100 Advanced & Numerical MCQs (Level -2) on Physical Optics, Physics (Unit-Wise MCQs Practice):

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

Whether you are preparing for Board examinations, chapter tests, college assessments, or competitive entrance examinations (MDCAT, ECAT, NUST, PIEAS, GIKI, UET, FAST, NTS, and other engineering or medical admission tests), this comprehensive Physical Optics MCQ Collection is designed to help you master one of the most important branches of wave optics. The questions are arranged progressively—from fundamental concepts to advanced numerical problems and higher-order thinking—ensuring complete and systematic preparation for every type of examination.

This chapter explains the wave nature of light and covers the fundamental principles of wavefronts, Huygens' Principle, interference, diffraction, polarization, Michelson Interferometer, diffraction grating, X-ray diffraction, and Bragg's Law. It strengthens conceptual understanding while developing analytical reasoning and numerical problem-solving skills required for both academic and competitive examinations.

This chapter-wise MCQ collection includes:

100 Basic MCQs Level-1 (1–100) – Covering fundamental concepts, wavefronts, Huygens' Principle, interference of light, Young's Double Slit Experiment, fringe formation, diffraction, polarization, diffraction grating, Michelson Interferometer, X-ray diffraction, and Bragg's Law.

100 Advanced & Numerical MCQs (101–200) – Focusing on fringe width calculations, path difference, thin-film interference, diffraction grating equations, Michelson Interferometer applications, Brewster's Law, Malus' Law, Bragg's Law, and conceptual numerical problems.

50 Higher-Order Thinking Skills (HOTS) MCQs (201–250) – Designed to strengthen analytical reasoning, conceptual understanding, assertion-reason questions, experimental analysis, application-based learning, and multi-concept problem-solving.

50 Challenging MCQs Quiz with Answers – A carefully selected mix of conceptual, numerical, and HOTS questions designed for quick revision, self-assessment, and complete exam preparation.

This MCQ collection covers:

Nature of light and electromagnetic waves
Wavefronts and Huygens' Principle
Construction of wavefronts using Huygens' Principle
Interference of light and conditions for interference
Young's Double Slit Experiment (YDSE)
Path difference, phase difference, and fringe width calculations
Thin-film interference and anti-reflection coatings
Michelson Interferometer and its applications
Diffraction at a single slit
Diffraction grating and grating equation
Resolving power of diffraction gratings
X-ray diffraction and Bragg's Law
Polarization of light and transverse nature of light
Polaroids and their applications
Brewster's Law and polarizing angle
Malus' Law and intensity calculations
Practical applications of interference, diffraction, and polarization
Experimental techniques and modern optical instruments
Numerical problems and real-life applications in physical optics

Every MCQ includes the correct answer along with a clear, concept-based explanation to strengthen understanding, improve problem-solving skills, and reinforce key principles of Physical Optics.

This question bank helps students to:

Build a strong conceptual foundation in Physical Optics
Master interference, diffraction, and polarization concepts
Improve numerical and analytical problem-solving skills
Understand wave optics and modern optical instruments
Develop logical reasoning through HOTS and application-based questions
Avoid common examination mistakes
Increase speed, accuracy, and confidence in objective-type questions
Prepare effectively for both Board examinations and competitive entrance tests

With 250 carefully selected MCQs arranged into 100 Basic, 100 Advanced & Numerical, and 50 Higher-Order Thinking Skills (HOTS) questions, along with a 50 Most Important MCQs Quiz, this all-in-one Physical Optics MCQ Bank provides complete preparation for the chapter. It is an excellent study resource for strengthening concepts, improving exam performance, and achieving success in Board examinations, MDCAT, ECAT, NUST, PIEAS, GIKI, UET, FAST, NTS, and other competitive engineering and medical entrance examinations.

