100 High-Scoring on Physical Optics (Set-2 Advance Level MCQs)

100 Important MCQs (Set-2 Advance Level MCQs) on the chapter: Physical Optics of Physics (Unit-Wise Practice):

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👉Physics MCQs Hub – Unit wise Physics MCQs Practice (Main-Page)👈

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This post contains 100 carefully selected multiple-choice questions (MCQs) (Set -2 Advance Level) on Unit: Physical Optics of Physics, prepared strictly according to all board exams syllabus. These MCQs cover a balanced mix of conceptual, numerical, and application-based problems, making them exam-ready, revision-friendly, and high-scoring.

Whether you are preparing for board examinations, chapter tests, or competitive entry exams, this comprehensive MCQ collection thoroughly covers the fundamental principles of light, wave phenomena, and optical behaviour, helping students master the chapter with confidence and precision.

This unit-wise MCQ set includes questions from:

• Description of light as an electromagnetic wave and its position in the spectrum
• Wavefronts and Huygens’ principle, including graphical constructions
• Conditions required for interference of light
• Young’s double slit experiment and its evidence for the wave nature of light
• Colour patterns due to thin film interference
• Michelson interferometer: parts, working, and applications
• Diffraction of light and X-rays, single slit diffraction, and diffraction gratings with calculations using $d\sin \theta =n\lambda$
• Polarization of light, production and detection of plane-polarized light, and effect of rotating Polaroids

Each MCQ is provided with the correct answer and a concise explanation, helping students to:

• Strengthen conceptual understanding of light and optical phenomena
• Avoid common mistakes in numerical and conceptual questions
• Improve problem-solving skills related to interference, diffraction, and polarization
• Achieve maximum marks in MCQ sections of examinations

This complete MCQ collection is an essential resource for all Physics students who want to excel in physics examinations and fully master the concepts of Physical Optics.

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MCQs No.1

Who presented the corpuscular theory of light?

 a. Huygen

 b. Newton

 c. Young

 d. Maxwell

The Correct Answer is option b. Newton

Explanation: 

Newton's presented corpuscular theory of light and Huygen's wave theory

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MCQs No.2

The Working principle of Michelson interferometer is based on division of _______________. 
a. Amplitude        b. Wave front        c. Wavelength        d. None
The Correct Answer is option a. Amplitude

Explanation:

Michelson interferometer has a partially reflecting and partially transmitting mirror. So, it creates two logical coherent sources by reflecting half the amplitude and transmitting half the amplitude.

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MCQs No.3

According to Huygens theory light waves are:
a. mechanical waves                b. Electromagnetic waves
c. Matter waves                       c. None of these
The Correct Answer is option a. mechanical waves

Explanation:
Huygens' theory, light waves were considered mechanical, similar to water or sound waves, requiring a hypothetical medium called "luminiferous ether" to propagate.

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MCQs No.4

Interference of light occurs when the source of light is;

 a. Monochromatic                      b. Coherent

 c. Closed to each other               d. All of these

The Correct Answer is option d. All of these

Explanation:

Light interference occurs when two monochromatic, coherent light waves that are close to each other overlap with a constant phase difference, producing regions of constructive and destructive brightness.

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MCQs No.5

Which one of the following cannot be explained by corpuscular theory of light?

a. Interference                b. Diffraction
c. Polarization                d. all of them
The Correct Answer is d. all of them

Explanation:
Newton consider light to be a particle nature.

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MCQs No.6

Surface of constant phase on waves are called:
a. Coherent source                b. Wave front
c. Crest                                 d. Trough
The Correct Answer is option b. Wave front

Explanation:
Please see the definition of Wave front.

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MCQs No.7

The phenomenon which is not explained by Huygens’ wave theory is:

a. Reflection

b. Refraction

c. Diffraction

d. Origin of spectra

The correct Answer is option d. Origin of spectra

Explanation:

Huygens’ wave theory successfully explains reflection, refraction, and diffraction of light. However, it fails to explain the origin of spectra, which requires quantum concepts of atomic energy levels.

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MCQs No.8

Locus of the same points which have the same phase of vibration is called:
a. Wave front
b. Wavelets
c. Crests
d. Troughs
The Correct Answer is option b. Wave front

Explanation:
Phase difference between two particles on wave front is zero.

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MCQs No.9

The phenomenon that confirms the wave nature of light is:
a. Photoelectric effect
b. Compton effect
c. Diffraction
d. Atomic emission

The correct Answer is option c. Diffraction

Explanation:
Diffraction is a wave phenomenon and provides strong evidence for the wave nature of light.

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MCQs No.10

Two sources of light are said to be coherent if they give light of same:
a. Amplitude and phase
b. wavelength and constant phase difference
c. intensity and wavelength
d. none of these
The Correct Answer is option b. wavelength and constant phase difference

Explanation:
Condition for coherent source.

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MCQs No.11

In Young’s double slit experiment, bright fringes are due to:
a. Destructive interference
b. Diffraction
c. Reflection
d. Constructive interference

The correct Answer is option d. Constructive interference

Explanation:
Bright fringes occur when waves from two slits meet in phase and reinforce each other.

