WAVE OPTICS MCQs for NEET UG PHYSICS — Medium

1 Medium

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.

2 Medium

2 of 104Question 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.

3 Medium

3 of 104Question 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.

4 Medium

4 of 104Question 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.

5 Medium

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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.

6 Medium

6 of 104Question 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.

7 Medium

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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.

8 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.

9 Medium

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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.

10 Medium

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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.

11 Medium

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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.

12 Medium

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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.

13 Medium

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Understanding +4 | -1

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

View answer and explanation

Correct option: 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.

14 Medium

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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.

15 Medium

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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.

16 Medium

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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.

17 Medium

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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.

18 Medium

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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.

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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.

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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.

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Analyzing +4 | -1

In YDSE, if one slit is covered by a transparent film so that it transmitts only one half the light intensity of other then choose the incorrect statement :

View answer and explanation

Correct option: B

Fringe pattern shifts away from the covered slit

Option B is the incorrect statement. When one slit is covered with a transparent film, the fringe pattern shifts towards the covered slit because it introduces an additional path difference.

Solution approach

Correct Answer + Core Reason: Fringe pattern shifts away from the covered slit. Covering one slit with a transparent film introduces an additional path difference, causing the fringe pattern to shift towards the covered slit.

Step-by-Step Solution:

Step 1: Understand the effect of the transparent film: The transparent film introduces an additional path difference, which affects the position of the central bright fringe.

Step 2: Determine the direction of the shift: The fringe pattern shifts towards the slit covered by the film.

Step 3: Analyze the effect on fringe brightness: The bright fringes become less bright, and the dark fringes become somewhat bright due to the intensity difference between the two slits.

Step 4: Analyze the fringe width: The fringe width remains unchanged as it depends on the wavelength, distance between the slits, and distance to the screen, none of which are affected by the film.

Why other options are wrong:
- Option A: The fringe pattern shifts *towards* the covered slit, not away from it.
- Option C: This statement is correct; the bright fringes become less bright, and the dark fringes become somewhat bright.
- Option D: This statement is correct; the fringe width remains unchanged.

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

Memory Aids:
- Remember that covering one slit introduces an extra path difference, causing a shift in the fringe pattern.

Quick Tips:
- Always consider the direction of the shift when a transparent film is introduced in YDSE.

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Application +4 | -1

In young’s double slit experiment, the fringe pattern is observed on screen placed at a distance D. The slits are illuminated by light of wavelength

\lambda

. The distance from the central point where intensity becomes

\frac{3}{4}th

of the maxima will be

View answer and explanation

Correct option: A

\frac{{\lambda D}}{{6d}}

Option A (

\frac{{\lambda D}}{{6d}}

) is correct because it is derived from the intensity formula in Young's double-slit experiment.

Solution approach

Correct Answer + Core Reason: $\frac{{\lambda D}}{{6d}}$ . The distance is calculated using the intensity formula in Young's double-slit experiment.

Step-by-Step Solution:

Step 1: The intensity at a point in Young's double-slit experiment is given by $I = I_0 \cos^2(\frac{\phi}{2})$ , where $I_0$ is the maximum intensity and $\phi$ is the phase difference.

Step 2: We are given that $I = \frac{3}{4} I_0$ . Therefore, $\frac{3}{4} I_0 = I_0 \cos^2(\frac{\phi}{2})$ .

Step 3: Simplify: $\cos^2(\frac{\phi}{2}) = \frac{3}{4}$ . Taking the square root, $\cos(\frac{\phi}{2}) = \frac{\sqrt{3}}{2}$ .

Step 4: This implies $\frac{\phi}{2} = \frac{\pi}{6}$ , so $\phi = \frac{\pi}{3}$ .

Step 5: The phase difference is related to the path difference by $\phi = \frac{2\pi}{\lambda} \Delta x$ , where $\Delta x$ is the path difference.