Best Budget Laptops for Students for Study, Coding & Office Work
(2026 Amazon Deals)

👉 Check Today’s Price on Amazon👈

👉 Check Acer Aspire 5 Price on Amazon
👉 Buy Lenovo IdeaPad Slim 3 on Amazon
👉 See HP Pavilion Latest Deal on Amazon
👉 View ASUS Vivobook 16 Details on Amazon
👉 Check MacBook Air Student Price on Amazon

Level-II – 100 Advanced & Numerical MCQs (MCQs 101–200)

MCQs No. 101

The fringe width in Young's Double Slit Experiment is given by:

a. $\frac{dD}{\lambda}$

b. $\frac{\lambda D}{d}$

c. $\frac{\lambda d}{D}$

d. $\frac{D}{\lambda d}$

The Correct Answer is option b. $\frac{\lambda D}{d}$

Explanation:

The fringe width (β) in Young's Double Slit Experiment is directly proportional to the wavelength (λ) and the screen distance (D), and inversely proportional to the slit separation (d).

MCQs No. 102

In Young's Double Slit Experiment, the wavelength of light is doubled while all other quantities remain unchanged. The fringe width will:

a. Become half

b. Remain unchanged

c. Double

d. Become four times

The Correct Answer is option c. Double

Explanation:

Since fringe width is directly proportional to wavelength (β ∝ λ), doubling the wavelength doubles the fringe width.

MCQs No. 103

The slit separation in Young's Double Slit Experiment is doubled. The fringe width becomes:

a. Double

b. Half

c. Four times

d. Unchanged

The Correct Answer is option b. Half

Explanation:

Fringe width is inversely proportional to slit separation. Increasing the slit separation reduces the distance between adjacent fringes.

MCQs No. 104

The distance between the slits is 0.5 mm, the screen is 2 m away, and the wavelength of light is 600 nm. The fringe width is:

a. 1.2 mm

b. 2.4 mm

c. 3.6 mm

d. 4.8 mm

The Correct Answer is option b. 2.4 mm

Explanation:

Using

β = λD/d

= (600 × 10⁻⁹ × 2)/(0.5 × 10⁻³)

= 2.4 × 10⁻³ m

= 2.4 mm

MCQs No. 105

If the screen distance is increased from 2 m to 4 m, the fringe width will:

a. Become half

b. Double

c. Remain unchanged

d. Become one-fourth

The Correct Answer is option b. Double

Explanation:

Fringe width is directly proportional to the distance between the slits and the screen.

MCQs No. 106

If the wavelength is 500 nm and the slit separation is 0.25 mm, what is the fringe width when the screen is 1.5 m away?

a. 2 mm

b. 3 mm

c. 4 mm

d. 5 mm

The Correct Answer is option b. 3 mm

Explanation:

= (500 × 10⁻⁹ × 1.5)/(0.25 × 10⁻³)

= 3 × 10⁻³ m

= 3 mm

MCQs No. 107

The phase difference corresponding to a path difference of one wavelength is:

a. π

b. π/2

c. 2π

d. 4π

The Correct Answer is option c. 2π

Explanation:

A complete wavelength corresponds to one complete cycle, or 2π radians.

MCQs No. 108

A path difference of λ/2 produces a phase difference of:

a. π

b. 2π

c. π/2

d. 3π

The Correct Answer is option a. π

Explanation:

Half a wavelength corresponds to 180° (π radians), producing destructive interference.

MCQs No. 109

The fringe width becomes maximum when:

a. Slit separation is maximum

b. Wavelength is minimum

c. Screen distance is maximum and slit separation is minimum

d. Frequency is maximum

The Correct Answer is option c. Screen distance is maximum and slit separation is minimum

Explanation:

Large screen distance and small slit separation produce wider interference fringes.

MCQs No. 110

If both wavelength and screen distance are doubled simultaneously, the fringe width becomes:

a. Double

b. Four times

c. Half

d. Unchanged

The Correct Answer is option b. Four times

Explanation:

Since β ∝ λD,

β' = (2λ)(2D)/d

= 4β

MCQs No. 111

If both wavelength and slit separation are doubled, the fringe width will:

a. Double

b. Half

c. Remain unchanged

d. Become four times

The Correct Answer is option c. Remain unchanged

Explanation:

Both quantities increase by the same factor, so their effects cancel.

MCQs No. 112

A student uses blue light instead of red light in Young's Double Slit Experiment. The fringe width will:

a. Increase

b. Decrease

c. Remain unchanged

d. Become zero

The Correct Answer is option b. Decrease

Explanation:

Blue light has a shorter wavelength than red light, producing narrower fringes.