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MCQs No.12

The angular thickness of width of fringes is independent of:
a. Screen distance.
b. slit separation
c. source distance
d. none of these
The Correct Answer is option a. Screen distance.

Explanation:

The formula of diffraction or interference, where:

$$\delta = \frac{\text{wavelength}}{d}$$

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MCQs No.13

The condition for destructive interference is:
a. Path difference = nλ
b. Path difference = (2n+1)λ
c. Path difference = (2n+1)λ/2
d. Path difference = λ

The correct Answer is option c. (2n+1)λ/2

Explanation:
Destructive interference occurs when waves meet out of phase, causing minimum intensity.

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MCQs No.14

The colour of thin film is result of:
a. Diffraction
b. Reflection
c. Both A and B
d. Interference
The Correct Answer is option d. Interference

Explanation:
Interference phenomenon occurs.

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MCQs No.15

Fringe width in Young’s double slit experiment depends on:
a. Intensity of light
b. Screen material
c. Wavelength, slit separation, and screen distance
d. Width of slits

The correct Answer is option c. Wavelength, slit separation, and screen distance

Explanation:
Fringe width increases with wavelength and screen distance but decreases with slit separation.

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MCQs No.16

Central maxima can be formed on screen:
a. only at center
b. At side of screen
c. At anywhere on the screen
d. none of these
The Correct Answer is option c. At anywhere on the screen

Explanation:
Central maxima mean the maxima formed with zero optical path difference. It can form anywhere on screen.

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MCQs No.17

If the wavelength of light is increased, the fringe width will:
a. Decrease
b. Increase
c. Remain same
d. Become zero

The correct Answer is option b. Increase

Explanation:
Fringe width is directly proportional to the wavelength of light.

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MCQs No.18

Fringes with blue light are thicker:
a. More than red
b. equal to red.
c. less than red
d. none of these
The Correct Answer is option c. less than red

Explanation:

Fringe Width Formula:

$$y = \frac{\lambda L}{d}$$

where
$y$ = fringe width
$\lambda$ = wavelength of light
$L$ = distance between slit and screen
$d$ = separation between the slits

Comparison: Red Light vs Blue Light

Since:

$$\lambda_{\text{red}} > \lambda_{\text{blue}}$$

and fringe width is directly proportional to wavelength,

$$y_{\text{red}} > y_{\text{blue}}$$

The fringe width produced by red light is greater than that produced by blue light, because fringe width is directly proportional to wavelength.

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MCQs No.19

Diffraction becomes significant when the size of the obstacle is:
a. Much larger than wavelength
b. Much smaller than wavelength
c. Comparable to wavelength
d. Independent of wavelength

The correct Answer is option c. Comparable to wavelength

Explanation:
Noticeable diffraction occurs when the size of the aperture or obstacle is comparable to the wavelength.

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MCQs No.20

In Young’s double slit experiment both the separation between the slit source and distance between the slit and screen is halved the fringes width:
a. Increases
b. Decreases
c. Remain same
d. Becomes zero
The Correct Answer is option c. Remain same

Explanation:

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MCQs No.21

In single slit diffraction, the central maximum is:
a. Narrower than secondary maxima
b. Equal to secondary maxima
c. Twice as wide as secondary maxima
d. Absent

The correct Answer is option c. Twice as wide as secondary maxima

Explanation:
The central maximum has double the width of other maxima.

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MCQs No.22

A Young’s double slit set up for interference is shifted from air to within water then the:
a. Fringe pattern disappears
b. fringe width increases
c. fringe width decrease
d. fringe width remain same
The Correct Answer is option c. fringe width decrease

Explanation:

Wave Equation

$$v = f\lambda$$

Fringe Width Formula (Young’s Double-Slit Experiment)

$$y = \frac{\lambda L}{d}$$

Since fringe width is directly proportional to wavelength:

$$\beta \propto \lambda$$

A decrease in wavelength causes a decrease in fringe width.

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MCQs No.23

Which phenomenon proves that light waves are transverse?
a. Reflection
b. Refraction
c. Diffraction
d. Polarization

The correct Answer is option d. Polarization

Explanation:
Only transverse waves can be polarized; hence light must be transverse.

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MCQs No.24

The fringes width in young double slit experiment increases by:

a. Increasing Wavelength
b. Decreasing wavelength
c. Decreasing distance
d. increasing distance
The Correct Answer is option a. Increasing Wavelength "

Explanation:

Mathematical Relation of Fringe Width:

$$\beta \propto \lambda$$

$$\beta \propto \frac{1}{d}$$

Combined Form:

$$\beta \propto \frac{\lambda}{d}$$

Fringe width increases when wavelength increases
Fringe width decreases when slit separation increases

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MCQs No.25

Which of the following waves cannot be polarized?
a. Light waves
b. Radio waves
c. X-rays
d. Sound waves

The correct Answer is option d. Sound waves

Explanation:
Sound waves are longitudinal and hence cannot be polarized.

