WAVE OPTICS MCQs for NEET UG PHYSICS — Mixed difficulty

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 Medium

2 of 143Question permalink

Remembering +4 | -1

H-polaroid is prepared by

View answer and explanation

Correct option: D

Stretching polyvinyl alcohol and then impregnation with iodine

Option D (Stretching polyvinyl alcohol and then impregnation with iodine) is correct because this is the standard method for creating H-polaroid.

Solution approach

Correct Answer + Core Reason: Stretching polyvinyl alcohol and then impregnation with iodine. This process aligns the polymer chains and then dopes them with iodine to create a dichroic material.

Step-by-Step Solution:

Step 1: Understand the principle of polarization. Polarization requires a material that absorbs light differently depending on its polarization direction.

Step 2: Consider the manufacturing process of H-polaroid. It involves stretching polyvinyl alcohol (PVA) to align its long-chain molecules.

Step 3: Impregnate the stretched PVA with iodine. The iodine atoms attach to the aligned PVA molecules, forming long chains that act as a dichroic absorber.

Step 4: The aligned iodine chains preferentially absorb light polarized parallel to their direction, while transmitting light polarized perpendicular to them.

Why other options are wrong:
- Option A: Orienting herapathite crystal in the same direction in nitrocellulose is a method for creating older types of polaroids, not H-polaroid.
- Option B: Using thin tourmaline crystals is a method for polarization, but it's not how H-polaroid is made. Tourmaline crystals are naturally dichroic, but they are not used in H-polaroid.
- Option C: Stretching polyvinyl alcohol and then heated with dehydration agent might change the properties of the material, but it does not create a polaroid. It does not introduce a dichroic absorber like iodine.

Key Concepts:
- Polarization of light
- Dichroism
- Polaroid manufacturing

Memory Aids:
- H-polaroid: PVA + Iodine

Quick Tips:
- Remember that H-polaroid uses PVA and iodine.

3 Medium

3 of 143Question permalink

Understanding +4 | -1

Evidence for the wave nature of light cannot be obtained from

View answer and explanation

Correct option: A

Reflection

Option A (Reflection) is correct because reflection can be explained by both wave and particle nature of light, and doesn't uniquely demonstrate wave nature.

Solution approach

Correct Answer + Core Reason: Reflection. Reflection can be explained by both wave and particle nature, not unique to wave nature.

Step-by-Step Solution:
Step 1: Identify the question type: Conceptual question about the nature of light.
Step 2: Recall the properties that demonstrate wave nature: Interference, diffraction, polarization, Doppler effect.
Step 3: Consider each option:
- Reflection: Can be explained by both wave and particle theories.
- Doppler effect: Demonstrates the change in frequency due to relative motion, a wave phenomenon.
- Interference: A wave phenomenon where waves superpose to form a resultant wave of greater or lower amplitude.
- Diffraction: The bending of waves around obstacles, a wave phenomenon.
Step 4: Conclusion: Reflection does not uniquely demonstrate the wave nature of light.

Why other options are wrong:
- Option B: Doppler effect is a wave phenomenon.
- Option C: Interference is a wave phenomenon.
- Option D: Diffraction is a wave phenomenon.

Key Concepts:
- Wave nature of light
- Particle nature of light
- Reflection, interference, diffraction, Doppler effect

Memory Aids:
- Wave properties: Interference, diffraction, polarization

Quick Tips:
- Remember that some phenomena can be explained by both wave and particle nature of light.

4 Medium

4 of 143Question permalink

Remembering +4 | -1

Which radiation in sunlight, causes heating effect

View answer and explanation

Correct option: B

Infrared

Option B (Infrared) is correct because infrared radiation is primarily responsible for the heating effect of sunlight.

Solution approach

Correct Answer + Core Reason: Infrared. Infrared radiation is responsible for the heating effect.

Step-by-Step Solution:

Step 1: Recall the electromagnetic spectrum. Sunlight consists of ultraviolet, visible, and infrared radiation.

Step 2: Identify the radiation responsible for heating. Infrared radiation is known for its thermal effects.

Why other options are wrong:
- Option A: Ultraviolet radiation causes skin damage and other biological effects, but its primary effect is not heating.
- Option C: Visible light allows us to see, but it is not the primary cause of the heating effect.
- Option D: While all contribute to some extent, infrared is the most significant for heating.

Key Concepts:
- Electromagnetic spectrum
- Infrared radiation
- Thermal radiation

Memory Aids:
- Infrared = heat.

Quick Tips:
- Remember the effects of different types of electromagnetic radiation.

5 Medium

5 of 143Question permalink

Understanding +4 | -1

In the propagation of light waves, the angle between the direction of vibration and plane of polarization is

View answer and explanation

Correct option: A

0^\circ

Option A is correct. The direction of vibration and the plane of polarization are always parallel, hence the angle between them is 0°.

Solution approach

Correct Answer + Core Reason: $0^\circ$ . The direction of vibration is parallel to the plane of polarization.

Step-by-Step Solution:

Step 1: Define plane of polarization. The plane of polarization is the plane in which the electric field vector of a polarized light wave oscillates.

