the question asks for the distance from the central maxima or the distance between two fringes.

5 Easy

5 of 23Question permalink

Application +4 | -1

A microwave of wavelength

2.0 \mathrm{~cm}

falls normally on a slit of width

4.0 \mathrm{~cm}

. The angular spread of the central maxima of the diffraction pattern obtained on a screen

1.5 \mathrm{~m}

away from the slit, will be

View answer and explanation

Correct option: C

60^{\circ}

Option C (

60^{\circ}

) is correct because the angular spread of the central maxima is given by

2\theta

, where

\sin \theta = \frac{\lambda}{a}

.

Solution approach

Correct Answer + Core Reason: $60^{\circ}$ . The angular spread is twice the angle where the first minimum occurs.

Step-by-Step Solution:

Step 1: Recall the condition for the first minimum in single-slit diffraction: $a \sin{\theta} = \lambda$ , where $a$ is the slit width, $\theta$ is the angle, and $\lambda$ is the wavelength.

Step 2: Substitute the given values: $4.0 \text{{ cm}} \times \sin{\theta} = 2.0 \text{{ cm}}$ .

Step 3: Solve for $\sin{\theta}$ : $\sin{\theta} = \frac{{2.0 \text{{ cm}}}}{{4.0 \text{{ cm}}}} = \frac{{1}}{{2}}$ .

Step 4: Find the angle $\theta$ : $\theta = \sin^{-1}{\frac{{1}}{{2}}} = 30^{\circ}$ .

Step 5: Calculate the angular spread: The angular spread of the central maxima is $2\theta = 2 \times 30^{\circ} = 60^{\circ}$ .

Why other options are wrong:
- Option A: $45^{\circ}$ is incorrect. It might be obtained by incorrectly applying the formula or making a calculation error.
- Option B: $15^{\circ}$ is incorrect. It might be obtained by dividing the correct angle by 2 instead of multiplying by 2 for the spread.
- Option D: $30^{\circ}$ is incorrect. This is the angle to the first minimum, but the question asks for the angular spread of the central maxima, which is twice this angle.

Key Concepts:
- Single-slit diffraction
- Angular spread of central maxima
- Condition for minima

Memory Aids:
- Remember that the angular spread is twice the angle to the first minimum.

Quick Tips:
- Always remember to double the angle to find the angular spread of the central maxima.

6 Easy

6 of 23Question permalink

Application +4 | -1

In Young's double slit experiment, the width of fringe obtained from a source of light of wavelength

5000\mathop {\text{A}}\limits^ \circ

is 3.9

\mathrm{mm}

. if the apparatus is immersed in a liquid of

\mu=1.3

, the fringe width will be

View answer and explanation

Correct option: A

3

The correct answer is A (3) because the fringe width is inversely proportional to the refractive index of the medium.

Solution approach

Correct Answer + Core Reason: 3 mm. The fringe width decreases when the apparatus is immersed in a liquid due to the change in wavelength.

Step-by-Step Solution:

Step 1: Identify the given information: Wavelength $\lambda = 5000 \mathop {\text{A}}\limits^ \circ$ , Fringe width $\beta = 3.9 \text{ mm}$ , Refractive index $\mu = 1.3$ .

Step 2: Recall the formula for fringe width in Young's double-slit experiment: $\beta = \frac{{\lambda D}}{d}$ , where $D$ is the distance to the screen and $d$ is the slit separation.

Step 3: When the apparatus is immersed in a liquid, the wavelength changes to $\lambda' = \frac{\lambda}{\mu}$ .

Step 4: The new fringe width is $\beta' = \frac{{\lambda' D}}{d} = \frac{{\lambda D}}{{\mu d}} = \frac{\beta}{\mu}$ .

Step 5: Substitute the given values: $\beta' = \frac{{3.9 \text{ mm}}}{{1.3}} = 3 \text{ mm}$ .

Why other options are wrong:
- Option B: This is incorrect as the fringe width changes with the refractive index of the medium.
- Option C: This is incorrect as the fringe width changes with the refractive index of the medium.
- Option D: This is incorrect as the fringe width changes with the refractive index of the medium.

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

Memory Aids:
- Remember that the fringe width is inversely proportional to the refractive index.

Quick Tips:
- Always remember to adjust the wavelength when the experiment is performed in a medium other than air or vacuum.

7 Easy

7 of 23Question permalink

Understanding +4 | -1

If the monochromatic source in Young's double slit experiment is replaced by white light, then

View answer and explanation

Correct option: B

There will be a central bright white fringe surrounded by a few coloured fringes

Option B is correct. With white light, the central fringe is white because all colors interfere constructively. Colored fringes appear on either side due to different wavelengths interfering at different locations.

Solution approach

Correct Answer + Core Reason: There will be a central bright white fringe surrounded by a few coloured fringes. White light consists of multiple wavelengths, leading to colored fringes around a central white fringe.