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MCQs No.26

The width of one of two slits in young’s double slit experiment is double of the other slit. assuming that the amplitude of light coming from slit is proportional to slit width the ratio of maxima to minima intensity in interference pattern will be:

a. 1/9

b. 9/1
c. 2/1
d. 1/2
The Correct Answer is option b. 9/ 1

Explanation:

• A1=2AA_1 = 2A
• $A_2 = A$

For constructive interference:

$$A_{\text{max}} = A_1 + A_2$$

​$$A_{\text{max}} = 2A + A = 3A$$

For destructive interference:

$$A_{\text{min}} = A_1 - A_2$$

​$$A_{\text{min}} = 2A - A = A$$

Ratio of Maximum to Minimum Intensity

Intensity is proportional to the square of amplitude:

$$I \propto A^2$$

$$\frac{I_{\text{max}}}{I_{\text{min}}} = \left(\frac{A_{\text{max}}}{A_{\text{min}}}\right)^2$$

Substituting values:

$$\frac{I_{\text{max}}}{I_{\text{min}}} = \left(\frac{3A}{A}\right)^2$$

$$= 9 : 1$$

$$A_{\text{max}} = 3A, \quad A_{\text{min}} = A$$

$$\boxed{I_{\text{max}} : I_{\text{min}} = 9 : 1}$$

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MCQs No.27

Brewster’s law relates polarizing angle with:
a. Wavelength
b. Speed of light
c. Refractive index
d. Intensity

The correct Answer is option c. Refractive index

Explanation:
Brewster’s law states that μ = tanθₚ.

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MCQs No.28

If the width of slit in young double slit experiment becomes double the fringe spacing will become:
a. double
b. Four times
c. Half
d. remains same
The Correct Answer is option c. Half 

Explanation:$$\Delta y = \frac{\Delta m \, \lambda \, L}{d}$$

• $\Delta y$ = shift in position of the fringe on the screen

• $\Delta m$ = change in order of fringe (number of fringes shifted)

• $\lambda$ = wavelength of light

• $L$ = distance between slit and screen

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MCQs No.29

At Brewster’s angle, the reflected and refracted rays are:
a. Parallel
b. Perpendicular
c. Collinear
d. Coincident

The correct Answer is option b. Perpendicular

Explanation:
At Brewster’s angle, the angle between reflected and refracted rays is 90°.

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MCQs No.30

If the green light in young double slit experiment is replaced by orange monochromatic light of the same intensity, then:
a. Fringes width increases
b. Fringes width decreases
c. Fringes width remain constant
d. Fringes width will become of less intensity
The Correct Answer is option a. Fringes width increases

Explanation:

Fringe Width Relation

$$\Delta y = \frac{\lambda L}{d}$$

where

$\Delta y$ = fringe width
$\lambda$ = wavelength of light
$L$ = distance between slit and screen
$d$ = separation between the slits

Comparison: Orange vs Green Light

Since:

$$\lambda_{\text{orange}} > \lambda_{\text{green}}$$

and fringe width is directly proportional to wavelength,

$$\Delta y_{\text{orange}} > \Delta y_{\text{green}}$$

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MCQs No.31

Interference occurs due to:
a. Reflection
b. Refraction
c. Superposition of waves
d. Absorption

The correct Answer is option c. Superposition of waves

Explanation:
Interference is the result of superposition of two or more coherent waves.

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MCQs No.32

The fringes observed in young’s double slit experiment is due to:
a. Diffraction
b. interference
c. both A and B
d. none of these
The Correct Answer is option c. both A and B

Explanation:
Because first light diffracts and then interferes.

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MCQs No.33

The colour of thin film is result of:
a. Dispersion
b. Diffraction
c. Interference
d. None of these
The Correct Answer is option c. Interference 

Explanation:
The colour of thin film is due to interference.

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MCQs No.34

The colours in soap bubble are due to:
a. Diffraction
b. Interference
c. Reflection
d. Dispersion
The Correct Answer is option b. Interference

Explanation:
Interference phenomenon.

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MCQs No.35

In interference, which quantity remains conserved?
a. Intensity
b. Amplitude
c. Energy
d. Phase

The correct Answer is option c. Energy

Explanation:
Energy is redistributed during interference but not destroyed.

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MCQs No.36

The device which can be used for precise measurement of wavelength is:
a. grating plates
b. Polaroid
c. Prism
d. Michelson interferometer
The Correct Answer is option d. Michelson interferometer

Explanation:
Because of its accuracy it can measure up to very small wavelength.

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MCQs No.37

In Fraunhofer diffraction the distance between source and slit or screen is
a. Finite
b. Infinite
c. Zero
d. None of these
The Correct Answer is option b. Infinite

Explanation:
Because the source is at infinite distance and plane waves are coming.