Step 2: Define direction of vibration. The direction of vibration is the direction in which the electric field vector oscillates.

Step 3: Relate the direction of vibration to the plane of polarization. The direction of vibration is *in* the plane of polarization. Therefore, the angle between them is 0 degrees.

Why other options are wrong:
- Option B: This is incorrect because the vibration is *in* the plane of polarization, not perpendicular to it.
- Option C: This is incorrect because the vibration is *in* the plane of polarization, not at an angle of 45 degrees to it.
- Option D: This is incorrect because the vibration is *in* the plane of polarization, not at an angle of 80 degrees to it.

Key Concepts:
- Polarization of light
- Plane of polarization
- Direction of vibration

Memory Aids:
- Remember that the electric field vector oscillates *in* the plane of polarization.

Quick Tips:
- Visualize the electric field vector oscillating in the plane of polarization.

6 Medium

6 of 143Question permalink

Understanding +4 | -1

In Young’s experiment, one slit is covered with a blue filter and the other (slit) with a yellow filter. Then the interference pattern

View answer and explanation

Correct option: D

Will not be formed

Option D (Will not be formed) is correct because interference requires coherent sources, and using different colored filters makes the sources incoherent.

Solution approach

Correct Answer + Core Reason: Will not be formed. Interference requires coherent sources, which are not achieved with different color filters.

Step-by-Step Solution:

Step 1: Identify the problem: The problem describes Young's experiment with different color filters on each slit.

Step 2: Key Concept: Interference requires coherent sources. Coherent sources have the same frequency and a constant phase difference.

Step 3: Explanation: Using a blue filter on one slit and a yellow filter on the other means that the light from the two slits will have different wavelengths (and therefore different frequencies). This makes the sources incoherent, and no sustained interference pattern will be observed.

Step 4: Conclusion: Since the sources are incoherent, no interference pattern will be formed.

Why other options are wrong:
- Option A: Will be blue: This is incorrect because there will be no interference pattern at all, not just a blue one.
- Option B: Will be yellow: This is incorrect because there will be no interference pattern at all, not just a yellow one.
- Option C: Will be green: This is incorrect because there will be no interference pattern at all. Green is a combination of blue and yellow, but no stable pattern forms.

Key Concepts:
- Interference of light
- Coherent sources
- Young's double-slit experiment

Memory Aids:
- Interference needs same frequency (color) light.

Quick Tips:
- Remember that different colors mean different wavelengths and frequencies, which lead to incoherence.

7 Easy

7 of 143Question 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.

8 Hard

8 of 143Question permalink

Remembering +4 | -1

In Young’s double slit experiment, a minimum is obtained when the phase difference of superimposing waves is

View answer and explanation

Correct option: B

\left( {2n - 1} \right)\pi

Option B is correct. A minimum (destructive interference) is obtained when the phase difference is an odd multiple of

\pi

, represented as

(2n - 1)\pi

.

Solution approach

Correct Answer + Core Reason: $\left( {2n - 1} \right)\pi$ . A minimum occurs when the phase difference is an odd multiple of $\pi$ .

Step-by-Step Solution:

Step 1: Define minimum (destructive interference): Destructive interference occurs when waves superpose out of phase, resulting in a smaller amplitude.

Step 2: Phase difference condition: The phase difference ( $\phi$ ) for destructive interference is given by $\phi = (2n - 1)\pi$ , where $n$ is an integer (1, 2, 3, ...).

Why other options are wrong:
- Option A: Incorrect because zero phase difference corresponds to constructive interference.
- Option C: Incorrect because $n\pi$ includes both constructive (even multiples of $\pi$ ) and destructive (odd multiples of $\pi$ ) interference.
- Option D: Incorrect because $(n + 1)\pi$ also includes both constructive and destructive interference depending on n.

Key Concepts:
- Destructive interference
- Phase difference

Memory Aids:
- Minimum = Odd Multiple of Pi

Quick Tips:
- Remember the phase difference condition for destructive interference.

9 Easy

9 of 143Question 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.

10 Hard

10 of 143Question permalink

Application +4 | -1

The sodium yellow doublet has wavelengths

5890 \mathrm{~A}^0

and

\lambda \mathrm{~A}^0

and resolving power of a grating to resolve these lines is 982, then value of

\lambda

is

View answer and explanation

Correct option: A

5896 \mathrm{~A}^0

Option A (

5896 \mathrm{~A}^0

) is correct because it is calculated using the resolving power formula for a grating.

Solution approach

Correct Answer + Core Reason: $5896 \mathrm{~A}^0$ . The wavelength is calculated using the resolving power formula.

Step-by-Step Solution:

Step 1: Identify the given information: $\lambda_1 = 5890 \mathrm{~A}^0$ , resolving power $R = 982$ .

Step 2: Recall the formula for resolving power: $R = \frac{\lambda}{\Delta \lambda}$ , where $\lambda$ is the average wavelength and $\Delta \lambda$ is the difference in wavelengths.

Step 3: Let $\lambda_2 = \lambda$ . Then $\Delta \lambda = |\lambda_2 - \lambda_1| = |\lambda - 5890|$ . Also, $\lambda \approx \lambda_1$ for this calculation.