Step 6: Therefore, $\frac{\pi}{3} = \frac{2\pi}{\lambda} \Delta x$ , which gives $\Delta x = \frac{\lambda}{6}$ .

Step 7: The path difference is also related to the distance $y$ from the central point by $\Delta x = \frac{yd}{D}$ , where $d$ is the slit separation and $D$ is the distance to the screen.

Step 8: Thus, $\frac{\lambda}{6} = \frac{yd}{D}$ , which gives $y = \frac{\lambda D}{6d}$ .

Why other options are wrong:
- **Option B ( $\frac{{\lambda D}}{d}$ ): This value is incorrect and corresponds to a different intensity level.
- Option C ( $\frac{{\lambda D}}{{2d}}$ ): This value is incorrect and corresponds to a different intensity level.
- Option D ( $\frac{{2\lambda D}}{{3d}}$ ): This value is incorrect and corresponds to a different intensity level.

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

Memory Aids:**
- Remember the intensity formula: $I = I_0 \cos^2(\frac{\phi}{2})$
- Relationship between phase difference and path difference: $\phi = \frac{2\pi}{\lambda} \Delta x$

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

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Analyzing +4 | -1

Assertion : Light ray passing through water droplet is partially polarized.
Reason : Water is a polar molecule.

View answer and explanation

Correct option: A

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

Option A is correct. Light scattering from water droplets is partially polarized due to the polar nature of water molecules, making both the assertion and reason true, and the reason the correct explanation.

Solution approach

Correct Answer + Core Reason: If both Assertion and Reason are true and the Reason is a correct explanation of the Assertion. The polarization of light passing through water droplets is related to the polar nature of water molecules.

Step-by-Step Solution:

Step 1: Analyze the Assertion: Light ray passing through a water droplet can be partially polarized due to scattering and refraction.

Step 2: Analyze the Reason: Water is a polar molecule, which affects how light interacts with it, leading to polarization.

Step 3: Connect the Assertion and Reason: The polar nature of water molecules causes the partial polarization of light.

Why other options are wrong:
- Option B: Incorrect because the Reason is the correct explanation of the Assertion.
- Option C: Incorrect because both Assertion and Reason are true.
- Option D: Incorrect because both Assertion and Reason are true.

Key Concepts:
- Polarization of light
- Polar molecules
- Light scattering

Memory Aids:
- Polar molecules can polarize light.

Quick Tips:
- Consider the molecular properties when analyzing light interactions.

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Understanding +4 | -1

Which one of the following property of light does not support wave theory of light?

View answer and explanation

Correct option: C

Light shows photoelectric effect

Option C (Light shows photoelectric effect) is correct because the photoelectric effect demonstrates the particle nature of light, not the wave nature.

Solution approach

Correct Answer + Core Reason: Light shows photoelectric effect. Photoelectric effect demonstrates particle nature, not 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 support the wave theory of light: Reflection, refraction, polarization, interference, diffraction.
Step 3: Recall the properties that support the particle theory of light: Photoelectric effect, Compton scattering.
Step 4: Analyze each option:
- Light obeys laws of reflection and refraction: Can be explained by both wave and particle theories.
- Light waves get polarized: A wave phenomenon.
- Light shows photoelectric effect: A particle phenomenon.
- Light shows interference: A wave phenomenon.
Step 5: Conclusion: Photoelectric effect does not support the wave theory of light.

Why other options are wrong:
- Option A: Reflection and refraction can be explained by both wave and particle theories.
- Option B: Polarization is a wave phenomenon.
- Option D: Interference is a wave phenomenon.

Key Concepts:
- Wave theory of light
- Particle theory of light
- Photoelectric effect
- Polarization, interference, reflection, refraction

Memory Aids:
- Wave properties: Interference, diffraction, polarization
- Particle properties: Photoelectric effect, Compton scattering

Quick Tips:
- Remember which phenomena support the wave and particle theories of light.