MCQs No. 113

Which colour of visible light produces the widest interference fringes?

a. Violet

b. Blue

c. Green

d. Red

The Correct Answer is option d. Red

Explanation:

Red light has the longest wavelength in the visible spectrum, resulting in the greatest fringe width.

MCQs No. 114

If coherent sources are replaced by incoherent sources, the interference pattern will:

a. Become brighter

b. Shift sideways

c. Disappear

d. Become wider

The Correct Answer is option c. Disappear

Explanation:

Only coherent sources maintain a constant phase difference necessary for a stable interference pattern.

MCQs No. 115

For the first bright fringe from the center, the path difference is:

a. λ/2

b. λ

c. 2λ

d. Zero

The Correct Answer is option b. λ

Explanation:

The first bright fringe occurs where the path difference is one wavelength.

MCQs No. 116

The first dark fringe from the center corresponds to a path difference of:

a. λ

b. λ/2

c. 2λ

d. Zero

The Correct Answer is option b. λ/2

Explanation:

A path difference of λ/2 produces destructive interference.

MCQs No. 117

If the wavelength is decreased while all other quantities remain constant, the fringe width:

a. Increases

b. Decreases

c. Remains unchanged

d. Doubles

The Correct Answer is option b. Decreases

Explanation:

Fringe width is directly proportional to wavelength.

MCQs No. 118

The SI unit of wavelength is:

a. Meter

b. Hertz

c. Newton

d. Joule

The Correct Answer is option a. Meter

Explanation:

Wavelength is a length and is measured in metres.

MCQs No. 119

The unit "nanometre" is equal to:

a. 10⁻⁶ m

b. 10⁻⁹ m

c. 10⁻³ m

d. 10⁻¹² m

The Correct Answer is option b. 10⁻⁹ m

Explanation:

One nanometre (nm) equals 10⁻⁹ metre.

MCQs No. 120

In Young's Double Slit Experiment, the interference pattern is observed on:

a. A prism

b. A screen

c. A mirror

d. A lens

The Correct Answer is option b. A screen

Explanation:

The alternating bright and dark fringes are formed on a screen placed at some distance from the slits.

MCQs No. 121

The intensity at a dark fringe is ideally:

a. Maximum

b. Half

c. Zero

d. Infinite

The Correct Answer is option c. Zero

Explanation:

Perfect destructive interference results in zero intensity at the dark fringe.

MCQs No. 122

Increasing the frequency of light while keeping the speed constant causes the wavelength to:

a. Increase

b. Decrease

c. Remain constant

d. Double

The Correct Answer is option b. Decrease

Explanation:

Since $v = f\lambda$ , for constant speed, wavelength decreases as frequency increases.

MCQs No. 123

The wavelength of visible light is usually measured in:

a. Millimetres

b. Centimetres

c. Nanometres

d. Kilometres

The Correct Answer is option c. Nanometres

Explanation:

Visible wavelengths are very small and are conveniently expressed in nanometres (nm).

MCQs No. 124

If the slit separation is reduced by half, the fringe width becomes:

a. Half

b. Double

c. Four times

d. Unchanged

The Correct Answer is option b. Double

Explanation:

Fringe width is inversely proportional to slit separation.

MCQs No. 125

A larger fringe width makes the interference pattern:

a. Less visible

b. Easier to observe and measure

c. Completely disappear

d. Independent of wavelength

The Correct Answer is option b. Easier to observe and measure

Explanation:

Wider fringes are more distinct and easier to measure accurately, reducing experimental errors.

MCQs No. 126

A soap film of uniform thickness is illuminated with monochromatic light. The observed bright and dark bands are produced due to:

a. Diffraction

b. Polarization

c. Thin film interference

d. Dispersion

The Correct Answer is option c. Thin film interference

Explanation:

Thin film interference occurs due to the superposition of light waves reflected from the upper and lower surfaces of a thin transparent film.

MCQs No. 127

The optical path difference for light reflected from a thin film depends mainly upon:

a. Density of the film

b. Thickness and refractive index of the film

c. Temperature of the film

d. Area of the film

The Correct Answer is option b. Thickness and refractive index of the film

Explanation:

The optical path difference is proportional to both the thickness and refractive index of the thin film.