Step 4: So, $R = \frac{\lambda_1}{|\lambda - 5890|}$

Step 5: Rearrange to solve for $|\lambda - 5890|$ : $|\lambda - 5890| = \frac{\lambda_1}{R} = \frac{5890}{982} \approx 6$

Step 6: Therefore, $\lambda - 5890 = 6$ or $\lambda - 5890 = -6$ . Since $\lambda$ must be greater than $\lambda_1$ , we take the positive value.

Step 7: $\lambda = 5890 + 6 = 5896 \mathrm{~A}^0$

Step 8: Final Answer: $5896 \mathrm{~A}^0$

Why other options are wrong:
- **Option B ( $5880 \mathrm{~A}^0$ ): This is incorrect because it is a result of incorrect calculation or using the wrong formula.
- Option C ( $5869 \mathrm{~A}^0$ ): This is incorrect because it is a result of incorrect calculation or using the wrong formula.
- Option D ( $5876 \mathrm{~A}^0$ ): This is incorrect because it is a result of incorrect calculation or using the wrong formula.

Key Concepts:
- Resolving Power of a Grating
- Wavelength and Resolving Power

Memory Aids:**
- Remember the resolving power formula: $R = \frac{\lambda}{\Delta \lambda}$

Quick Tips:
- Always check the units before performing calculations.

11 Medium

11 of 143Question permalink

Remembering +4 | -1

Which of the following is a dichroic crystal?

View answer and explanation

Correct option: B

Tourmaline

Tourmaline is a dichroic crystal, meaning it exhibits different colors when viewed from different directions due to selective absorption of light.

Solution approach

Correct Answer + Core Reason: Tourmaline. Tourmaline is a dichroic crystal, meaning it absorbs light differently depending on the polarization and direction of the light.

Step-by-Step Solution:

Step 1: Define Dichroism: Dichroism is the property of a material that causes it to absorb light differently depending on the polarization and direction of the light.

Step 2: Identify Dichroic Crystals: Tourmaline is a well-known example of a dichroic crystal.

Step 3: Explain the Phenomenon: When unpolarized light passes through a dichroic crystal, one polarization component is absorbed more strongly than the other, resulting in polarized light.

Why other options are wrong:
- Option A: Quartz is not a dichroic crystal.
- Option C: Mica is not a dichroic crystal.
- Option D: Selenite is not a dichroic crystal.

Key Concepts:
- Dichroism
- Polarization
- Crystal Optics

Memory Aids:
- Remember that Tourmaline is a classic example of a dichroic crystal.

Quick Tips:
- Dichroism is related to the selective absorption of light based on its polarization.

12 Medium

12 of 143Question permalink

Analyzing +4 | -1

Assertion: We cannot observe diffraction pattern from a wide slit iluminated by monochromatic light.

Reason: In diffraction pattern, all the bright bands are not of the same intensity.

View answer and explanation

Correct option: B

Both Assertion and Reason are true and Reason is not the correct explanation of Assertion

Option B (Both Assertion and Reason are true and Reason is not the correct explanation of Assertion) is correct because both statements are true, but the reason does not explain the assertion.

Solution approach

Correct Answer + Core Reason: Both Assertion and Reason are true and Reason is not the correct explanation of Assertion. The assertion is true, and the reason is also true, but they are not related.

Step-by-Step Solution:

Step 1: Analyze the Assertion: We cannot observe a diffraction pattern from a wide slit illuminated by monochromatic light. This statement is true. If the slit is too wide compared to the wavelength of light, the diffraction effects become negligible, and no clear diffraction pattern is observed.

Step 2: Analyze the Reason: In a diffraction pattern, all the bright bands are not of the same intensity. This statement is also true. The central bright fringe is the most intense, and the intensity decreases as you move away from the center.

Step 3: Determine the relationship: The reason does not explain why we cannot observe a diffraction pattern from a wide slit. The reason describes a property of diffraction patterns in general, but it doesn't explain the condition for observing a diffraction pattern in the first place.

Step 4: Final Answer: Both Assertion and Reason are true and Reason is not the correct explanation of Assertion

Why other options are wrong:
- Option A: The reason does not explain the assertion, so this option is incorrect.
- Option C: The assertion is true, so this option is incorrect.
- Option D: The assertion is true, so this option is incorrect.

Key Concepts:
- Single Slit Diffraction
- Diffraction Pattern Intensity

Memory Aids:
- Diffraction is significant when the wavelength is comparable to the slit width.

Quick Tips:
- Consider the relative sizes of the wavelength and the slit width when analyzing diffraction.

13 Easy

13 of 143Question 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.

14 Medium

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Remembering +4 | -1

The resolving limit of healthy eye is about

View answer and explanation

Correct option: A

1'

or

{\left( {\frac{1}{{60}}} \right)^0}

Option A is correct. The resolving limit of a healthy human eye is approximately 1 minute of arc, or (1/60) of a degree.

Solution approach

Correct Answer + Core Reason: $1'$ or ${\left( {\frac{{1}}{{{{60}}}}} \right)^0}$ . The resolving limit of the human eye is about 1 arcminute.

