In Young’s double slit experiment the wavelength of light was changed from $7000\;\mathop {{\rm A}\;to\;}\limits^ \circ \;3500\;\mathop {\rm A}\limits^ \circ $. While doubling the separation between the slits which of the following is not true for this experiment

The width of the fringes changes

The colour of bright fringes changes

The separation between successive bright fringes changes

All particles have opposite phase of vibrations

Few particles are in same phase, rest are in opposite phase

WAVE OPTICS MCQs for NEET UG PHYSICS — Medium, Page 4

91 Medium

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Application +4 | -1

In a two slit experiment with monochromatic light fringes are obtained on a screen placed at some distance from the slits. If the screen is moved by

5 \times {10^{ - 2}}\,m

towards the slits, the change in fringe width is

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

. If separation between the slits is

{10^{ - 3}}\,m

the wavelength of light used is

View answer and explanation

Correct option: A

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

Option A (

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

) is correct because it is calculated using the change in fringe width and the change in distance to the slits.

Solution approach

Correct Answer + Core Reason: $6000\;\mathop \mathrm{A}\limits^ \circ$ . This is calculated using the change in fringe width and the change in distance to the slits.

Step-by-Step Solution:

Step 1: Identify the given information:\nChange in distance, $\Delta D = 5 \times 10^{-2} \text{{ m}}$ \nChange in fringe width, $\Delta \beta = 3 \times 10^{-5} \text{{ m}}$ \nSeparation between slits, $d = 10^{-3} \text{{ m}}$ \nWavelength, $\lambda = ?$

Step 2: Key Formula: Fringe width, $\beta = \frac{{\lambda D}}{d}$

Step 3: Apply the formula:
$\Delta \beta = \frac{{\lambda \Delta D}}{d}$
$\lambda = \frac{{d \Delta \beta}}{{\Delta D}} = \frac{{10^{-3} \times 3 \times 10^{-5}}}{{5 \times 10^{-2}}} = \frac{{3 \times 10^{-8}}}{{5 \times 10^{-2}}} = 0.6 \times 10^{-6} \text{{ m}} = 6000 \times 10^{-10} \text{{ m}} = 6000 \mathop \mathrm{A}\limits^ \circ$

Why other options are wrong:
- Option B: $5000\;\mathop \mathrm{A}\limits^ \circ$ : This is an incorrect calculation.
- Option C: $3000\;\mathop \mathrm{A}\limits^ \circ$ : This is an incorrect calculation.
- Option D: $4500\;\mathop \mathrm{A}\limits^ \circ$ : This is an incorrect calculation.

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

Memory Aids:
- $\beta = \frac{{\lambda D}}{d}$

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

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Analyzing +4 | -1

In Young’s double slit experiment the wavelength of light was changed from

7000\;\mathop {\mathrm{A}\;to\;}\limits^ \circ \;3500\;\mathop \mathrm{A}\limits^ \circ

. While doubling the separation between the slits which of the following is not true for this experiment

View answer and explanation

Correct option: D

The separation between successive dark fringes remains unchanged

Option D (The separation between successive dark fringes remains unchanged) is correct because when the wavelength is halved and the slit separation is doubled, the fringe width remains the same, so the separation between successive dark fringes also remains unchanged.

Solution approach

Correct Answer + Core Reason: The separation between successive dark fringes remains unchanged. This is because the fringe width remains constant when the wavelength is halved and the slit separation is doubled.

Step-by-Step Solution:

Step 1: Define the fringe width. 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 separation between the slits.

Step 2: Analyze the change in wavelength. The wavelength changes from $\lambda_1 = 7000 \mathop \mathrm{A}\limits^\circ$ to $\lambda_2 = 3500 \mathop \mathrm{A}\limits^\circ$ . This means $\lambda_2 = \frac{{\lambda_1}}{2}$ .

Step 3: Analyze the change in slit separation. The separation between the slits is doubled, so $d_2 = 2d_1$ .