MCQs No. 128

A thin film has a thickness of 400 nm and a refractive index of 1.5. The optical path length through the film is:

a. 400 nm

b. 500 nm

c. 600 nm

d. 800 nm

The Correct Answer is option c. 600 nm

Explanation:

Optical path length = n × t

= 1.5 × 400 nm

= 600 nm

MCQs No. 129

The colours seen in an oil film on water change because:

a. The refractive index changes continuously

b. The thickness of the film is not uniform

c. The wavelength of sunlight changes

d. The oil absorbs different colours

The Correct Answer is option b. The thickness of the film is not uniform

Explanation:

Different thicknesses produce different path differences, causing different wavelengths to undergo constructive interference.

MCQs No. 130

An anti-reflection coating on a camera lens works on the principle of:

a. Reflection

b. Diffraction

c. Destructive interference

d. Refraction

The Correct Answer is option c. Destructive interference

Explanation:

The coating is designed so that reflected waves cancel each other, greatly reducing reflected light.

MCQs No. 131

The primary purpose of the Michelson Interferometer is to measure:

a. Electric current

b. Magnetic field

c. Very small distances and wavelengths

d. Pressure

The Correct Answer is option c. Very small distances and wavelengths

Explanation:

The Michelson Interferometer is an extremely precise optical instrument used for measuring wavelength, refractive index, and very small distances.

MCQs No. 132

In a Michelson Interferometer, moving one mirror by 0.25 μm changes the optical path by:

a. 0.25 μm

b. 0.50 μm

c. 1.00 μm

d. 2.00 μm

The Correct Answer is option b. 0.50 μm

Explanation:

Since the light travels to the mirror and back, the optical path changes by twice the mirror displacement.

MCQs No. 133

If a mirror in a Michelson Interferometer is moved by one wavelength, the optical path difference changes by:

a. λ/2

b. λ

c. 2λ

d. 4λ

The Correct Answer is option c. 2λ

Explanation:

The light travels twice the mirror displacement, so moving the mirror by λ changes the optical path by 2λ.

MCQs No. 134

The beam splitter in a Michelson Interferometer divides the incoming beam into:

a. Three beams

b. Two coherent beams

c. Two incoherent beams

d. Four beams

The Correct Answer is option b. Two coherent beams

Explanation:

The beam splitter produces two coherent beams that travel along different paths before recombining.

MCQs No. 135

Fringe shifting in a Michelson Interferometer is caused by:

a. Increasing the intensity of light

b. Changing the optical path difference

c. Changing the colour of the mirrors

d. Using ordinary light

The Correct Answer is option b. Changing the optical path difference

Explanation:

Any change in the optical path difference shifts the interference fringes.

MCQs No. 136

If one mirror in a Michelson Interferometer is moved by 300 nm, the optical path difference changes by:

a. 300 nm

b. 450 nm

c. 600 nm

d. 900 nm

The Correct Answer is option c. 600 nm

Explanation:

Optical path difference = 2 × mirror displacement

= 2 × 300 nm

= 600 nm

MCQs No. 137

The Michelson Interferometer is most suitable for measuring:

a. Large distances

b. Extremely small displacements

c. Electric potential

d. Temperature

The Correct Answer is option b. Extremely small displacements

Explanation:

The instrument can detect displacements much smaller than the wavelength of light.

MCQs No. 138

If the wavelength of light is increased, the number of fringes observed for the same mirror displacement will:

a. Increase

b. Decrease

c. Remain unchanged

d. Become infinite

The Correct Answer is option b. Decrease

Explanation:

Larger wavelengths produce fewer fringes for the same change in optical path difference.

MCQs No. 139

The Michelson Interferometer operates on the principle of:

a. Diffraction only

b. Interference of coherent light

c. Polarization only

d. Reflection only

The Correct Answer is option b. Interference of coherent light

Explanation:

The instrument divides and recombines coherent light beams to produce interference fringes.

MCQs No. 140

Which quantity is directly determined by counting the number of fringe shifts in a Michelson Interferometer?

a. Electric current

b. Optical path difference

c. Pressure

d. Temperature

The Correct Answer is option b. Optical path difference

Explanation:

Each fringe shift corresponds to a specific change in optical path difference, allowing precise measurements.