Step-by-Step Solution:

Step 1: Define resolving limit. The resolving limit is the smallest angular separation between two objects that can be distinguished as separate.

Step 2: Recall the approximate resolving limit of a healthy human eye. The resolving limit of a healthy human eye is approximately 1 arcminute (1').

Step 3: Convert arcminutes to degrees. 1 arcminute is equal to 1/60 of a degree, so $1' = (\frac{1}{60})^{\circ}$ .

Why other options are wrong:
- Option B: 1 arcsecond (1'') is much smaller than the resolving limit of the human eye. 1 arcsecond is 1/60 of an arcminute.
- Option C: 1 degree (1°) is much larger than the resolving limit of the human eye.
- Option D: $\frac{1}{60}$ arcseconds is much smaller than the resolving limit of the human eye.

Key Concepts:
- Resolving limit
- Angular resolution
- Arcminutes and arcseconds

Memory Aids:
- Remember that 1 degree = 60 arcminutes, and 1 arcminute = 60 arcseconds.

Quick Tips:
- Be familiar with the typical angular resolution of the human eye.

15 Medium

15 of 143Question permalink

Understanding +4 | -1

A Young’s double slit set up for interference is shifted from air to within water, then the fringe width

View answer and explanation

Correct option: B

Decreases

When a Young's double-slit setup is moved from air to water, the wavelength of light decreases, causing the fringe width to decrease as well. The refractive index of water is greater than air.

Solution approach

Correct Answer + Core Reason: Decreases. The wavelength of light decreases in water, leading to a decrease in fringe width.

Step-by-Step Solution:

Step 1: Understand the Effect of Medium: When light travels from air to water, its speed and wavelength decrease, while its frequency remains constant.

Step 2: Relate Wavelength to Refractive Index: The wavelength in a medium with refractive index n is given by $\lambda' = \frac{\lambda}{n}$ , where $\lambda$ is the wavelength in air.

Step 3: Recall the Formula: The fringe width ( $\beta$ ) in Young's double-slit experiment is given by $\beta = \frac{\lambda D}{d}$ .

Step 4: Analyze the Change in Fringe Width: Since the wavelength decreases in water, the fringe width also decreases: $\beta' = \frac{\lambda' D}{d} = \frac{\lambda D}{nd} = \frac{\beta}{n}$ .

Why other options are wrong:
- Option A: The fringe width becomes smaller, not infinite.
- Option C: The fringe width decreases, not increases.
- Option D: The fringe width changes; it does not remain unchanged.

Key Concepts:
- Young's Double Slit Experiment (YDSE)
- Fringe Width
- Wavelength
- Refractive Index

Memory Aids:
- Remember that the wavelength decreases when light enters a medium with a higher refractive index.

Quick Tips:
- The fringe width is inversely proportional to the refractive index of the medium.

16 Medium

16 of 143Question permalink

Remembering +4 | -1

The coherent formula for fringe visibility is

View answer and explanation

Correct option: A

V = \frac{{{I_{\max }} - {I_{\min }}}}{{{I_{\max }} + {I_{\min }}}}

Option A is correct. Fringe visibility is defined as

V = \frac{{{I_{\max }} - {I_{\min }}}}{{{I_{\max }} + {I_{\min }}}}

which quantifies the contrast between the brightest and darkest fringes.

Solution approach

Correct Answer + Core Reason: $V = \frac{{{I_{\max }} - {I_{\min }}}}{{{I_{\max }} + {I_{\min }}}}$ . This formula defines the visibility of fringes in an interference pattern.

Step-by-Step Solution:

Step 1: Define fringe visibility: Fringe visibility (V) is a measure of the contrast between the bright and dark fringes in an interference pattern.

Step 2: Recall the formula: The formula for fringe visibility is given by $V = \frac{{{I_{\max }} - {I_{\min }}}}{{{I_{\max }} + {I_{\min }}}}$ , where $I_{\max}$ is the maximum intensity and $I_{\min}$ is the minimum intensity.

Why other options are wrong:
- Option B: Incorrect because it's the inverse of the correct formula.
- Option C: Incorrect because it only considers the ratio of maximum to minimum intensity, not the difference.
- Option D: Incorrect because it only considers the ratio of minimum to maximum intensity, not the difference.

Key Concepts:
- Fringe visibility
- Interference
- Intensity

Memory Aids:
- Visibility = (Max - Min) / (Max + Min)

Quick Tips:
- Remember the formula for fringe visibility to quantify the contrast of the interference pattern.

17 Medium

17 of 143Question permalink

Understanding +4 | -1

Light waves travel in vacuum along the

y - \mathrm{axis}

. Which of the following may represent the wavefront?

View answer and explanation

Correct option: A

y = constant

Option A (

y = constant

) is correct because a wavefront is a surface of constant phase, and if light travels along the y-axis, the wavefront is perpendicular to it.

Solution approach

Correct Answer + Core Reason: $y = constant$ . Wavefronts are perpendicular to the direction of propagation.

Step-by-Step Solution:

Step 1: Define wavefront: A wavefront is a surface of constant phase.

Step 2: Direction of propagation: Light waves travel along the y-axis.