Step-by-Step Solution:

Step 1: Understand white light: White light is composed of a continuous spectrum of wavelengths.

Step 2: Analyze the central fringe: At the central point, the path difference is zero for all wavelengths. Therefore, all colors interfere constructively, resulting in a bright white fringe.

Step 3: Analyze the fringes away from the center: Away from the center, different wavelengths will have different path differences, leading to constructive interference at different locations. This results in colored fringes.

Why other options are wrong:
- Option A: Incorrect because the central fringe is bright, not dark.
- Option C: Incorrect because the fringe width depends on the wavelength, so they will not be of equal width.
- Option D: Incorrect because interference still occurs, but the pattern is more complex due to the multiple wavelengths.

Key Concepts:
- Young's double-slit experiment
- White light
- Interference
- Wavelength

Memory Aids:
- White Light = Central White Fringe + Colored Fringes

Quick Tips:
- Remember that white light produces a central white fringe in interference experiments.

8 Easy

8 of 23Question permalink

Understanding +4 | -1

Statement-1: In interference experiment, the fringe system will shift, if a glass plate is interposed in the path of the one of the interfering beams and the shift will be on the side of the glass plate.
Statement-2: The velocity of light in glass is smaller than in air.

View answer and explanation

Correct option: A

Statement-1 is true, Statement-2 is true, Statement-2 is the correct explanation for Statement-1

Option A is correct. Statement-1 is true because the introduction of a glass plate shifts the fringe system. Statement-2 is true because light travels slower in glass than in air, and it is the correct explanation for the shift.

Solution approach

Correct Answer + Core Reason: Statement-1 is true, Statement-2 is true, Statement-2 is the correct explanation for Statement-1. The shift in the fringe system is due to the change in the optical path length caused by the glass plate.

Step-by-Step Solution:

Step 1: Analyze Statement-1: Introducing a glass plate increases the optical path length of one of the interfering beams, causing a shift in the fringe system towards the side of the glass plate. Thus, Statement-1 is true.

Step 2: Analyze Statement-2: The velocity of light is lower in glass than in air due to the refractive index of glass being greater than 1. Thus, Statement-2 is true.

Step 3: Connect the statements: The shift in the fringe system is directly caused by the reduced speed of light in the glass plate, which increases the optical path length. Therefore, Statement-2 is the correct explanation for Statement-1.

Why other options are wrong:
- Option B: Incorrect because Statement-2 is the correct explanation for Statement-1.
- Option C: Incorrect because Statement-2 is true.
- Option D: Incorrect because Statement-1 is true.

Key Concepts:
- Interference
- Optical path length
- Refractive index

Memory Aids:
- Glass Plate = Shift in Fringes

Quick Tips:
- Remember that the shift in the fringe system is due to the change in optical path length.

9 Easy

9 of 23Question permalink

Application +4 | -1

A parallel beam of monochromatic unpolarised
light is incident on a transparent plate of
refractive index

\sqrt{3}.

The reflected beam is
completely polarized. The angle of incidence
is

View answer and explanation

Correct option: B

60^\circ

Option B (

60^\circ

) is correct because at the Brewster angle, the reflected light is completely polarized, and the tangent of this angle equals the refractive index.

Solution approach

Correct Answer + Core Reason: $60^\circ$ . The angle of incidence is the Brewster angle, where $\tan i = \mu$ .

Step-by-Step Solution:

Step 1: Identify the given information: Refractive index $\mu = \sqrt{3}$ . We need to find the angle of incidence $i$ .

Step 2: Recall Brewster's Law: The angle of incidence $i$ for which the reflected light is completely polarized is given by $\tan i = \mu$ .

Step 3: Apply Brewster's Law: $\tan i = \sqrt{3}$

Step 4: Solve for $i$ : $i = \arctan(\sqrt{3})$

Step 5: Calculate the angle: $i = 60^\circ$

Step 6: Final Answer: $60^\circ$

Why other options are wrong:
- **Option A ( $30^\circ$ ):** This is incorrect because $\tan 30^\circ = \frac{1}{\sqrt{3}}$ , not $\sqrt{3}$ .
- **Option C ( $45^\circ$ ):** This is incorrect because $\tan 45^\circ = 1$ , not $\sqrt{3}$ .
- **Option D ( $75^\circ$ ):** This is incorrect because $\tan 75^\circ = 2 + \sqrt{3}$ , not $\sqrt{3}$ .

Key Concepts:
- Brewster's Law
- Polarization of Light
- Refractive Index

Memory Aids:
- Remember that at the Brewster angle, the reflected and refracted rays are perpendicular.

Quick Tips:
- Always remember Brewster's Law: $\tan i = \mu$ when the reflected light is completely polarized.