MCQs No. 141

An interference pattern becomes less distinct when:

a. Coherent light is used

b. Monochromatic light is used

c. The coherence between the sources decreases

d. Slit separation decreases

The Correct Answer is option c. The coherence between the sources decreases

Explanation:

Loss of coherence causes the bright and dark fringes to fade and eventually disappear.

MCQs No. 142

The visibility of interference fringes depends mainly upon:

a. The coherence of the light sources

b. Atmospheric pressure

c. Screen material

d. Room temperature

The Correct Answer is option a. The coherence of the light sources

Explanation:

Stable and coherent sources produce sharp, high-contrast interference fringes.

MCQs No. 143

The SI unit of optical path length is:

a. Metre

b. Joule

c. Newton

d. Pascal

The Correct Answer is option a. Metre

Explanation:

Optical path length is a length quantity and is measured in metres.

MCQs No. 144

The main advantage of an interferometer over an ordinary measuring scale is its:

a. Simplicity

b. Extremely high precision

c. Low cost

d. Small size

The Correct Answer is option b. Extremely high precision

Explanation:

Interferometers can measure distances comparable to the wavelength of light with exceptional accuracy.

MCQs No. 145

A change in refractive index inside one arm of a Michelson Interferometer will:

a. Produce no effect

b. Change the optical path difference

c. Stop interference

d. Destroy the mirrors

The Correct Answer is option b. Change the optical path difference

Explanation:

Changing the refractive index changes the optical path length, resulting in fringe movement.

MCQs No. 146

The wavelength of monochromatic light can be determined accurately using:

a. A thermometer

b. Michelson Interferometer

c. Ammeter

d. Voltmeter

The Correct Answer is option b. Michelson Interferometer

Explanation:

The Michelson Interferometer is one of the most accurate instruments for measuring wavelength.

MCQs No. 147

The greater the mirror displacement, the ______ observed.

a. fewer fringe shifts are

b. more fringe shifts are

c. lower the intensity is

d. smaller the wavelength is

The Correct Answer is option b. more fringe shifts are

Explanation:

A larger mirror displacement causes a larger optical path difference, producing more fringe shifts.

MCQs No. 148

Interference techniques are widely used in modern science because they provide:

a. High electrical power

b. Highly precise measurements

c. Better colour vision

d. Stronger magnetic fields

The Correct Answer is option b. Highly precise measurements

Explanation:

Interference methods are capable of measuring extremely small distances and changes with remarkable accuracy.

MCQs No. 149

A laser is often used in interferometers because it produces:

a. Incoherent light

b. Highly coherent monochromatic light

c. White light

d. Infrared radiation only

The Correct Answer is option b. Highly coherent monochromatic light

Explanation:

Lasers provide stable, coherent, and monochromatic light, making them ideal for producing clear interference patterns.

MCQs No. 150

Which of the following instruments is most suitable for measuring extremely small changes in length with very high accuracy?

a. Vernier calipers

b. Screw gauge

c. Michelson Interferometer

d. Meter rule

The Correct Answer is option c. Michelson Interferometer

Explanation:

The Michelson Interferometer can detect changes in length much smaller than the wavelength of light, making it one of the most precise measuring instruments in optics.

MCQs No. 151

The angular width of the central maximum in single-slit diffraction increases when the slit width:

a. Increases

b. Decreases

c. Remains constant

d. Becomes infinite

The Correct Answer is option b. Decreases

Explanation:

The angular width of the central maximum is inversely proportional to the slit width. A narrower slit produces greater diffraction and a wider central maximum.

MCQs No. 152

A slit is made narrower while using light of the same wavelength. The diffraction pattern will:

a. Become narrower

b. Become wider

c. Disappear

d. Remain unchanged

The Correct Answer is option b. Become wider

Explanation:

Reducing the slit width increases diffraction, causing the diffraction pattern to spread over a larger angle.

MCQs No. 153

The wavelength of light is doubled while the slit width remains constant. The diffraction pattern will:

a. Become narrower

b. Remain unchanged

c. Become wider

d. Disappear

The Correct Answer is option c. Become wider

Explanation:

Diffraction increases with wavelength. Therefore, doubling the wavelength increases the angular width of the diffraction pattern.