Step 3: Wavefront orientation: The wavefront must be perpendicular to the direction of propagation. Therefore, the equation of the wavefront is $y = constant$ .

Why other options are wrong:
- Option B: $x = constant$ would be a wavefront if light traveled along the x-axis.
- Option C: $z = constant$ would be a wavefront if light traveled along the z-axis.
- Option D: $x + y + z = constant$ represents a plane that is not perpendicular to the y-axis.

Key Concepts:
- Wavefront
- Direction of propagation
- Perpendicularity of wavefront and direction

Memory Aids:
- Wavefront is always perpendicular to the direction of light.

Quick Tips:
- Visualize the wavefront as a plane perpendicular to the direction of light.

18 Easy

18 of 143Question 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.

19 Medium

19 of 143Question permalink

Understanding +4 | -1

The phase difference between incident wave and reflected wave is

180^\circ

when light ray

View answer and explanation

Correct option: A

Enters into glass from air

Option A (Enters into glass from air) is correct because when light travels from a rarer to a denser medium, it undergoes a phase change of 180 degrees upon reflection.

Solution approach

Correct Answer + Core Reason: Enters into glass from air. Phase change of 180 degrees occurs when light reflects from a denser medium.

Step-by-Step Solution:
Step 1: Recall the concept of phase change upon reflection.
Step 2: When light travels from a rarer medium to a denser medium, it undergoes a phase change of $\pi$ (180 degrees) upon reflection.
Step 3: When light travels from a denser medium to a rarer medium, there is no phase change upon reflection.
Step 4: Analyze the options:
- Air to glass: Rarer to denser, phase change of 180 degrees.
- Glass to air: Denser to rarer, no phase change.
- Diamond to glass: Denser to rarer (since refractive index of diamond > glass), no phase change.
- Glass to water: Denser to rarer (since refractive index of glass > water), no phase change.
Step 5: Conclusion: The phase difference is 180 degrees only when light enters glass from air.

Why other options are wrong:
- Option B: Light entering air from glass does not have a phase change of 180 degrees.
- Option C: Light entering glass from diamond does not have a phase change of 180 degrees.
- Option D: Light entering water from glass does not have a phase change of 180 degrees.

Key Concepts:
- Phase change upon reflection
- Refractive index
- Rarer and denser medium

Memory Aids:
- DR (Denser Reflection): Phase change occurs when light reflects from a Denser medium.

Quick Tips:
- Remember the rule about phase change upon reflection based on the refractive indices of the media.

20 Medium

20 of 143Question permalink

Understanding +4 | -1

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

View answer and explanation

Correct option: A

Wavelength increases

The fringe width in Young's double-slit experiment increases when the wavelength increases. The fringe width is directly proportional to the wavelength of light.

Solution approach

Correct Answer + Core Reason: Wavelength increases. The fringe width is directly proportional to the wavelength of light.

Step-by-Step Solution:

Step 1: Recall the Formula: The fringe width ( $\beta$ ) in Young's double-slit experiment is given by $\beta = \frac{\lambda D}{d}$ , 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 2: Analyze the Relationship: From the formula, it is clear that the fringe width is directly proportional to the wavelength ( $\lambda$ ).

Why other options are wrong:
- Option B: The fringe width is *inversely* proportional to the distance between the slits (d).
- Option C: The fringe width is *directly* proportional to the distance between the source and the screen (D), so decreasing the distance would decrease the fringe width.
- Option D: The width of the slits does not directly affect the fringe width.

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

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.

21 Medium

21 of 143Question permalink

Remembering +4 | -1

Light is an electromagnetic wave. Its speed in vacuum is given by the expression

View answer and explanation

Correct option: D

\frac{1}{{\sqrt {{\mu _o}{\varepsilon _o}} }}

Option D is correct. The speed of light in a vacuum is given by the formula c = 1/√(μ₀ε₀).

Solution approach

Correct Answer + Core Reason: $\frac{1}{{\sqrt {{\mu _o}{\varepsilon _o}} }}$ . The speed of light in vacuum is given by this formula.

Step-by-Step Solution:

Step 1: Recall the relationship between the speed of light, permeability of free space, and permittivity of free space.

Step 2: State the formula for the speed of light in vacuum: $c = \frac{1}{\sqrt{\mu_0 \varepsilon_0}}$ , where $c$ is the speed of light, $\mu_0$ is the permeability of free space, and $\varepsilon_0$ is the permittivity of free space.

Why other options are wrong:
- Option A: This option is incorrect because it represents the reciprocal of the speed of light.
- Option B: This option is incorrect because it represents the square root of the ratio of permeability to permittivity, which is related to the impedance of free space, not the speed of light.
- Option C: This option is incorrect because it represents the square root of the ratio of permittivity to permeability, which is not directly related to the speed of light.

Key Concepts:
- Speed of light in vacuum
- Permeability of free space ( $\mu_0$ )
- Permittivity of free space ( $\varepsilon_0$ )

Memory Aids:
- Remember the formula: "Speed of light is one over the square root of mu naught epsilon naught."

Quick Tips:
- Be familiar with the fundamental constants and their relationship to the speed of light.