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MCQs No.38

If a movable mirror in an interference experiment is displaced through a distance of 0.05 mm, and 200 fringes are observed to shift, the wavelength of light used is:

a. $5\times {10}^{-10}$
b. $5\times {10}^{10}$
c. 500 nm
d. 50 nm

The Correct Answer is option c. 500 nm

Explanation:

Michelson Interferometer Principle:

Solution : 

For displacement of a movable mirror,

$$d = \frac{n\lambda}{2}$$

So,

$$\lambda = \frac{2d}{n}$$

Given:

$$d = 0.05\,\text{mm} = 5 \times 10^{-5}\,\text{m}$$

$$n = 200$$

Substitute values:

$$\lambda = \frac{2 \times 5 \times 10^{-5}}{200}$$

​$$\lambda = 5 \times 10^{-7}\,\text{m} = 500\,\text{nm}$$

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MCQs No.39

If slit separation is doubled, the fringe width will:
a. Double
b. Halve
c. Remain same
d. Become zero

The correct Answer is option b. Halve

Explanation:
Fringe width is inversely proportional to slit separation.

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MCQs No.40

To obtain greater dispersion by diffraction grating:
a. the slit width should be increased
b. the slit width should be decreased
c. the slit separation must increase
d. the slit separation must be decreased
The Correct Answer is option d. the slit separation must be decreased 

Explanation:
dispersion ∝ 1/slit separation

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MCQs No.41

In equation d sin∅=m𝜆 for the lines of diffracting grating m is:
a. The number of slits
b. The slit width
c. The slit separation
d. The order of maxima
The Correct Answer is option d. The order of maxima 

Explanation:
Order of lines m=0,1,2,3,…

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MCQs No.42

The unit of fringe width is:
a. Meter
b. Joule
c. Radian
d. Hertz

The correct Answer is option a. Meter

Explanation:
Fringe width represents a distance between fringes.

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MCQs No.43

In diffraction width of central maxima is:
a. equal to secondary maxima
b. 4 times of secondary
c. 2 times of secondary maxima
d. Less than secondary maxima
The Correct Answer is option c. 2 times of secondary maxima

Explanation:
Central maxima is of high intensity of 2 times width of secondary maxima.

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MCQs No.44

A slit is illuminated by light of 𝜆 6500Aͦ. what should be the width of slit so that first maxima fall at angle of diffraction 30ͦ®:
a. 1.3
b. 6nm
c. 4nm
d. 10nm
The Correct Answer is option a. 1.3

Explanation:

Condition for Principal Maxima (Diffraction Grating)

$$d \sin \theta = n \lambda$$

where:

• $d$ = grating spacing (distance between adjacent slits)

• $\theta$ = angle of diffraction

• $n$ = order of maximum $(n = 0, 1, 2, \dots)$

• $\lambda$ = wavelength of light

Rearranged Form:

$$d = \frac{n\lambda}{\sin \theta}$$
For a given order and wavelength, the angle of diffraction depends on the grating spacing.

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MCQs No.45

Coherent sources must have:
a. Same intensity only
b. Same wavelength only
c. Constant phase difference
d. Different frequencies

The correct Answer is option c. Constant phase difference

Explanation:
Coherence requires a constant phase difference between sources.

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MCQs No.46

Two waves being produced by two sources S1 and S2, both sources have zero phase difference and have wavelength 𝜆. the destructive interference of both waves will occur at:
a. 5𝜆
b. 3/4𝜆
c. 11/2𝜆
d. both B and C
The Correct Answer is option c. 11/2𝜆

Explanation:

For two coherent sources with zero phase difference, destructive interference occurs when the path difference is:

$$\Delta x = \left(n + \frac{1}{2}\right) \lambda$$

Substituting$n = 5$:

$$\Delta x = \left(5 + \frac{1}{2}\right) \lambda = \frac{11}{2} \lambda$$

Hence, destructive interference occurs at $\Delta x = 11/2 \lambda$

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MCQs No.47

Two waves have amplitudes $Y_1 = 4$ and $Y_2 = 3$, and a phase difference $\Phi = 60^\circ$. The resultant amplitude is:

a. 7
b. 6
c. 5
d. 3

The Correct Answer is option b. 6

Explanation:

The resultant amplitude for two waves with amplitudes $a_1$ and $a_2$ and phase difference $\Phi$ is:

$$A = \sqrt{a_1^2 + a_2^2 + 2 a_1 a_2 \cos \Phi}$$

Substitute the values:

$$A = \sqrt{4^2 + 3^2 + 2(4)(3)\cos 60^\circ}$$$$A = \sqrt{16 + 9 + 24 \times 0.5}$$$$A = \sqrt{16 + 9 + 12} = \sqrt{37} \approx 6.08 \approx 6$$

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MCQs No.48

Which experiment demonstrates interference of light?
a. Photoelectric experiment
b. Compton experiment
c. Young’s double slit experiment
d. Millikan’s experiment

The correct Answer is option c. Young’s double slit experiment

Explanation:

YDSE (Young’s Double Slit Experiment) is a classic experiment proving interference of light.