10 Easy

10 of 23Question permalink

Application +4 | -1

Two light waves of the same intensity are interfering. The intensity of light at a bright fringe compared to the intensity of one of the waves will

View answer and explanation

Correct option: C

Be four times

Option C (Be four times) is correct because the intensity of light is proportional to the square of the amplitude, and in constructive interference, the amplitudes add up.

Solution approach

Correct Answer + Core Reason: Be four times. Intensity is proportional to the square of the amplitude, and amplitudes add in constructive interference.

Step-by-Step Solution:

Step 1: Identify the given information:\nTwo light waves have the same intensity, $I_1 = I_2 = I$ \nThey are interfering at a bright fringe (constructive interference).

Step 2: Key Concepts:
- Intensity is proportional to the square of the amplitude: $I \propto A^2$
- For constructive interference, the amplitudes add up: $A_{resultant} = A_1 + A_2$

Step 3: Explanation:\nSince the intensities are the same, the amplitudes are also the same: $A_1 = A_2 = A$ \nAt a bright fringe, the amplitudes add up: $A_{resultant} = A + A = 2A$ \nThe resultant intensity is: $I_{resultant} \propto (2A)^2 = 4A^2$ \nSince $I \propto A^2$ , $I_{resultant} = 4I$

Step 4: Conclusion: The intensity of light at a bright fringe is four times the intensity of one of the waves.

Why other options are wrong:
- Option A: Become two times: This would be true for the amplitudes adding linearly, but intensity is proportional to the square of the amplitude.
- Option B: Be three times: This is an incorrect calculation.
- Option D: Be eight times: This is an incorrect calculation.

Key Concepts:
- Interference of light
- Intensity of light
- Amplitude of light

Memory Aids:
- $I \propto A^2$

Quick Tips:
- Remember that intensity is proportional to the square of the amplitude.

11 Easy

11 of 23Question permalink

Understanding +4 | -1

In young's double slit experiment with a source of light of wavelength

6320\,\mathop {\text{A}}\limits^ \circ

, the first maxima will occur when

View answer and explanation

Correct option: C

Path difference is

6320\,\mathop {\text{A}}\limits^ \circ

Option C is correct. The first maxima occurs when the path difference is equal to one wavelength, which is

6320\,\mathop {\text{A}}\limits^ \circ

.

Solution approach

Correct Answer + Core Reason: Path difference is $6320\,\mathop {\text{A}}\limits^ \circ$ . The first maxima occurs when the path difference is equal to the wavelength.

Step-by-Step Solution:

Step 1: Define first maxima: The first maxima (or first bright fringe) in Young's double-slit experiment corresponds to $n = 1$ in the constructive interference condition.

Step 2: Apply the condition: For constructive interference, the path difference ( $\Delta x$ ) is given by $\Delta x = n\lambda$ . For the first maxima ( $n = 1$ ), $\Delta x = \lambda$ .

Step 3: Substitute the given wavelength: Given $\lambda = 6320\,\mathop {\text{A}}\limits^ \circ$ , the path difference for the first maxima is $\Delta x = 6320\,\mathop {\text{A}}\limits^ \circ$ .

Why other options are wrong:
- Option A: Incorrect because $9480\,\mathop {\text{A}}\limits^ \circ$ is not equal to the wavelength.
- Option B: Incorrect because a phase difference of 3 radians does not correspond to the first maxima. The phase difference should be $2\pi$ for the first maxima.
- Option D: Incorrect because a phase difference of $\pi$ radians corresponds to a minimum (destructive interference).

Key Concepts:
- Young's double-slit experiment
- Constructive interference
- Path difference
- Wavelength

Memory Aids:
- First Maxima = One Wavelength Path Difference

Quick Tips:
- Remember the path difference condition for constructive interference: $\Delta x = n\lambda$ .

12 Easy

12 of 23Question permalink

Understanding +4 | -1

Assertion: Distance between position of bright and dark fringe remain same in YDSE
Reason: Fringe width

\beta=\frac{\lambda D}{d}

View answer and explanation

Correct option: A

If both Assertion and Reason are correct and Reason is the correct explanation of Assertion

Both the Assertion and Reason are correct, and the Reason is the correct explanation of the Assertion because the fringe width determines the distance between bright and dark fringes in YDSE.

Solution approach

Correct Answer + Core Reason: If both Assertion and Reason are correct and Reason is the correct explanation of Assertion. The fringe width determines the spacing between bright and dark fringes in YDSE.

Step-by-Step Solution:

Step 1: Understand YDSE: In Young's Double Slit Experiment (YDSE), interference fringes are formed on a screen due to the superposition of light waves from two coherent sources.

Step 2: Define Fringe Width: The fringe width ( $\beta$ ) is the distance between two consecutive bright or dark fringes.

Step 3: State the Formula: The fringe width is given by the formula $\beta = \frac{\lambda D}{d}$ , where $\lambda$ is the wavelength of light, D is the distance between the slits and the screen, and d is the distance between the two slits.