22 Medium

22 of 143Question permalink

Application +4 | -1

Light propagates

2\;cm

distance in glass of refractive index 1.5 in time

{t_0}

. In the same time

{t_0}

, light propagates a distance of

2.25\;cm

in a medium. The refractive index of the medium is

View answer and explanation

Correct option: A

4/3

Option A (

4/3

) is correct because it is calculated using the relationship between refractive index, speed of light, and distance traveled.

Solution approach

Correct Answer + Core Reason: $4/3$ . This is calculated using the relationship between refractive index, speed of light, and distance traveled.

Step-by-Step Solution:

Step 1: Identify the given information:\nDistance in glass, $d_1 = 2 \text{{ cm}}$ \nRefractive index of glass, $n_1 = 1.5$ \nTime taken, $t_0$ \nDistance in medium, $d_2 = 2.25 \text{{ cm}}$ \nRefractive index of the medium, $n_2 = ?$

Step 2: Key Concepts:
- Refractive index, $n = \frac{c}{v}$ , where $c$ is the speed of light in vacuum and $v$ is the speed of light in the medium.
- Speed, $v = \frac{d}{t}$

Step 3: Apply the formulas:\nIn glass: $v_1 = \frac{{d_1}}{{t_0}} = \frac{2}{{t_0}}$ \nIn medium: $v_2 = \frac{{d_2}}{{t_0}} = \frac{{2.25}}{{t_0}}$
$n_1 = \frac{c}{v_1} \Rightarrow v_1 = \frac{c}{{n_1}}$
$n_2 = \frac{c}{v_2} \Rightarrow v_2 = \frac{c}{{n_2}}$
$\frac{{v_1}}{{v_2}} = \frac{{d_1}}{{d_2}} = \frac{2}{{2.25}} = \frac{2}{{\frac{9}{4}}} = \frac{8}{9}$
$\frac{{\frac{c}{{n_1}}}}{{\frac{c}{{n_2}}}} = \frac{{n_2}}{{n_1}} = \frac{8}{9}$
$n_2 = n_1 \times \frac{8}{9} = 1.5 \times \frac{8}{9} = \frac{3}{2} \times \frac{8}{9} = \frac{{24}}{{18}} = \frac{4}{3}$

Why other options are wrong:
- Option B: $3/2$ : This is the refractive index of the glass, not the medium.
- Option C: $8/3$ : This is an incorrect calculation.
- Option D: None of these: The correct answer is 4/3, which is among the options.

Key Concepts:
- Refractive index
- Speed of light

Memory Aids:
- $n = \frac{c}{v}$

Quick Tips:
- Remember the relationship between refractive index, speed of light, and distance traveled.

23 Medium

23 of 143Question permalink

Application +4 | -1

Critical angle for certain medium is

{\sin ^{ - 1}}\left( {0.6} \right).

The polarizing angle of that medium is

View answer and explanation

Correct option: C

{\tan ^{ - 1}}\left[ {1.6667} \right]

The correct answer is C (

{\tan ^{ - 1}}\left[ {1.6667} \right]

) because it is the Brewster's angle calculated from the refractive index, which is derived from the critical angle.

Solution approach

Correct Answer + Core Reason: ${\tan ^{ - 1}}\left[ {1.6667} \right]$ . The polarizing angle is calculated using the relationship between the refractive index and the critical angle.

Step-by-Step Solution:

Step 1: The critical angle is given by $\sin C = \frac{1}{\mu}$ , where $\mu$ is the refractive index.

Step 2: Given $\sin C = 0.6$ , so $\mu = \frac{1}{0.6} = \frac{10}{6} = \frac{5}{3} \approx 1.6667$ .

Step 3: The polarizing angle (Brewster's angle) $i_p$ is given by $\tan i_p = \mu$ .

Step 4: Therefore, $i_p = \tan^{-1} (\mu) = \tan^{-1} \left( \frac{5}{3} \right) = \tan^{-1} (1.6667)$ .

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

Key Concepts:
- Critical angle
- Refractive index
- Brewster's angle

Memory Aids:
- Remember the relationship between critical angle and refractive index: $\sin C = \frac{1}{\mu}$ .
- Remember Brewster's law: $\tan i_p = \mu$ .

Quick Tips:
- Make sure to use the correct formula for Brewster's angle.

24 Hard

24 of 143Question permalink

Application +4 | -1

For what distance is ray optics a good approximation when the aperture is 4mm wide and the wave length is 500 nm

View answer and explanation

Correct option: A

32 m

Option A (32 m) is correct because it is calculated using the Fresnel distance formula.

Solution approach

Correct Answer + Core Reason: 32 m. The distance is calculated using the Fresnel distance formula.

Step-by-Step Solution:

Step 1: Identify the given information: Aperture width $a = 4\,mm = 4 \times 10^{-3}\,m$ , wavelength $\lambda = 500\,nm = 500 \times 10^{-9}\,m$ .