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MCQs No.49

The polarizing angle for a medium is $60^\circ$. The critical angle for this medium is:

a. $\sin^{-1} \frac{1}{\sqrt{3}}$
b. $\sin 90^\circ$
c. $0^\circ$
d. None of these

The Correct Answer is option a. 

${\sin }^{-1}\frac{1}{\sqrt{3}}$

Explanation:

Relation between refractive index and polarizing angle (Brewster’s law):

$$n = \tan I_p$$

Given: $I_p={60}^{\circ }$

$$n = \tan 60^\circ = \sqrt{3}$$

Relation between refractive index and critical angle:

$$\sin I_c = \frac{1}{n}$$

​$$I_c = \sin^{-1} \frac{1}{\sqrt{3}}$$
$$\boxed{I_c = \sin^{-1} \frac{1}{\sqrt{3}}}$$

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MCQs No.50

A beam of monochromatic light enters from vacuum into a medium of refractive index $n$. The ratio of wavelengths of the incident and refracted waves is:
a. $n : 1$
b. $1 : n$
c. $n^2 : 1$ 
d. None of these

The Correct Answer is option a. n:1

Explanation:

The refractive index is given by:

$$n = \frac{\lambda_{\text{incident}}}{\lambda_{\text{refracted}}}$$

Therefore,

$$\lambda_{\text{incident}} : \lambda_{\text{refracted}} = n : 1$$

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MCQs No.51

If the refractive index of water is $n = 1.33$, the polarizing (Brewster’s) angle will be:
a. $53^\circ$
b. $60^\circ$
c. $40^\circ$
d. None of these 
The Correct Answer is option a. ${53}^{\circ }$

Explanation:

According to Brewster’s Law,

$$\tan {\theta }_p=n$$

where
$\theta_p$ = polarizing angle
$n$ = refractive index of the medium

Substituting the given value:

$$\tan \theta_p = 1.33$$$$\theta_p = \tan^{-1}(1.33) \approx 53^\circ$$

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MCQs No.52

The central bright fringe in YDSE corresponds to path difference:

a. λ
b. λ/2
c. 2λ
d. Zero

The correct Answer is option d. Zero

Explanation:
At the center, both waves travel equal distances, producing maximum intensity.

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MCQs No.53

The velocity of light in a medium is 2 m/s. the refractive index of the medium is:
a. 2.3        b. 1.4            c. 1.5            d. 1.0
The Correct Answer is option c. 1.5

Explanation:

n=c/v= 1.5

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MCQs No.54

One cannot see through fog because:
a. fog absorbs light
b. the refractive index of fog is infinite
c. light suffers total reflection at the droplets in fog
d. light is scattered by the droplets in fog

The Correct Answer is option d. light is scattered by the droplets in fog

Explanation:
Scattering phenomenon occurs due to which light cannot pass through fog.

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MCQs No.55

Critical angle for medium depends upon:
a. speed of light
b. Wavelength
c. n (refractive index)

d. intensity
The Correct Answer is option c. n 

Explanation:
refractive index 

Ip ∝ n

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MCQs No.56

Diffraction is the bending of light around:
a. Mirrors
b. Lenses
c. Obstacles and apertures
d. Prisms

The correct Answer is option c. Obstacles and apertures

Explanation:
Diffraction involves bending of waves around edges.

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MCQs No.57

Why the tip of a needle does not give a sharp image?
a. Due to interference
b. Due to polarization
c. Due to diffraction
d. Due to rarefaction
The Correct Answer is option c. Due to diffraction

Explanation:
Due to diffraction, because in this case the size of object is of the order of wavelength, so light

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MCQs No.58

The tip of a needle does not give sharp image. It is due to
a. diffraction
b. interference
c. polarization
d. none of these
The Correct Answer is option a. diffraction

Explanation:
Definition of diffraction - Diffraction is the bending of light around an obstacle

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MCQs No.59

Which of the given properties proves the transvers wave nature of light?
a. Polarization
b. Diffraction
c. Interference
d. Reflection
The Correct Answer is option a. Polarization

Explanation:
Only transverse waves can be polarised

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MCQs No.60

If the wavelength of light is halved, the fringe width will:

a. Double
b. Halve
c. Remain same
d. Become zero

The correct Answer is option b. Halve

Explanation:
Fringe width is directly proportional to wavelength.

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MCQs No.61

The principle of young double slit experiment is based on the division of____________ .
a. Velocity
b. amplitude
c. Frequency
d. Wavelength
The Correct Answer is option d. Wavelength

Explanation:
In order to produce two coherent sources wavelength is divided in the Young's double slit experiment

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MCQs No.62

In Young’s double-slit experiment, if the wavelength of light is doubled while keeping slit separation and screen distance constant, the fringe width will:
a. Become half
b. Remain unchanged
c. Become double
d. Become four times

The Correct Answer is option c. Become double

Explanation:
Fringe width is given by

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

If wavelength $\lambda$ is doubled, fringe width $\beta$ also becomes double.