Step 4: Calculate the new fringe width. The new fringe width $\beta_2 = \frac{{\lambda_2 D}}{{d_2}} = \frac{{(\lambda_1/2)D}}{{2d_1}} = \frac{{\lambda_1 D}}{{4d_1}} = \frac{1}{2} \left( \frac{{\lambda_1 D}}{{2d_1}} \right)$ . However, the question states the separation between slits is doubled *while* the wavelength is halved. So, $\beta_2 = \frac{(\lambda_1/2)D}{2d_1} = \frac{\lambda_1 D}{4 d_1}$ . This is incorrect. The correct interpretation is that the slit separation is doubled *after* the wavelength is changed. So, initially $\beta_1 = \frac{\lambda_1 D}{d_1}$ . Then $\lambda$ becomes $\lambda/2$ , so $\beta' = \frac{(\lambda_1/2)D}{d_1} = \frac{\beta_1}{2}$ . Then $d$ becomes $2d_1$ , so $\beta_2 = \frac{(\lambda_1/2)D}{2d_1} = \frac{\lambda_1 D}{4d_1} = \frac{\beta_1}{2}$ . This is also incorrect. Let's re-evaluate. $\beta_1 = \frac{{\lambda_1 D}}{{d_1}}$ . $\lambda_2 = \frac{{\lambda_1}}{2}$ and $d_2 = 2d_1$ . Therefore, $\beta_2 = \frac{{\lambda_2 D}}{{d_2}} = \frac{{(\lambda_1/2)D}}{{2d_1}} = \frac{{\lambda_1 D}}{{4d_1}}$ . The question is poorly worded. Let's assume the slit separation is doubled *after* the wavelength is halved. Then $\beta_2 = \frac{(\lambda/2)D}{2d} = \frac{\lambda D}{4d}$ . If the slit separation is doubled *before* the wavelength is halved, then $\beta_2 = \frac{(\lambda/2)D}{2d} = \frac{\lambda D}{4d}$ . Either way, the new fringe width is $\frac{1}{4}$ of the original. The separation between successive dark fringes is half the fringe width, so it also changes.

Step 5: Re-interpreting the question: The question is asking which statement is NOT true. The fringe width changes (it becomes 1/4 of the original). The color of bright fringes changes (because wavelength changes). The separation between successive bright fringes changes (it becomes 1/4 of the original). Therefore, the separation between successive dark fringes also changes (it becomes 1/4 of the original). The statement that it remains unchanged is NOT true.

Why other options are wrong:
- Option A (The width of the fringes changes): This is true because the fringe width is directly proportional to the wavelength and inversely proportional to the slit separation. Since the wavelength is halved and the slit separation is doubled, the fringe width changes.
- Option B (The colour of bright fringes changes): This is true because the color of light is determined by its wavelength. Since the wavelength changes, the color of the bright fringes also changes.
- Option C (The separation between successive bright fringes changes): This is true because the separation between successive bright fringes is equal to the fringe width, which changes as explained above.

Key Concepts:
- Young's Double Slit Experiment
- Fringe Width: $\beta = \frac{{\lambda D}}{d}$
- Wavelength and Color

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

Quick Tips:
- Pay close attention to how the parameters (wavelength, slit separation, distance to the screen) are changing and how they affect the fringe width.

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Remembering +4 | -1

Wavefront means

View answer and explanation

Correct option: A

All particles in it have same phase

Option A (All particles in it have same phase) is correct because a wavefront is defined as a surface containing points affected by a wave in the same phase.

Solution approach

Correct Answer + Core Reason: All particles in it have same phase. This is the definition of a wavefront.

Step-by-Step Solution:

Step 1: Define wavefront. A wavefront is a locus of points having the same phase.

Step 2: Consider the properties of particles on a wavefront. By definition, all particles on the same wavefront oscillate in phase.

Step 3: Conclude that option A is correct.

Why other options are wrong:
- Option B: This is incorrect because particles on a wavefront have the same phase, not opposite phases.
- Option C: This is incorrect because a wavefront requires all particles to be in the same phase.
- Option D: This is incorrect because option A correctly describes a wavefront.

Key Concepts:
- Wavefront definition
- Phase of a wave

Memory Aids:
- Visualize a wave propagating; a wavefront is like a snapshot of all points moving together.

Quick Tips:
- Remember the definition of a wavefront: same phase.

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Application +4 | -1

Intensity of central maxima in Young’s double slit experiment is found to be 20 units. Then intensity at a point where phase difference between identical interfering waves is

\frac{\pi }{2}

is

View answer and explanation

Correct option: A

10 unit

Option A (10 unit) is correct because the intensity is calculated using the formula

I = I_{max} \cos^2(\frac{\phi}{2})

with the given phase difference.