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Application +4 | -1

In Young’s double slit experiment, when violet light of wavelength

4358\mathop \mathrm{A}\limits^ \circ

is used, the 84 fringe are seen in the field of view, but when sodium light of certain wavelength is used, then 62 fringes are seen in the field of view, the wavelength of sodium light is

View answer and explanation

Correct option: B

5904\mathop \mathrm{A}\limits^ \circ

Option B (

5904\mathop \mathrm{A}\limits^ \circ

) is correct because the ratio of the wavelengths is inversely proportional to the number of fringes observed.

Solution approach

Correct Answer + Core Reason: $5904\mathop \mathrm{A}\limits^ \circ$ . The wavelength is calculated using the inverse relationship between wavelength and number of fringes.

Step-by-Step Solution:

Step 1: Identify the given information: Wavelength of violet light $\lambda_1 = 4358\mathop \mathrm{A}\limits^ \circ$ , number of fringes for violet light $n_1 = 84$ , number of fringes for sodium light $n_2 = 62$ .

Step 2: Since the field of view is the same, the product of the wavelength and the number of fringes is constant: $n_1 \lambda_1 = n_2 \lambda_2$

Step 3: Solve for the wavelength of sodium light $\lambda_2$ : $\lambda_2 = \frac{n_1 \lambda_1}{n_2}$

Step 4: Substitute the given values: $\lambda_2 = \frac{84 \times 4358}{62}$

Step 5: Calculate the wavelength: $\lambda_2 = \frac{366072}{62} = 5904.387 \approx 5904\mathop \mathrm{A}\limits^ \circ$

Step 6: Final Answer: $5904\mathop \mathrm{A}\limits^ \circ$

Why other options are wrong:
- **Option A ( $6893\mathop \mathrm{A}\limits^ \circ$ ): This is incorrect because it is a result of incorrect calculation or using the wrong formula.
- Option C ( $5523\mathop \mathrm{A}\limits^ \circ$ ): This is incorrect because it is a result of incorrect calculation or using the wrong formula.
- Option D ( $6429\mathop \mathrm{A}\limits^ \circ$ ): This is incorrect because it is a result of incorrect calculation or using the wrong formula.

Key Concepts:
- Young's Double Slit Experiment
- Wavelength and Number of Fringes

Memory Aids:
- Remember that the number of fringes is inversely proportional to the wavelength.

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

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Remembering +4 | -1

The wave theory of light was given by

View answer and explanation

Correct option: C

Huygen

Option C (Huygen) is correct because Christian Huygens is credited with proposing the wave theory of light.

Solution approach

Correct Answer + Core Reason: Huygen. Huygens proposed the wave theory of light.

Step-by-Step Solution:

Step 1: Recall the history of the wave theory of light.

Step 2: Identify the originator of the wave theory.

Why other options are wrong:
- Option A: Maxwell developed the electromagnetic theory of light.
- Option B: Planck developed quantum theory.
- Option D: Young performed the double-slit experiment, supporting the wave theory, but he didn't originate it.

Key Concepts:
- Wave theory of light
- History of physics

Memory Aids:
- Huygens = wave theory.

Quick Tips:
- Know the key figures in the development of light theories.

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Analyzing +4 | -1

Two monochromatic coherent point sources

S_1

and

S_2

are separated by a distance

L

. Each source emits light of wavelength

\lambda

, where

L >> \lambda

. The line

S_1 S_2

when extended meets a screen perpendicular to it at a point

A.

View answer and explanation

Correct option: A

The interference fringes on the screen are circular in shape

Option A (The interference fringes on the screen are circular in shape) is correct because the path difference remains constant along circles centered at the midpoint of

S_1

and

S_2

.

Solution approach

Correct Answer + Core Reason: The interference fringes on the screen are circular in shape. The path difference is constant along circles centered at the midpoint of $S_1$ and $S_2$ .

Step-by-Step Solution:

Step 1: Consider the geometry of the setup. Two coherent point sources, $S_1$ and $S_2$ , are separated by a distance $L$ . A screen is placed perpendicular to the line $S_1 S_2$ at point $A$ .