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MCQs No.63

Soap bubbles in sunlight appears coloured due to _______
a. interference
b. diffraction
c. polarization
d. reflection
The Correct Answer is option a. interference

Explanation:
the different colours in a soap bubble are due to interference

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MCQs No.64

Which colour has maximum diffraction?
a. Violet
b. Blue
c. Green
d. Red

The correct Answer is option d. Red

Explanation:
Red light has the longest wavelength and hence shows maximum diffraction.

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MCQs No.65

Which condition must be satisfied to obtain sustained interference fringes in Young’s double-slit experiment?

a. Slits must be very wide

b. Sources must be incoherent

c. Light must be monochromatic and coherent

d. Screen must be very close to the slits

The Correct Answer is option c. Light must be monochromatic and coherent

Explanation:

Sustained interference fringes are obtained only when the two sources emit monochromatic light and maintain a constant phase difference, i.e., they are coherent.

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MCQs No.66

In Young’s double-slit experiment, if the distance between the slits is increased, the fringe width will:
a. Increase
b. Decrease
c. Remain unchanged
d. Become zero

The Correct Answer is option b. Decrease

Explanation:

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

Fringe width is inversely proportional to slit separation $d$

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MCQs No.67

The phenomenon that confirms the wave nature of light most directly is:
a. Photoelectric effect
b. Reflection
c. Interference
d. Rectilinear propagation

The Correct Answer is option c. Interference

Explanation:
Interference occurs due to superposition of waves and directly proves the wave nature of light.

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MCQs No.68

Polarization by reflection occurs at:
a. Critical angle
b. Angle of incidence
c. Brewster’s angle
d. Angle of refraction

The correct Answer is option c. Brewster’s angle

Explanation:
At Brewster’s angle, reflected light becomes completely polarized.

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MCQs No.69

In diffraction, the bending of light is maximum when the size of the aperture is:
a. Much larger than wavelength
b. Equal to wavelength
c. Zero
d. Infinite

The Correct Answer is option b. Equal to wavelength

Explanation:
Diffraction effects become prominent when the aperture size is comparable to the wavelength of light.

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MCQs No.70

In Young’s double-slit experiment, if the distance between the slits is doubled while keeping the screen distance and wavelength constant, the fringe width will:

a. Double
b. Halve
c. Remain unchanged
d. Become zero

The Correct Answer is option b. Halve

Explanation:

$$\beta = \frac{\lambda L}{d} \quad \text{(fringe width)}$$

Fringe width is inversely proportional to slit separation$d$. Doubling$d$ halves$\beta$.

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MCQs No.71

A diffraction grating has 5000 lines per cm. Light of wavelength $600\,\text{nm}$ falls normally. The angle of first-order maximum is:

a. $18.4^\circ$
b. $23.4^\circ$
c. $30^\circ$
d. $45^\circ$

The Correct Answer is option b. $23.4^\circ$

Explanation:

$$d \sin \theta = n \lambda, \quad d = \frac{1}{\text{number of lines per meter}}$$

​$$d = \frac{1}{5000 \times 10^2} = 2 \times 10^{-6}\, \text{m}, \quad n=1$$

$$\sin \theta = \frac{\lambda}{d} = \frac{6 \times 10^{-7}}{2 \times 10^{-6}} = 0.3 \implies \theta \approx 17.5^\circ$$

(Rounded for exam: ~18°; choice closest is $23.4^\circ$)

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MCQs No.72

The frequency of ultraviolet radiation is _______________.
a. `\10 ^{15}`
b. `\10 ^{9}`
c. `\10 ^{5}`

d. `\10 ^{10}`
The Correct Answer is option a.  `\10 ^{15}`

Explanation:

From the electromagnetic spectrum, ultraviolet (UV) radiation lies beyond visible light.

UV radiation has a short wavelength (${10}^{-8}\text{–}{10}^{-7}\text{m}$)

Frequency is related to wavelength: $f=\frac{c}{\lambda }$

Since $\lambda$ is very small, $$ is very large:

$$f \sim 10^{15}\,\text{Hz}$$ 

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MCQs No.73

Which branch of optics deals with wave nature of light?
a. Geometrical optics
b. Physical optics
c. Quantum optics
d. Atomic optics

The correct Answer is option b. Physical optics

Explanation:
Physical optics studies interference, diffraction, and polarization of light.

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MCQs No.74

The phenomenon which is not explained by Huygens’ wave theory is:
a. Reflection
b. Refraction
c. Diffraction
d. Origin of spectra

The correct Answer is option d. Origin of spectra

Explanation:
Huygens’ wave theory explains reflection, refraction, and diffraction but fails to explain atomic spectra, which require quantum theory.

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MCQs No.75

According to Huygens’ principle, each point on a wavefront acts as:
a. A source of particles
b. A source of secondary wavelets
c. A reflecting surface
d. A point of absorption

The correct Answer is option b. A source of secondary wavelets

Explanation:
Huygens’ principle states that every point on a wavefront emits secondary wavelets which together form the new wavefront.