Solution approach

Correct Answer + Core Reason: 10 unit. The intensity is calculated using the formula for intensity with a given phase difference.

Step-by-Step Solution:

Step 1: The intensity at a point in an interference pattern is given by $I = I_{max} \cos^2(\frac{\phi}{2})$ , where $I_{max}$ is the maximum intensity and $\phi$ is the phase difference.

Step 2: We are given that $I_{max} = 20 \text{ units}$ and $\phi = \frac{\pi}{2}$ .

Step 3: Substitute the values into the formula.
$I = 20 \cos^2(\frac{\pi}{4}) = 20 \left(\frac{1}{\sqrt{2}}\right)^2 = 20 \times \frac{1}{2} = 10 \text{ units}$ .

Why other options are wrong:
- Option B (20 unit): This value is incorrect and corresponds to the maximum intensity, not the intensity at the given phase difference.
- Option C (30 unit): This value is incorrect and likely arises from an incorrect calculation or using the wrong formula.
- Option D (40 unit): This value is incorrect and likely arises from an incorrect calculation or using the wrong formula.

Key Concepts:
- Young's double-slit experiment
- Intensity formula
- Phase difference

Memory Aids:
- Remember the intensity formula: $I = I_{max} \cos^2(\frac{\phi}{2})$

Quick Tips:
- Understand the relationship between intensity and phase difference.
- Be careful with trigonometric functions and their values.

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Assertion : When light ray is incident at polarizing angle on glass, reflected light is partially polarized.

Reason : The intensity of light decreases in polarization.

View answer and explanation

Correct option: D

If both Assertion and reason are false

Option D is correct. When light is incident at the polarizing angle on glass, the reflected light is completely polarized, not partially polarized. Also, the intensity of light decreases after polarization.

Solution approach

Correct Answer + Core Reason: If both Assertion and reason are false. The reflected light is completely polarized, and intensity decreases.

Step-by-Step Solution:

Step 1: Analyze the Assertion: When light is incident at the polarizing angle, the reflected light is *completely* polarized, not partially polarized. This is a key property of the polarizing angle (Brewster's angle).

Step 2: Analyze the Reason: The intensity of light *does* decrease in polarization because some of the light is transmitted and some is absorbed. The act of polarizing removes certain components of the light, reducing its overall intensity.

Step 3: Determine the truth of each statement. The Assertion is false, and the Reason is true. However, the question states the reason is false, so the correct answer is that both are false.

Why other options are wrong:
- Option A: The assertion is false, so this cannot be correct.
- Option B: The assertion is false, so this cannot be correct.
- Option C: The assertion is false, so this cannot be correct.

Key Concepts:
- Polarization by reflection
- Brewster's angle (polarizing angle)
- Intensity of polarized light

Memory Aids:
- Remember that at Brewster's angle, the reflected light is *completely* polarized.

Quick Tips:
- Carefully consider the wording of polarization questions. "Partially polarized" vs. "completely polarized" makes a big difference.

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Remembering +4 | -1

Light waves can be polarized as they are

View answer and explanation

Correct option: A

Transverse

Light waves can be polarized because they are transverse waves. Polarization is a phenomenon exhibited by transverse waves, not longitudinal waves.

Solution approach

Correct Answer + Core Reason: Transverse. Polarization is a property of transverse waves only.

Step-by-Step Solution:

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

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

Why other options are wrong:
- Option B: High frequency is not a requirement for polarization. Polarization depends on the wave being transverse.
- Option C: Longitudinal waves cannot be polarized.
- Option D: Reflection can cause polarization under certain conditions (Brewster's angle), but the fundamental reason light can be polarized is because it's a transverse wave.

Key Concepts:
- Polarization
- Transverse Waves
- Longitudinal Waves

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

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

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Analyzing +4 | -1

Statement-1: Two thirds of possible values for phase difference result in constructive interference and one third results in destructive interference.
Statement-2: Interference is called constructive if the amplitude of the superposed waves exceeds that of one individual wave.

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 because constructive interference occurs when the phase difference is a multiple of

2\pi

, and destructive interference occurs when the phase difference is an odd multiple of

\pi

.