Step 4: Analyze the Assertion: The distance between the position of bright and dark fringes remains the same because the fringe width is constant.

Step 5: Analyze the Reason: The reason correctly states the formula for fringe width, which explains why the distance between bright and dark fringes remains the same.

Why other options are wrong:
- Option B: The Reason *is* the correct explanation of the Assertion.
- Option C: The Reason is correct, so this option is incorrect.
- Option D: Both Assertion and Reason are correct, so this option is incorrect.

Key Concepts:
- Young's Double Slit Experiment (YDSE)
- Fringe Width
- Interference

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

Quick Tips:
- The fringe width is directly proportional to the wavelength and the distance to the screen, and inversely proportional to the slit separation.

13 Easy

13 of 23Question permalink

Remembering +4 | -1

According to Huygen’s theory of light

View answer and explanation

Correct option: B

Light is a wave

Option B is correct. Huygens' theory of light describes light as a wave, not as a stream of photons or particles.

Solution approach

Correct Answer + Core Reason: Light is a wave. Huygens' principle describes light propagation as a wave phenomenon.

Step-by-Step Solution:

Step 1: Recall Huygens' theory: Huygens' principle states that every point on a wavefront is a source of secondary spherical wavelets. These wavelets interfere to create the next wavefront.

Step 2: Relate to the options: This description aligns with light being a wave.

Why other options are wrong:
- Option A: Incorrect because light as a stream of photons is a quantum mechanical description, not Huygens'.
- Option C: Incorrect because while light exhibits particle-like behavior in some phenomena, Huygens' theory specifically describes it as a wave.
- Option D: Incorrect because Huygens' theory does describe light.

Key Concepts:
- Huygens' principle
- Wave nature of light

Memory Aids:
- Huygens = Waves

Quick Tips:
- Remember that Huygens' principle is a wave theory.

14 Easy

14 of 23Question permalink

Application +4 | -1

An objective lens of a telescope has diameter 200 cm. The limit of resolution of telescope for wavelength 600 nm is nearly

View answer and explanation

Correct option: B

3.7\mathrm{ } \times \mathrm{ }{10^{-7}}rad

Option B (

3.7\mathrm{ } \times \mathrm{ }{10^{-7}}rad

) is correct because it is calculated using the formula for the limit of resolution of a telescope.

Solution approach

Correct Answer + Core Reason: $3.7\mathrm{ } \times \mathrm{ }{10^{-7}}rad$ . The limit of resolution is calculated using the formula $\theta = 1.22 \frac{\lambda}{D}$ .

Step-by-Step Solution:

Step 1: Identify the given values: Diameter $D = 200 \text{ cm} = 2 \text{ m}$ , Wavelength $\lambda = 600 \text{ nm} = 600 \times 10^{-9} \text{ m}$ .

Step 2: Use the formula for the limit of resolution: $\theta = 1.22 \frac{\lambda}{D}$ .

Step 3: Substitute the values: $\theta = 1.22 \times \frac{600 \times 10^{-9}}{2}$ .

Step 4: Calculate the result: $\theta = 1.22 \times 300 \times 10^{-9} = 366 \times 10^{-9} = 3.66 \times 10^{-7} \text{ rad}$ .

Step 5: Round to the nearest value: $\theta \approx 3.7 \times 10^{-7} \text{ rad}$ .

Why other options are wrong:
- Option A: $9\mathrm{ } \times \mathrm{ }{10^{ - 7}}rad$ : This is incorrect as it doesn't use the correct formula or substitution.
- Option C: $3.7\mathrm{ } \times \mathrm{ }{10^{-8}}rad$ : This is incorrect because there is a miscalculation in the power of 10.
- Option D: $3.7\mathrm{ } \times \mathrm{ }{10^{-8}}rad$ : This is incorrect because there is a miscalculation in the power of 10.

Key Concepts:
- Limit of resolution of a telescope
- Rayleigh criterion

Memory Aids:
- Remember the formula: $\theta = 1.22 \frac{\lambda}{D}$

Quick Tips:
- Ensure that the units are consistent before substituting into the formula.
- Remember the factor of 1.22 in the Rayleigh criterion.

15 Easy

15 of 23Question permalink

Application +4 | -1

The diffraction pattern of a light of wavelength

400 \mathrm{~nm}

diffracting from a salt of width

0.2 \mathrm{~mm}

is focused on the focal plane of a convex lens of focal length

100 \mathrm{~cm}

. The width of the

1^{\text {st }}

secondary

View answer and explanation

Correct option: B

2 \mathrm{~mm}

The correct answer is B (2 mm) because the width of the first secondary maximum in a single-slit diffraction pattern is twice the distance from the central maximum to the first minimum.

Solution approach

Correct Answer + Core Reason: 2 mm. The width of the first secondary maximum is calculated using the formula for the position of the first minimum in single-slit diffraction.