Step 2: Recall the Fresnel distance formula: $Z = \frac{a^2}{\lambda}$

Step 3: Substitute the given values: $Z = \frac{(4 \times 10^{-3})^2}{500 \times 10^{-9}}$

Step 4: Calculate the distance: $Z = \frac{16 \times 10^{-6}}{5 \times 10^{-7}} = \frac{160}{5} = 32\,m$

Step 5: Final Answer: 32 m

Why other options are wrong:
- Option B (64 m): This is incorrect because it is a result of incorrect calculation or using the wrong formula.
- Option C (16 m): This is incorrect because it is a result of incorrect calculation or using the wrong formula.
- Option D (8 m): This is incorrect because it is a result of incorrect calculation or using the wrong formula.

Key Concepts:
- Fresnel Distance
- Ray Optics Approximation

Memory Aids:
- Remember the Fresnel distance formula: $Z = \frac{a^2}{\lambda}$

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

25 Medium

25 of 143Question permalink

Application +4 | -1

Two point white does are

\mathrm{1}\,\,\mathrm{mm}

apart on a black paper. They are viewed by eye of pupil diameter

\mathrm{3}\,\,\mathrm{mm}

. Approximately, what is the maximum distance at which these dots can be resolved by the eye? [Take wavelength of light

\mathrm{ = 500}\,\,\mathrm{nm}

]

View answer and explanation

Correct option: A

\mathrm{5}\,\,\mathrm{m}

Option A (

\mathrm{5}\,\mathrm{m}

) is correct because it is calculated using the resolution limit formula.

Solution approach

Correct Answer + Core Reason: $\mathrm{5}\,\mathrm{m}$ . The maximum distance is calculated using the resolution limit formula.

Step-by-Step Solution:

Step 1: Identify the given information: Distance between the dots $d = 1\,mm = 1 \times 10^{-3}\,m$ , pupil diameter $D = 3\,mm = 3 \times 10^{-3}\,m$ , wavelength of light $\lambda = 500\,nm = 500 \times 10^{-9}\,m$ .

Step 2: Recall the Rayleigh criterion for resolution: $\theta = 1.22 \frac{\lambda}{D}$ , where $\theta$ is the angular separation between the two dots.

Step 3: Let $x$ be the maximum distance at which the dots can be resolved. Then $\theta = \frac{d}{x}$ .

Step 4: Equate the two expressions for $\theta$ : $\frac{d}{x} = 1.22 \frac{\lambda}{D}$

Step 5: Solve for $x$ : $x = \frac{d D}{1.22 \lambda}$

Step 6: Substitute the given values: $x = \frac{1 \times 10^{-3} \times 3 \times 10^{-3}}{1.22 \times 500 \times 10^{-9}}$

Step 7: Calculate the distance: $x = \frac{3 \times 10^{-6}}{1.22 \times 5 \times 10^{-7}} = \frac{3 \times 10^{-6}}{6.1 \times 10^{-7}} = \frac{30}{6.1} \approx 4.92 \, m \approx 5 \, m$

Step 8: Final Answer: $\mathrm{5}\,\mathrm{m}$

Why other options are wrong:
- **Option B ( $\mathrm{1}\,\mathrm{m}$ ): This is incorrect because it is a result of incorrect calculation or using the wrong formula.
- Option C ( $\mathrm{6}\,\mathrm{m}$ ): This is incorrect because it is a result of incorrect calculation or using the wrong formula.
- Option D ( $\mathrm{3}\,\mathrm{m}$ ): This is incorrect because it is a result of incorrect calculation or using the wrong formula.

Key Concepts:
- Resolving Power of the Eye
- Rayleigh Criterion
- Angular Separation

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

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

26 Medium

26 of 143Question permalink

Remembering +4 | -1

Fringe pattern in YDSE is formed due to the phenomenon of

View answer and explanation

Correct option: C

Interference of light

Option C (Interference of light) is correct because the fringe pattern in YDSE is a direct result of the interference of light waves.

Solution approach

Correct Answer + Core Reason: Interference of light. The fringe pattern in YDSE arises from the superposition of light waves.

Step-by-Step Solution:

Step 1: Recall the principle of YDSE: Two coherent light sources create an interference pattern on a screen.

Step 2: Define Interference: Interference is the phenomenon where two or more waves superpose to form a resultant wave of greater, lower, or the same amplitude.

Step 3: Relate to YDSE: The bright and dark fringes are formed due to constructive and destructive interference, respectively.

Why other options are wrong:
- Option A: Particle nature of light explains phenomena like the photoelectric effect, not interference patterns.
- Option B: Polarization is related to the direction of oscillation of light waves, not the formation of interference patterns.
- Option D: Rectilinear propagation explains why light travels in straight lines, but not the fringe pattern in YDSE.

Key Concepts:
- Interference of light
- Young's Double Slit Experiment (YDSE)
- Superposition principle

Memory Aids:
- Interference creates the fringes in YDSE.

Quick Tips:
- YDSE is a classic example of wave interference.

27 Medium

27 of 143Question permalink

Understanding +4 | -1

Which one of the following phenomena is not explained by Huygen’s construction of wavefront

View answer and explanation

Correct option: D

Origin of spectra

Option D (Origin of spectra) is correct because Huygens' principle primarily explains wave propagation, reflection, refraction, and diffraction, not the origin of spectra which is related to atomic structure and quantum mechanics.