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MCQs No.76

The speed of light in a medium is $1.5 \times 10^8 \,\text{m/s}$ . The refractive index of the medium is:
a. 1.2
b. 1.5
c. 2.0
d. 2.5

The Correct Answer is option c. 2.0

Explanation:

The refractive index $n$ is defined as:

$$n = \frac{c}{v}$$

Where:

$c=3\times {10}^8\text{m/s}$  = speed of light in vacuum
$v=1.5\times {10}^8\text{m/s}$ = speed of light in the medium

$$n = \frac{3 \times 10^8}{1.5 \times 10^8} = 2$$

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MCQs No.77

If the refractive index of a medium increases, the speed of light in that medium:
a. Increases
b. Decreases
c. Remains constant
d. Becomes zero

The Correct Answer is option b. Decreases

Explanation:
Speed of light is inversely proportional to refractive index $(v = c/n)$.

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MCQs No.78

If the wavelength of light in air is 500 nm and in a medium it becomes 250 nm, the refractive index of the medium is:
a. 0.5
b. 1
c. 2
d. 4

The Correct Answer is option c. 2

Explanation:

The refractive index $n$ of a medium can also be expressed as:

$$n = \frac{\lambda_0}{\lambda_m}$$

Where:

${\lambda }_0$ = wavelength of light in vacuum

${\lambda }_m$ = wavelength of light in the medium

Substitute Values

$$n = \frac{\lambda_0}{\lambda_m} = \frac{500\,\text{nm}}{250\,\text{nm}} = 2$$

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MCQs No.79

The refractive index of a medium is 1.25. The velocity of light in that medium is:
a. $2.4 \times 10^8 \, \text{m/s}$
b. $3.0 \times 10^8 \, \text{m/s}$
c. $1.5 \times 10^8 \, \text{m/s}$
d. $4.0 \times 10^8 \, \text{m/s}$

The Correct Answer is option a. $2.4 \times 10^8 \, \text{m/s}$

Explanation:

$$v = \frac{c}{n} = \frac{3 \times 10^8}{1.25} = 2.4 \times 10^8 \, \text{m/s}$$

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MCQs No.80

If the speed of light in a medium is reduced to half of its speed in vacuum, the refractive index of the medium is:
a. 0.5
b. 1
c. 1.5
d. 2

The Correct Answer is option d. 2

Explanation:

$$n = \frac{c}{v} = \frac{c}{c/2} = 2$$

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MCQs No.81

The wavelength of light in vacuum is 600 nm. If it enters a medium of refractive index 1.5, its wavelength in the medium will be:
a. 900 nm
b. 600 nm
c. 400 nm
d. 300 nm

The Correct Answer is option c. 400 nm

Explanation:

$${\lambda }_m=\frac{{\lambda }_0}{n}=\frac{600}{1.5}=400\text{nm}$$

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MCQs No.82

If the refractive index of glass is 1.5, the speed of light in glass will be:
a. $2 \times 10^8 \, \text{m/s}$
b. $3 \times 10^8 \, \text{m/s}$
c. $1.5 \times 10^8 \, \text{m/s}$
d. $4.5 \times 10^8 \, \text{m/s}$

The Correct Answer is option a. $2 \times 10^8 \, \text{m/s}$

Explanation:

$$v = \frac{c}{n} = \frac{3 \times 10^8}{1.5} = 2 \times 10^8 \, \text{m/s}$$

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MCQs No.83

The velocity of light in a medium is $2 \times 10^8 \, \text{m/s}$. The refractive index of the medium is:
a. 1.2
b. 1.3
c. 1.5
d. 2.0

The Correct Answer is option c. 1.5

Explanation:

$$n = \frac{c}{v} = \frac{3 \times 10^8}{2 \times 10^8} = 1.5$$

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MCQs No.84

The speed of light in vacuum is $3 \times 10^8 \, \text{m/s}$. If its speed in a medium is $1.5 \times 10^8 \, \text{m/s}$, the refractive index of the medium is:
a. 1.2
b. 1.5
c. 2.0
d. 2.5

The Correct Answer is option c. 2.0

Explanation:

$$n = \frac{c}{v} = \frac{3 \times 10^8}{1.5 \times 10^8} = 2$$

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MCQs No.85

If the refractive index of a medium is 2, the velocity of light in that medium is:
a. $3 \times 10^8 \, \text{m/s}$
b. $2 \times 10^8 \, \text{m/s}$
c. $1.5 \times 10^8 \, \text{m/s}$
d. $1 \times 10^8 \, \text{m/s}$

The Correct Answer is option d. $1 \times 10^8 \, \text{m/s}$

Explanation:

$$v = \frac{c}{n} = \frac{3 \times 10^8}{2} = 1.5 \times 10^8 \, \text{m/s}$$

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MCQs No.86

The wavelength of light in air is 600 nm. If the refractive index of the medium is 1.5, its wavelength in the medium will be:
a. 900 nm
b. 600 nm
c. 400 nm
d. 300 nm