Step-by-Step Solution:

Step 1: Identify the given information: Wavelength $\lambda = 400 \text{ nm} = 400 \times 10^{-9} \text{ m}$ , Slit width $a = 0.2 \text{ mm} = 0.2 \times 10^{-3} \text{ m}$ , Focal length $f = 100 \text{ cm} = 1 \text{ m}$ .

Step 2: The position of the first minimum in single-slit diffraction is given by $\sin \theta = \frac{\lambda}{a}$ .

Step 3: Since the angle is small, we can approximate $\sin \theta \approx \theta$ . Therefore, $\theta = \frac{\lambda}{a}$ .

Step 4: The distance from the central maximum to the first minimum on the screen is $y = f \theta = f \frac{\lambda}{a} = 1 \times \frac{{400 \times 10^{-9}}}{{0.2 \times 10^{-3}}} = 2 \times 10^{-3} \text{ m} = 2 \text{ mm}$ .

Step 5: The width of the first secondary maximum is twice the distance from the central maximum to the first minimum, so the width is $2y = 2 \times 2 \text{ mm} = 4 \text{ mm}$ . However, the question asks for the width of the *first secondary*, which is from the first minima on one side to the first minima on the other side. The distance from the central maxima to the first minima is $y = 2 \text{ mm}$ . Therefore, the width is $2 \text{ mm}$ .

Why other options are wrong:
- Option A: This is incorrect as it is the slit width, not the width of the first secondary maximum.
- Option C: This is incorrect as it is not the correct width of the first secondary maximum.
- Option D: This is incorrect as it is not the correct width of the first secondary maximum.

Key Concepts:
- Single-slit diffraction
- Position of minima

Memory Aids:
- Remember the formula for the position of the first minimum in single-slit diffraction.

Quick Tips:
- Pay attention to whether the question asks for the position of the minimum or the width of the secondary maximum.

16 Easy

16 of 23Question permalink

Application +4 | -1

In a Young’s double slit experiment, a student observes 8 fringes in a certain segment of screen when a monochromatic light of 600 nm wavelength is used. If the wavelength of light is changed to 400 nm, then the number of fringes he would observe in the same region of the screen is:

View answer and explanation

Correct option: A

12

Option A (12) is correct because the number of fringes is inversely proportional to the wavelength.

Solution approach

Correct Answer + Core Reason: 12. The number of fringes is inversely proportional to the wavelength of light.

Step-by-Step Solution:

Step 1: Let $n_1$ be the number of fringes observed with wavelength $\lambda_1$ , and $n_2$ be the number of fringes observed with wavelength $\lambda_2$ .

Step 2: We are given $n_1 = 8$ when $\lambda_1 = 600 \text{ nm}$ , and we want to find $n_2$ when $\lambda_2 = 400 \text{ nm}$ .

Step 3: Since the fringe width is proportional to the wavelength, and the segment of the screen is the same, the number of fringes is inversely proportional to the wavelength. Therefore, $n_1 \lambda_1 = n_2 \lambda_2$ .

Step 4: Substitute the given values: $8 \times 600 = n_2 \times 400$ .

Step 5: Solve for $n_2$ : $n_2 = \frac{8 \times 600}{400} = \frac{4800}{400} = 12$ .

Why other options are wrong:
- Option B: 6: This is incorrect as it calculates $n_2 = \frac{8 \times 400}{600}$ instead of $n_2 = \frac{8 \times 600}{400}$ .
- Option C: 8: This is incorrect as it assumes the number of fringes remains the same, which is not true when the wavelength changes.
- Option D: 9: This is incorrect as it is not the correct calculation based on the inverse proportionality.

Key Concepts:
- Young's double-slit experiment
- Fringe width and wavelength relationship

Memory Aids:
- Remember that fringe width is directly proportional to wavelength: $\beta \propto \lambda$ .
- Number of fringes in a given region is inversely proportional to wavelength.

Quick Tips:
- In Young's double-slit experiment, if the region of the screen is constant, the number of fringes is inversely proportional to the wavelength.

17 Easy

17 of 23Question permalink

Application +4 | -1

If the free space between the plates is replaced by a transparent liquid of refractive index

1.5

and the wavelength of light remains

4200 \mathop A\limits^ \circ

the number of fringes seen will be:

View answer and explanation

Correct option: A

45

Option A (45) is correct because the number of fringes is directly proportional to the refractive index of the medium.

Solution approach

Correct Answer + Core Reason: 45. The number of fringes is directly proportional to the refractive index.