Solution approach

Correct Answer + Core Reason: Origin of spectra. Huygens' principle explains wave propagation, not the origin of spectra.

Step-by-Step Solution:
Step 1: Recall Huygens' principle: Every point on a wavefront is a source of secondary spherical wavelets.
Step 2: Huygens' principle is used to explain:
- Wave propagation
- Reflection
- Refraction
- Diffraction
Step 3: The origin of spectra is related to the energy levels of atoms and the emission or absorption of photons when electrons transition between these levels.
Step 4: Conclusion: Huygens' principle does not explain the origin of spectra.

Why other options are wrong:
- Option A: Refraction can be explained by Huygens' principle.
- Option B: Reflection can be explained by Huygens' principle.
- Option C: Diffraction can be explained by Huygens' principle.

Key Concepts:
- Huygens' principle
- Wavefront
- Reflection, refraction, diffraction
- Atomic spectra

Memory Aids:
- Huygens' principle: Wavelets from a wavefront.

Quick Tips:
- Remember the applications of Huygens' principle.

28 Medium

28 of 143Question permalink

Understanding +4 | -1

The phenomenon which does not take place in sound waves is

View answer and explanation

Correct option: D

Polarisation

Polarization is a phenomenon that only occurs with transverse waves, such as light waves. Sound waves are longitudinal waves and cannot be polarized.

Solution approach

Correct Answer + Core Reason: Polarisation. Sound waves are longitudinal and cannot be polarized.

Step-by-Step Solution:

Step 1: Understand Wave Types: There are two main types of waves: transverse and longitudinal.

Step 2: Define Polarization: Polarization is the phenomenon in which the oscillations of a wave are confined to a single plane.

Step 3: 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.

Step 4: Identify Sound Waves: Sound waves are longitudinal waves.

Why other options are wrong:
- Option A: Scattering occurs in sound waves (e.g., scattering by obstacles).
- Option B: Diffraction occurs in sound waves (e.g., sound bending around corners).
- Option C: Interference occurs in sound waves (e.g., constructive and destructive interference of sound waves).

Key Concepts:
- Polarization
- Transverse Waves
- Longitudinal Waves
- Sound Waves

Memory Aids:
- Remember that only transverse waves can be polarized, and sound waves are longitudinal.

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

29 Easy

29 of 143Question 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.

30 Hard

30 of 143Question permalink

Understanding +4 | -1

Ratio of intensities in consecutive maxima in a diffraction pattern due to a single slit is

View answer and explanation

Correct option: D

1: \frac{4}{9 \pi^2}: \frac{4}{25 \pi^2}

Option D (

1: \frac{4}{9 \pi^2}: \frac{4}{25 \pi^2}

) is correct because the intensities of the secondary maxima in single-slit diffraction decrease rapidly and are proportional to the inverse square of the sine of the angle.

Solution approach

Correct Answer + Core Reason: $1: \frac{{4}}{{9 \pi^2}}: \frac{{4}}{{25 \pi^2}}$ . This represents the ratio of intensities in single-slit diffraction.

Step-by-Step Solution:

Step 1: Recall the intensity distribution in single-slit diffraction: The intensity is given by $I = I_0 \left( \frac{{\sin{\alpha}}}{{\alpha}} \right)^2$ , where $\alpha = \frac{{\pi a \sin{\theta}}}{{\lambda}}$ .

Step 2: Intensity of secondary maxima: The secondary maxima occur approximately at $\alpha = \frac{{3\pi}}{{2}}, \frac{{5\pi}}{{2}}, \ldots$

Step 3: Calculate the ratios: The intensity of the first secondary maximum is $I_1 = I_0 \left( \frac{{\sin{\frac{{3\pi}}{{2}}}}}{\frac{{3\pi}}{{2}}} \right)^2 = I_0 \left( \frac{{1}}{\frac{{9\pi^2}}{{4}}} \right) = I_0 \frac{{4}}{{9\pi^2}}$ .\nThe intensity of the second secondary maximum is $I_2 = I_0 \left( \frac{{\sin{\frac{{5\pi}}{{2}}}}}{\frac{{5\pi}}{{2}}} \right)^2 = I_0 \left( \frac{{1}}{\frac{{25\pi^2}}{{4}}} \right) = I_0 \frac{{4}}{{25\pi^2}}$ .

Step 4: Find the ratio: The ratio of intensities is $I_0 : I_0 \frac{{4}}{{9\pi^2}} : I_0 \frac{{4}}{{25\pi^2}} = 1 : \frac{{4}}{{9\pi^2}} : \frac{{4}}{{25\pi^2}}$ .

Why other options are wrong:
- Option A: This is a simple arithmetic progression and doesn't represent the intensity ratios in diffraction.
- Option B: This is a ratio of squares, but the values are incorrect for diffraction intensities.
- Option C: The values are incorrect for diffraction intensities.

Key Concepts:
- Single-slit diffraction
- Intensity distribution
- Secondary maxima

Memory Aids:
- The intensity of secondary maxima decreases rapidly in diffraction.

Quick Tips:
- Remember the intensity formula for single-slit diffraction.

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