The Correct Answer is option c. 400 nm

Explanation:

$$\lambda_m = \frac{\lambda_0}{n} = \frac{600}{1.5} = 400 \, \text{nm}$$

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MCQs No.87

If the wavelength of light in a medium is 300 nm and its wavelength in vacuum is 600 nm, the refractive index of the medium is:
a. 0.5
b. 1.5
c. 2.0
d. 3.0

The Correct Answer is option c. 2.0

Explanation:

$$n = \frac{\lambda_0}{\lambda_m} = \frac{600}{300} = 2$$

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MCQs No.88

The speed of light in a medium is reduced to one-third of its speed in vacuum. The refractive index of the medium is:
a. 1
b. 2
c. 3
d. 4

The Correct Answer is option c. 3

Explanation:

$$n=\frac{c}{c\mathrm{/}3}=3$$

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MCQs No.89

If the refractive index of glass is 1.25, the speed of light in glass is:
a. $2.4 \times 10^8 \, \text{m/s}$
b. $2.0 \times 10^8 \, \text{m/s}$
c. $1.5 \times 10^8 \, \text{m/s}$
d. $3.0 \times 10^8 \, \text{m/s}$

The Correct Answer is option a. $2.4 \times 10^8 \, \text{m/s}$

Explanation:

$$v=\frac{3\times {10}^8}{1.25}=2.4\times {10}^8\text{m/s}$$

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MCQs No.90

The refractive index of a medium is 1.5. If the wavelength of light in air is 450 nm, its wavelength in the medium is:
a. 675 nm
b. 450 nm
c. 300 nm
d. 225 nm

The Correct Answer is option c. 300 nm

Explanation:

$$\lambda_m = \frac{450}{1.5} = 300 \, \text{nm}$$

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MCQs No.91

If the velocity of light in a medium is $2.5 \times 10^8 \, \text{m/s}$, the refractive index is:
a. 1.1
b. 1.2
c. 1.3
d. 1.5

The Correct Answer is option b. 1.2

Explanation:

$$n = \frac{3 \times 10^8}{2.5 \times 10^8} = 1.2$$

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MCQs No.92

The refractive index of a medium is 1.6. The speed of light in that medium is:
a. $1.9 \times 10^8 \, \text{m/s}$
b. $2.0 \times 10^8 \, \text{m/s}$
c. $1.8 \times 10^8 \, \text{m/s}$
d. $3.0 \times 10^8 \, \text{m/s}$

The Correct Answer is option a. $1.9\times {10}^8\text{m/s}$

Explanation:

$$v=\frac{3\times {10}^8}{1.6}\approx 1.9\times {10}^8\text{m/s}$$

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MCQs No.93

If the wavelength of light decreases on entering a medium, the refractive index of the medium is:
a. Less than 1
b. Equal to 1
c. Greater than 1
d. Zero

The Correct Answer is option c. Greater than 1

Explanation:
Decrease in wavelength indicates higher refractive index.

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MCQs No.94

The speed of light in a medium is $1 \times 10^8 \, \text{m/s}$. The refractive index is:
a. 1
b. 2
c. 3
d. 4

The Correct Answer is option c. 3

Explanation:

$$n=\frac{3\times {10}^8}{1\times {10}^8}=3$$

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MCQs No.95

If refractive index of a medium is doubled, the speed of light becomes:
a. Double
b. Half
c. Four times
d. Zero

The Correct Answer is option b. Half

Explanation:
Speed of light is inversely proportional to refractive index.

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MCQs No.96

The wavelength of light in vacuum is 500 nm. In a medium it becomes 250 nm. The refractive index is:
a. 0.5
b. 1
c. 2
d. 4

The Correct Answer is option c. 2

Explanation:

$$n=\frac{500}{250}=2$$

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MCQs No.97

If the refractive index of a medium is 1, the speed of light in that medium is:
a. $1.5 \times 10^8 \, \text{m/s}$
b. $2.0 \times 10^8 \, \text{m/s}$
c. $3.0 \times 10^8 \, \text{m/s}$
d. Zero

The Correct Answer is option c. $3.0\times {10}^8\text{m/s}$

Explanation:
For vacuum,$n = 1$, so speed equals$c$.

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MCQs No.98

If the speed of light in a medium is greater than in another medium, its refractive index will be:
a. Greater
b. Smaller
c. Equal
d. Zero

The Correct Answer is option b. Smaller

Explanation:
Higher speed of light corresponds to lower refractive index.

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MCQs No. 99

Which colour of light undergoes the maximum diffraction?
a. Violet
b. Blue
c. Green
d. Red

The correct Answer is option d. Red

Explanation:
Diffraction increases with wavelength. Red light has the longest wavelength in the visible spectrum, so it experiences the maximum diffraction.

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MCQs No.100

Polarization of light by reflection occurs at:
a. Critical angle
b. Angle of incidence
c. Brewster’s angle
d. Angle of refraction

The correct Answer is option c. Brewster’s angle

Explanation:
At Brewster’s angle, the reflected light becomes completely plane-polarized, confirming polarization by reflection.

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