Step-by-Step Solution:

Step 1: Identify the given information:\nRefractive index of air, $n_1 = 1$ \nNumber of fringes in air, $n_{f1} = 30$ \nRefractive index of liquid, $n_2 = 1.5$ \nWavelength, $\lambda = 4200 \mathop A\limits^\circ$ \nNumber of fringes in liquid, $n_{f2} = ?$

Step 2: Key Concept: The number of fringes is directly proportional to the refractive index.
$n_f \propto n$
$\frac{{n_{f1}}}{{n_1}} = \frac{{n_{f2}}}{{n_2}}$

Step 3: Apply the formula:
$\frac{{30}}{1} = \frac{{n_{f2}}}{{1.5}}$
$n_{f2} = 30 \times 1.5 = 45$

Step 4: Conclusion: The number of fringes seen will be 45.

Why other options are wrong:
- Option B: 0: This is incorrect because the number of fringes changes with the refractive index.
- Option C: 0: This is incorrect because the number of fringes changes with the refractive index.
- Option D: 0: This is incorrect because the number of fringes changes with the refractive index.

Key Concepts:
- Interference
- Fringes
- Refractive index

Memory Aids:
- $n_f \propto n$

Quick Tips:
- Remember the direct relationship between the number of fringes and the refractive index.

18 Easy

18 of 23Question permalink

Application +4 | -1

Monochromatic light of wavelength

600 \mathrm{~nm}

is incident from air on a glass surface (refractive index of glass is

1.5

). The frequency of the refracted light will be

View answer and explanation

Correct option: A

5 \times 10^{14} \mathrm{~Hz}

Option A (

5 \times 10^{14} \mathrm{~Hz}

) is correct because the frequency of light remains constant when it travels from one medium to another; it only depends on the source.

Solution approach

Correct Answer + Core Reason: $5 \times 10^{14} \mathrm{~Hz}$ . The frequency of light remains constant when it refracts from one medium to another.

Step-by-Step Solution:

Step 1: Identify the given values: Wavelength in air, $\lambda = 600 \, \mathrm{nm} = 600 \times 10^{-9} \, \mathrm{m}$ .

Step 2: Use the formula relating speed, frequency, and wavelength: $c = f \lambda$ , where $c$ is the speed of light in vacuum, and $f$ is the frequency.

Step 3: Calculate the frequency: $f = \frac{c}{\lambda} = \frac{3 \times 10^8 \, \mathrm{m/s}}{600 \times 10^{-9} \, \mathrm{m}} = \frac{3 \times 10^8}{6 \times 10^{-7}} = 0.5 \times 10^{15} \, \mathrm{Hz} = 5 \times 10^{14} \, \mathrm{Hz}$ .

Why other options are wrong:
- **Option B ( $4 \times 10^{14} \mathrm{~Hz}$ ): This is incorrect because it might be calculating the frequency using the wavelength in the glass, which is not needed since frequency remains constant.
- Option C ( $2.5 \times 10^{14} \mathrm{~Hz}$ ): This is incorrect because it might be dividing the frequency by the refractive index of the glass, which is not the correct approach since frequency remains constant.
- Option D ( $2 \times 10^{14} \mathrm{~Hz}$ ): This is incorrect because it might be dividing the speed of light by the refractive index and then dividing by the wavelength, which is not the correct approach.

Key Concepts:
- Refraction of light
- Frequency of light
- Relationship between speed, frequency, and wavelength

Memory Aids:
- Remember that the frequency of light remains constant during refraction.

Quick Tips:**
- The frequency of light depends only on the source and does not change when the light travels from one medium to another.

19 Easy

19 of 23Question permalink

Application +4 | -1

Two glass plates are touching at one end and separated by a thin wire at the other end. When a monochromatic parallel beam of wavelength

4200 \mathop A\limits^ \circ

incident normally on the glass plates is reflected, an interference pattern of 30 fringes are observed. If the wavelength of light used is taken

7000 \mathop A\limits^ \circ

instead of

4200 \mathop A\limits^ \circ

, the number of fringes observed will be:

View answer and explanation

Correct option: A

18

Option A (18) is correct because the number of fringes is inversely proportional to the wavelength of light used.

Solution approach

Correct Answer + Core Reason: 18. The number of fringes is inversely proportional to the wavelength.

Step-by-Step Solution:

Step 1: Identify the given information:\nWavelength 1, $\lambda_1 = 4200 \mathop A\limits^\circ$ \nNumber of fringes 1, $n_1 = 30$ \nWavelength 2, $\lambda_2 = 7000 \mathop A\limits^\circ$ \nNumber of fringes 2, $n_2 = ?$

Step 2: Key Concept: The number of fringes is inversely proportional to the wavelength.
$n \propto \frac{1}{\lambda}$
$n\lambda = \text{{constant}}$

Step 3: Apply the formula:
$n_1\lambda_1 = n_2\lambda_2$
$30 \times 4200 = n_2 \times 7000$
$n_2 = \frac{{30 \times 4200}}{{7000}} = \frac{{30 \times 42}}{{70}} = \frac{{30 \times 6}}{{10}} = 3 \times 6 = 18$

Step 4: Conclusion: The number of fringes observed will be 18.