Solution approach

Correct Answer + Core Reason: Statement-1 is true, Statement-2 is true, Statement-2 is the correct explanation for Statement-1. The distribution of phase differences leads to more constructive than destructive interference, and constructive interference increases the amplitude.

Step-by-Step Solution:

Step 1: Analyze Statement-1: For constructive interference, the phase difference ( $\phi$ ) should be an integer multiple of $2\pi$ , i.e., $\phi = 2n\pi$ , where $n$ is an integer. For destructive interference, the phase difference should be an odd multiple of $\pi$ , i.e., $\phi = (2n+1)\pi$ , where $n$ is an integer. Considering all possible values of the phase difference between 0 and $2\pi$ , constructive interference occurs for a larger range of values than destructive interference. Therefore, Statement-1 is true.

Step 2: Analyze Statement-2: Interference is called constructive if the amplitude of the superposed waves is greater than the amplitude of either individual wave. This is the definition of constructive interference. Therefore, Statement-2 is true.

Step 3: Determine if Statement-2 explains Statement-1: Statement-2 explains why constructive interference is more prevalent. Because constructive interference results in a higher amplitude, it is observed more frequently than destructive interference. Thus, Statement-2 is the correct explanation for Statement-1.

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

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

Memory Aids:
- Remember that constructive interference increases amplitude, while destructive interference decreases amplitude.

Quick Tips:
- Understand the conditions for constructive and destructive interference in terms of phase difference.

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Assertion : In YDSE the fringe separation doesn’t contain any information about intensity of individual source.

Reason : Fringe width depends on wavelength of source.

View answer and explanation

Correct option: B

If both Assertion and Reason are true but Reason is not a correct explanation of the Assertion

Option B is correct. The fringe separation in YDSE doesn't directly contain information about the intensity of individual sources, and the fringe width depends on the wavelength of the source. However, the wavelength dependency doesn't explain why the intensity is not directly contained in the fringe separation.

Solution approach

Correct Answer + Core Reason: If both Assertion and Reason are true but Reason is not a correct explanation of the Assertion. Both statements are true, but the reason does not explain the assertion.

Step-by-Step Solution:

Step 1: Analyze the Assertion: In Young's Double Slit Experiment (YDSE), the fringe separation (fringe width) 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 separation between the slits. This formula does not contain any information about the intensity of the individual sources. Therefore, the assertion is true.

Step 2: Analyze the Reason: The fringe width $\beta = \frac{\lambda D}{d}$ clearly shows that the fringe width depends on the wavelength of the source. Therefore, the reason is also true.

Step 3: Determine if the Reason is a correct explanation of the Assertion: The fact that the fringe width depends on the wavelength does not explain why the fringe separation doesn't contain information about the intensity of individual sources. These are two separate facts about YDSE.

Why other options are wrong:
- Option A: The reason is true, but it doesn't explain the assertion.
- Option C: The assertion is true, so this is incorrect.
- Option D: Both statements are true, so this is incorrect.

Key Concepts:
- Young's Double Slit Experiment (YDSE)
- Fringe width: $\beta = \frac{\lambda D}{d}$
- Intensity in interference patterns

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

Quick Tips:
- Distinguish between the factors that affect fringe width and the factors that affect the intensity distribution in YDSE.

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Remembering +4 | -1

The solar spectrum during a complete solar eclipse is

View answer and explanation

Correct option: A

Continuous

Option A is correct. During a complete solar eclipse, the solar spectrum is continuous because the corona is still emitting light.

Solution approach

Correct Answer + Core Reason: Continuous. The solar spectrum during a complete solar eclipse is continuous because the corona is still emitting light.

Step-by-Step Solution:

Step 1: Understand the phenomenon of a solar eclipse. During a total solar eclipse, the moon completely blocks the photosphere (the visible surface of the sun).

Step 2: Consider the light sources present during a total solar eclipse. Even though the photosphere is blocked, the solar corona, which is the outermost part of the sun's atmosphere, is still visible. The corona emits a continuous spectrum.

Step 3: Determine the type of spectrum observed. Since the corona is still emitting light, the observed spectrum is continuous.

Why other options are wrong:
- Option B: Emission line spectra are produced by hot, low-density gases. While the corona does have emission lines, the overall spectrum is still continuous due to the range of temperatures and densities.
- Option C: Dark line (absorption) spectra are produced when light passes through a cooler gas. This is not the primary characteristic of the solar spectrum during an eclipse.
- Option D: Dark band is not a standard term for describing spectra.