Step 2: For any point on the screen, the path difference between the waves from $S_1$ and $S_2$ determines whether constructive or destructive interference occurs.

Step 3: Let's consider points on the screen equidistant from the midpoint of $S_1$ and $S_2$ . The locus of points with a constant path difference from two fixed points is a hyperbola. However, since the screen is far away ( $L >> \lambda$ ), the hyperbolas approximate circles centered at the midpoint of $S_1$ and $S_2$ .

Step 4: Therefore, the interference fringes on the screen will be circular in shape.

Why other options are wrong:
- **Option B (The interference fringes on the screen are straight lines perpendicular to the line $S_1 S_2 A$ ): Straight fringes are observed in Young's double-slit experiment when the screen is far away and the slits are close together, but in this case, the fringes are circular.
- Option C (The point $A$ is an intensity maxima if $L=n \lambda / 2$ ):** The point A is an intensity maxima if the path difference between $S_1$ and $S_2$ is an integer multiple of the wavelength. Since the path difference at A is L, A is a maxima if $L = n\lambda$ , not $L = n\lambda/2$ .
- **Option D (The point $A$ is always an intensity maxima for any separation $L$ ): The intensity at point A depends on the path difference, which is equal to L. If L is not an integer multiple of the wavelength, then A will not be an intensity maxima.

Key Concepts:
- Interference of Light Waves
- Path Difference
- Coherent Sources
- Hyperbolic Fringes

Memory Aids:
- Remember that the locus of points with a constant path difference from two fixed points is a hyperbola, which approximates a circle when the screen is far away.

Quick Tips:**
- Visualize the geometry of the setup to determine the shape of the interference fringes.

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Remembering +4 | -1

The resolving power of a telescope depends on

View answer and explanation

Correct option: D

Diameter of the objective lens

Option D is correct. The resolving power of a telescope is directly proportional to the diameter of the objective lens.

Solution approach

Correct Answer + Core Reason: Diameter of the objective lens. Resolving power is proportional to the diameter of the objective lens.

Step-by-Step Solution:

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

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

Step 3: Identify the factor that the resolving power depends on. From the formula, it's clear that the resolving power depends on the diameter of the objective lens.

Why other options are wrong:
- Option A: The focal length of the eye lens affects the magnification, not the resolving power.
- Option B: The focal length of the objective lens affects the magnification, not the resolving power directly. While it contributes to magnification, the diameter is the key factor for resolution.
- Option C: The length of the telescope is related to the focal lengths of the objective and eyepiece, but it doesn't directly determine the resolving power.

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

Memory Aids:
- Remember that a larger objective lens allows for better resolution.

Quick Tips:
- The larger the diameter of the objective lens, the better the telescope's ability to distinguish between closely spaced objects.

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Remembering +4 | -1

The condition for diffraction of

{m^{th}}

order minima is

View answer and explanation

Correct option: A

d\sin {\mathrm{\theta }_m} = m\lambda ,\;m = 1,2,3, \ldots

Option A is correct. The condition for diffraction minima is given by dsinθ = mλ, where m = 1, 2, 3,...

Solution approach

Correct Answer + Core Reason: $d\sin {\mathrm{\theta }_m} = m\lambda ,\;m = 1,2,3, \ldots$ . This is the condition for diffraction minima.

Step-by-Step Solution:

Step 1: Recall the condition for diffraction minima in single-slit diffraction.

Step 2: State the formula for diffraction minima: $d \sin \theta_m = m \lambda$ , where $d$ is the width of the slit, $\theta_m$ is the angle of the $m^{th}$ minima, $\lambda$ is the wavelength of light, and $m$ is an integer ( $m = 1, 2, 3, ...$ ).