Why other options are wrong:
- Option B: 0: This is incorrect because the number of fringes changes with the wavelength.
- Option C: 0: This is incorrect because the number of fringes changes with the wavelength.
- Option D: 0: This is incorrect because the number of fringes changes with the wavelength.

Key Concepts:
- Interference
- Fringes
- Wavelength

Memory Aids:
- $n\lambda = \text{{constant}}$

Quick Tips:
- Remember the inverse relationship between the number of fringes and the wavelength.

20 Easy

20 of 23Question permalink

Understanding +4 | -1

A thin air film is formed by putting the convex surface of a plane-convex lens over a plane glass plate. With monochromatic light, this film gives an interference pattern due to light reflected from the top (convex) surface and the bottom (glass plate) surface of the film.

Statement I : When light reflects from the air-glass plate interface, the reflected wave suffers a phase change of

\pi

.

Statement II : The centre of the interference pattern is dark.

View answer and explanation

Correct option: B

Statement I is true, Statement II is true; Statement II is not the correct explanation for Statement I

Option B is correct. Statement I is true because light reflecting from a denser medium (air-glass) undergoes a phase change of

\pi

. Statement II is also true because the center is dark due to this phase change, but the phase change is not the explanation for statement I itself.

Solution approach

Correct Answer + Core Reason: Statement I is true, Statement II is true; Statement II is not the correct explanation for Statement I. Both statements are correct, but they are not directly related as cause and effect.

Step-by-Step Solution:

Step 1: Analyze Statement I: When light reflects from a medium with a higher refractive index (air to glass), there is a phase change of $\pi$ (or 180 degrees). Thus, Statement I is true.

Step 2: Analyze Statement II: At the center of the interference pattern, the air film has zero thickness. Due to the $\pi$ phase change upon reflection from the glass plate, the interference is destructive, resulting in a dark spot. Thus, Statement II is true.

Step 3: Determine the relationship: While both statements are true, the dark center is a consequence of the phase change, but the phase change itself doesn't explain why Statement I is true. Statement I is true because of the reflection at a denser medium.

Why other options are wrong:
- Option A: Incorrect because Statement II is not the correct explanation for Statement I.
- Option C: Incorrect because Statement II is true.
- Option D: Incorrect because Statement I is true.

Key Concepts:
- Phase change upon reflection
- Interference
- Thin films

Memory Aids:
- Reflection at Denser Medium = Pi Phase Change

Quick Tips:
- Remember the phase change rule for reflection at interfaces.

21 Easy

21 of 23Question permalink

Analyzing +4 | -1

Statement-1: In YDSE, if separation between slits is less than wavelength of light, then no interference pattern could be observed.
Statement-2: For interference pattern to be observed, light sources have to be coherent.

View answer and explanation

Correct option: B

Statement-1 is true, Statement-2 is true, Statement-2 is not the correct explanation for Statement-1

Option B is correct. Statement-1 is true because if the slit separation is less than the wavelength, no distinct interference pattern is formed. Statement-2 is also true because coherent sources are necessary for sustained interference, but the coherence is not the direct reason for the absence of a pattern when slit separation is less than the wavelength.

Solution approach

Correct Answer + Core Reason: Statement-1 is true, Statement-2 is true, Statement-2 is not the correct explanation for Statement-1. The slit separation being less than the wavelength prevents a proper interference pattern, and coherence is a general requirement for interference.

Step-by-Step Solution:

Step 1: Analyze Statement-1: In YDSE, the condition for observing an interference pattern is that the path difference between the waves from the two slits should be comparable to the wavelength of light. If the separation between the slits ( $d$ ) is less than the wavelength ( $\lambda$ ), the path difference will be too small to produce distinct fringes. Therefore, no clear interference pattern will be observed. Statement-1 is true.

Step 2: Analyze Statement-2: For interference to occur, the light sources must be coherent, meaning they must have a constant phase difference. This is a fundamental requirement for observing a sustained interference pattern. Statement-2 is true.

Step 3: Determine if Statement-2 explains Statement-1: While coherence is necessary for interference, the reason why no interference pattern is observed when the slit separation is less than the wavelength is primarily due to the small path difference, not the lack of coherence. Coherence is a separate requirement. Therefore, Statement-2 is not the correct explanation for Statement-1.

Why other options are wrong:
- Option A: Statement-2 is not the correct explanation for Statement-1.
- Option C: Statement-2 is true.
- Option D: Statement-1 is true.

Key Concepts:
- Young's Double Slit Experiment (YDSE)
- Interference of Light
- Coherence
- Path Difference
- Wavelength

Memory Aids:
- Remember that coherence is a necessary but not sufficient condition for observing an interference pattern.

Quick Tips:
- Understand the conditions required for observing a clear interference pattern in YDSE.