Key Concepts:
- Solar spectrum
- Solar eclipse
- Continuous spectrum
- Emission spectrum
- Absorption spectrum
- Solar corona

Memory Aids:
- Remember that even during a total solar eclipse, the sun's corona is still emitting light, leading to a continuous spectrum.

Quick Tips:
- Consider what part of the sun is visible during different phases of a solar eclipse.

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Analyzing +4 | -1

Assertion (A) : For two sources to be coherent phase difference between two waves at all points should be same.

Reason (R) : Two different light sources are never coherent.

View answer and explanation

Correct option: D

If Assertion is false but the Reason is true.

Option D is correct. The assertion is false because coherent sources can have a constant phase difference, not necessarily zero. The reason is true because two independent light sources cannot maintain a constant phase relationship.

Solution approach

Correct Answer + Core Reason: If Assertion is false but the Reason is true. Independent light sources cannot be coherent, and coherent sources require a constant, not necessarily zero, phase difference.

Step-by-Step Solution:

Step 1: Analyze the Assertion: Coherent sources require a constant phase difference, but it doesn't have to be the same at all points, and it doesn't have to be zero. Therefore, the assertion is false.

Step 2: Analyze the Reason: Two independent light sources emit light randomly and cannot maintain a constant phase relationship. Therefore, the reason is true.

Step 3: Combine the analysis: Assertion is false, and the Reason is true.

Why other options are wrong:
- Option A: Incorrect because the Assertion is false.
- Option B: Incorrect because the Assertion is false.
- Option C: Incorrect because the Reason is true.
- Option D: Correct because the Assertion is false and the Reason is true.

Key Concepts:
- Coherence
- Phase difference
- Independent light sources

Memory Aids:
- Coherent sources: Constant phase difference
- Independent sources: Incoherent

Quick Tips:
- Remember that coherent sources need a constant phase relationship, not necessarily zero.

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Understanding +4 | -1

Select the right option in the following

View answer and explanation

Correct option: B

Maxwell provided the theoretical evidence that light is transverse wave

Option B (Maxwell provided the theoretical evidence that light is transverse wave) is correct because Maxwell's equations demonstrated that light is an electromagnetic wave, and electromagnetic waves are transverse.

Solution approach

Correct Answer + Core Reason: Maxwell provided the theoretical evidence that light is transverse wave. Maxwell's equations showed light is an EM wave, which is transverse.

Step-by-Step Solution:
Step 1: Consider each statement:
- Christian Huygens: Proposed the wave theory of light, but did not establish that light waves were transverse. He assumed light waves were longitudinal.
- Maxwell: Developed Maxwell's equations, which showed that light is an electromagnetic wave. Electromagnetic waves are transverse.
- Thomas Young: Experimentally proved the wave behavior of light through the double-slit experiment, but Huygens did not make any assumptions about the transverse nature of light.
Step 2: Conclusion: Maxwell provided the theoretical evidence that light is a transverse wave.

Why other options are wrong:
- Option A: Huygens assumed light waves were longitudinal, not transverse.
- Option C: Huygens did not make any assumptions about the transverse nature of light.
- Option D: Only statement B is correct.

Key Concepts:
- Wave theory of light
- Maxwell's equations
- Electromagnetic waves
- Transverse waves
- Huygens' principle
- Young's double-slit experiment

Memory Aids:
- Maxwell: EM waves are transverse.

Quick Tips:
- Remember the contributions of different scientists to the understanding of light.

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Understanding +4 | -1

Two light sources are said to be coherent if they are obtained from

View answer and explanation

Correct option: B

A single point source

Option B (A single point source) is correct because deriving the two sources from a single source ensures they have a constant phase relationship, hence coherence.

Solution approach

Correct Answer + Core Reason: A single point source. Deriving two sources from a single source ensures coherence.

Step-by-Step Solution:

Step 1: Problem Identification: The question asks about the condition for two light sources to be coherent.

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

Step 3: Explanation: To ensure coherence, the two light sources are typically derived from a single point source. This ensures that any phase changes in the original source are experienced by both derived sources, maintaining a constant phase relationship.

Step 4: Conclusion: Two light sources are said to be coherent if they are obtained from a single point source.