Why other options are wrong:
- Option B: This option is incorrect because it has $\frac{m\lambda}{2}$ which represents a different condition, not the standard minima.
- Option C: This option is incorrect because it has $\left( {m + 1} \right)\frac{\lambda }{2}$ which represents a different condition, not the standard minima.
- Option D: This option is incorrect because it has $\left( {m - 1} \right)\frac{\lambda }{2}$ which represents a different condition, not the standard minima.

Key Concepts:
- Single-slit diffraction
- Diffraction minima
- Wavelength of light

Memory Aids:
- Remember the formula for minima: "d sine theta equals m lambda."

Quick Tips:
- Distinguish between the conditions for maxima and minima in diffraction.

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Application +4 | -1

A parallel beam of light of intensity

{I_0}

is incident on a glass plate,

25\%

of light is reflected by upper surface and

50\%

of light is reflected from lower surface. The ratio of maximum to minimum intensity in interference region of reflected rays is

View answer and explanation

Correct option: A

{\left( {\frac{{\frac{1}{2} + \sqrt {\frac{3}{8}} }}{{\frac{1}{2} - \sqrt {\frac{3}{8}} }}} \right)^2}

The correct answer is A (

{\left( {\frac{{\frac{1}{2} + \sqrt {\frac{3}{8}} }}{{\frac{1}{2} - \sqrt {\frac{3}{8}} }}} \right)^2}

) because it is the ratio of maximum to minimum intensity calculated from the reflection coefficients.

Solution approach

Correct Answer + Core Reason: ${\left( {\frac{{\frac{1}{2} + \sqrt {\frac{3}{8}} }}{{\frac{1}{2} - \sqrt {\frac{3}{8}} }}} \right)^2}$ . The ratio of maximum to minimum intensity is calculated from the amplitudes of the reflected waves.

Step-by-Step Solution:

Step 1: Identify the given information: $I_0$ is the incident intensity, $25\% = \frac{1}{4}$ is reflected from the upper surface, and $50\% = \frac{1}{2}$ is reflected from the lower surface.

Step 2: The amplitude of the wave reflected from the upper surface is $A_1 = \sqrt{\frac{1}{4} I_0} = \frac{1}{2} \sqrt{I_0}$ .

Step 3: The intensity of the light transmitted to the lower surface is $I_t = I_0 - \frac{1}{4} I_0 = \frac{3}{4} I_0$ .

Step 4: The amplitude of the wave reflected from the lower surface is $A_2 = \sqrt{\frac{1}{2} I_t} = \sqrt{\frac{1}{2} \times \frac{3}{4} I_0} = \sqrt{\frac{3}{8}} \sqrt{I_0}$ .

Step 5: $I_{\text{max}} = (A_1 + A_2)^2 = \left( \frac{1}{2} \sqrt{I_0} + \sqrt{\frac{3}{8}} \sqrt{I_0} \right)^2 = I_0 \left( \frac{1}{2} + \sqrt{\frac{3}{8}} \right)^2$ .

Step 6: $I_{\text{min}} = (A_1 - A_2)^2 = \left( \frac{1}{2} \sqrt{I_0} - \sqrt{\frac{3}{8}} \sqrt{I_0} \right)^2 = I_0 \left( \frac{1}{2} - \sqrt{\frac{3}{8}} \right)^2$ .

Step 7: $\frac{{I_{\text{max}}}}{{I_{\text{min}}}} = \frac{{\left( \frac{1}{2} + \sqrt{\frac{3}{8}} \right)^2}}{{\left( \frac{1}{2} - \sqrt{\frac{3}{8}} \right)^2}} = \left( \frac{{\frac{1}{2} + \sqrt{\frac{3}{8}}}}{{\frac{1}{2} - \sqrt{\frac{3}{8}}}} \right)^2$ .

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

Key Concepts:
- Interference
- Intensity ratio
- Reflection and transmission

Memory Aids:
- Remember to account for the intensity loss due to reflection at each surface.

Quick Tips:
- Be careful with the intensities and amplitudes of the reflected waves.

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