22 Easy

22 of 23Question permalink

Understanding +4 | -1

Assertion: Incoming light reflected by earth is partially polarized.
Reason: Atmospheric particle polarize the light.

View answer and explanation

Correct option: A

If both Assertion and Reason are true and Reason is the correct explanation of Assertion

Both the Assertion and Reason are true, and the Reason is the correct explanation of the Assertion. Atmospheric particles scatter sunlight, and this scattering causes the reflected light to be partially polarized.

Solution approach

Correct Answer + Core Reason: If both Assertion and Reason are true and Reason is the correct explanation of Assertion. Atmospheric scattering causes polarization of reflected light.

Step-by-Step Solution:

Step 1: Understand Polarization by Scattering: When light interacts with particles in the atmosphere, it is scattered in different directions.

Step 2: Explain Polarization: This scattering process can cause the light to become partially polarized, especially at certain angles.

Step 3: Relate to Earth's Reflected Light: The light reflected by the Earth's surface and atmosphere is therefore partially polarized due to this scattering effect.

Why other options are wrong:
- Option B: The Reason *is* the correct explanation of the Assertion.
- Option C: The Reason is correct, so this option is incorrect.
- Option D: Both Assertion and Reason are true, so this option is incorrect.

Key Concepts:
- Polarization
- Scattering of Light
- Atmospheric Optics

Memory Aids:
- Remember that scattering of light by atmospheric particles can cause polarization.

Quick Tips:
- Light reflected from surfaces or scattered by particles is often partially polarized.

23 Easy

23 of 23Question permalink

Application +4 | -1

A glass plate

12 \times 10^{-3} \mathrm{~mm}

thick is placed in the path of one of the interfering beams in a Young's double slit interference arrangement using monochromatic light of wavelength

6000 \mathop A\limits^ \circ

. If the central band shifts a distance equal to width of 10 bands. What is the thickness (in

\mu \mathrm{m}

) of the plate of diamond of refractive index

2.5

that has to be introduced in the path of second beam to bring the central band to original position?

View answer and explanation

Correct option: A

4

Option A (4) is correct because the optical path difference introduced by the glass plate must be compensated by the diamond plate.

Solution approach

Correct Answer + Core Reason: 4. The optical path difference introduced by the glass plate must be compensated by the diamond plate.

Step-by-Step Solution:

Step 1: Identify the given information:\nThickness of glass plate, $t_g = 12 \times 10^{-3} \text{{ mm}} = 12 \times 10^{-6} \text{{ m}}$ \nWavelength of light, $\lambda = 6000 \mathop A\limits^\circ = 6000 \times 10^{-10} \text{{ m}}$ \nShift of central band = width of 10 bands\nRefractive index of diamond, $n_d = 2.5$ \nRefractive index of glass (assumed), $n_g = 1.5$ (typical value for glass)

Step 2: Calculate the optical path difference introduced by the glass plate:\nOptical path difference, $\Delta x_g = (n_g - 1)t_g$ \nSince the central band shifts by 10 fringe widths, $\Delta x_g = 10\lambda$
$(n_g - 1)t_g = 10\lambda$
$(n_g - 1) \times 12 \times 10^{-6} = 10 \times 6000 \times 10^{-10}$
$n_g - 1 = \frac{{10 \times 6000 \times 10^{-10}}}{{12 \times 10^{-6}}} = \frac{{6 \times 10^{-6}}}{{12 \times 10^{-6}}} = 0.5$
$n_g = 1.5$

Step 3: Calculate the thickness of the diamond plate required to compensate for the shift:\nOptical path difference introduced by the diamond plate, $\Delta x_d = (n_d - 1)t_d$ \nTo bring the central band back to its original position, $\Delta x_d = \Delta x_g$
$(n_d - 1)t_d = (n_g - 1)t_g$
$(2.5 - 1)t_d = (1.5 - 1) \times 12 \times 10^{-6}$
$1.5t_d = 0.5 \times 12 \times 10^{-6}$
$t_d = \frac{{0.5 \times 12 \times 10^{-6}}}{{1.5}} = \frac{{6 \times 10^{-6}}}{{1.5}} = 4 \times 10^{-6} \text{{ m}}$

Step 4: Convert the thickness to micrometers:
$t_d = 4 \times 10^{-6} \text{{ m}} = 4 \mu \text{{m}}$

Why other options are wrong:
- Option B: 0: This is incorrect because a diamond plate of non-zero thickness is required to compensate for the shift.
- Option C: 0: This is incorrect because a diamond plate of non-zero thickness is required to compensate for the shift.
- Option D: 0: This is incorrect because a diamond plate of non-zero thickness is required to compensate for the shift.

Key Concepts:
- Optical path difference
- Interference
- Young's double-slit experiment

Memory Aids:
- Optical path difference = (refractive index - 1) * thickness

Quick Tips:
- Remember to convert all units to be consistent before calculating.

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