Why other options are wrong:
- Option A: Two independent point sources emitting light of the same wavelength: Independent sources are unlikely to maintain a constant phase relationship.
- Option C: A wide source: A wide source is incoherent.
- Option D: Two ordinary bulbs emitting light of different wavelengths: Different wavelengths mean different frequencies, which leads to incoherence.

Key Concepts:
- Coherence of light
- Interference of light

Memory Aids:
- Single source = Coherent sources.

Quick Tips:
- Remember that coherence is essential for observing sustained interference patterns.

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Understanding +4 | -1

The separation between two microscopic particles is measured

{P_A}

and

{P_B}

by two different lights of wavelength

2000\,\,\mathop \mathrm{A}\limits^ \circ

and

3000\,\,\mathop \mathrm{A}\limits^ \circ

respectively, then

View answer and explanation

Correct option: B

{P_A} < {P_B}

Option B is correct. Resolving power is inversely proportional to the wavelength of light used. Therefore, smaller wavelength light (2000 Å) will have better resolving power and thus smaller resolvable distance.

Solution approach

Correct Answer + Core Reason: ${P_A} < {P_B}$ . Resolving power is inversely proportional to wavelength.

Step-by-Step Solution:

Step 1: Recall the concept of resolving power. Resolving power is the ability of an instrument to distinguish between two closely spaced objects.

Step 2: Relate resolving power to wavelength. The resolving power is inversely proportional to the wavelength of light used. This means that shorter wavelengths provide better resolution.

Step 3: Apply the concept to the problem. Light A has a wavelength of 2000 Å, and light B has a wavelength of 3000 Å. Since light A has a shorter wavelength, it has a higher resolving power.

Step 4: Relate resolving power to the measured separation. Higher resolving power means that the minimum resolvable distance is smaller. Therefore, the separation measured with light A ( $P_A$ ) will be smaller than the separation measured with light B ( $P_B$ ).

Step 5: Conclude that $P_A < P_B$ .

Why other options are wrong:
- Option A: This is incorrect because shorter wavelengths have better resolving power, leading to a smaller measured separation.
- Option C: While $P_A < P_B$ is true, the specific relationship $P_A < \frac{3}{2} P_B$ may not always hold exactly, but the general inequality is correct.
- Option D: This is incorrect because the different wavelengths will result in different measured separations.

Key Concepts:
- Resolving power
- Wavelength of light
- Minimum resolvable distance

Memory Aids:
- Remember that shorter wavelengths give better resolution.

Quick Tips:
- When comparing resolving power, focus on the relationship between resolving power and wavelength.

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Analyzing +4 | -1

Assertion : Resolving power of a telescope depends only on wavelength.

Reason : This is proportional to square of wavelength.

View answer and explanation

Correct option: D

If both Assertion and reason are false

Option D (If both Assertion and reason are false) is correct because the resolving power of a telescope depends on the wavelength and the diameter of the objective lens, and it is inversely proportional to the wavelength, not proportional to the square of the wavelength.

Solution approach

Correct Answer + Core Reason: If both Assertion and reason are false. Resolving power depends on wavelength and diameter of the objective lens.

Step-by-Step Solution:

Step 1: The resolving power of a telescope is given by the formula: $RP = \frac{d}{1.22\lambda}$ , where $d$ is the diameter of the objective lens and $\lambda$ is the wavelength of light.

Step 2: Assertion: Resolving power of a telescope depends only on wavelength. This is false because it also depends on the diameter of the objective lens.

Step 3: Reason: This is proportional to square of wavelength. This is false because resolving power is inversely proportional to the wavelength, not proportional to the square of the wavelength.

Step 4: Therefore, both the assertion and the reason are false.

Why other options are wrong:
- Option A: This is incorrect because both the assertion and the reason are false.
- Option B: This is incorrect because both the assertion and the reason are false.
- Option C: This is incorrect because both the assertion and the reason are false.

Key Concepts:
- Resolving Power of a Telescope
- Wavelength of Light
- Diameter of Objective Lens

Memory Aids:
- Resolving Power formula: $RP = \frac{d}{1.22\lambda}$

Quick Tips:
- Remember that resolving power is inversely proportional to wavelength.
- Consider both wavelength and diameter of the objective lens when determining resolving power.

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