MCQs No. 154

A diffraction grating produces sharper spectra than a prism because it has:

a. Lower refractive index

b. A large number of closely spaced slits

c. Greater thickness

d. Higher mass

The Correct Answer is option b. A large number of closely spaced slits

Explanation:

Thousands of equally spaced slits produce very sharp principal maxima due to the constructive interference of many waves.

MCQs No. 155

The grating element of a diffraction grating is the:

a. Width of each slit only

b. Distance between two successive slits

c. Thickness of the grating

d. Length of the slit

The Correct Answer is option b. Distance between two successive slits

Explanation:

The grating element (d) is the distance from the center of one slit to the center of the next slit.

MCQs No. 156

A diffraction grating has 5000 lines per centimetre. The grating element is approximately:

a. $2 \times 10^{-6}$

b. $2 \times 10^{-5}$

c. $5 \times 10^{-6}$

d. $5 \times 10^{-5}$

The Correct Answer is option a. $2 \times 10^{-6}$

Explanation:

5000 lines/cm = 500000 lines/m

Grating element,

d = 1/500000 = 2 × 10⁻⁶ m

MCQs No. 157

The condition for principal maxima in a diffraction grating is:

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

b. $2d\sin\theta = n\lambda$

c. $d\cos\theta = n\lambda$

d. $2d\cos\theta = n\lambda$

The Correct Answer is option a. $d\sin\theta = n\lambda$

Explanation:

The diffraction grating equation gives the angles at which principal maxima are observed.

MCQs No. 158

For a given grating, the diffraction angle increases when:

a. Wavelength decreases

b. Wavelength increases

c. Slit width increases

d. Screen distance decreases

The Correct Answer is option b. Wavelength increases

Explanation:

According to the grating equation, larger wavelengths are diffracted through larger angles.

MCQs No. 159

Which colour deviates through the largest angle in a diffraction grating?

a. Violet

b. Blue

c. Green

d. Red

The Correct Answer is option d. Red

Explanation:

Red light has the longest wavelength among visible colours and therefore diffracts through the largest angle.

MCQs No. 160

A diffraction grating is mainly used to:

a. Increase light intensity

b. Measure wavelengths accurately

c. Polarize light

d. Produce reflection

The Correct Answer is option b. Measure wavelengths accurately

Explanation:

Diffraction gratings are widely used in spectrometers for accurate wavelength determination.

MCQs No. 161

Bragg's Law is expressed as:

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

b. $2d\sin\theta=n\lambda$

c. $d=n\lambda$

d. $2d=n\sin\theta$

The Correct Answer is option b. $2d\sin\theta=n\lambda$

Explanation:

Bragg's Law gives the condition for constructive interference of X-rays reflected from crystal planes.

MCQs No. 162

Bragg's Law is used to determine the:

a. Frequency of radio waves

b. Crystal structure of solids

c. Speed of sound

d. Electric field intensity

The Correct Answer is option b. Crystal structure of solids

Explanation:

Bragg's Law helps determine the spacing between crystal planes using X-ray diffraction.

MCQs No. 163

The spacing between crystal planes is 0.25 nm. X-rays of wavelength 0.10 nm produce first-order diffraction. The diffraction angle is closest to:

a. 11.5°

b. 23.6°

c. 45°

d. 60°

The Correct Answer is option a. 11.5°

Explanation:

Using Bragg's Law,

2d sinθ = nλ

2(0.25) sinθ = 0.10

sinθ = 0.20

θ ≈ 11.5°

MCQs No. 164

Increasing the wavelength of X-rays while keeping crystal spacing constant causes the Bragg angle to:

a. Decrease

b. Increase

c. Remain constant

d. Become zero

The Correct Answer is option b. Increase

Explanation:

According to Bragg's Law, a larger wavelength requires a larger diffraction angle.

MCQs No. 165

X-ray diffraction is possible because the spacing between crystal planes is comparable to the:

a. Amplitude of X-rays

b. Wavelength of X-rays

c. Frequency of X-rays

d. Intensity of X-rays

The Correct Answer is option b. Wavelength of X-rays

Explanation:

Diffraction occurs when the wavelength is comparable to the spacing between scattering centres.

MCQs No. 166

The first-order spectrum corresponds to:

a. n = 0

b. n = 1

c. n = 2

d. n = 3

The Correct Answer is option b. n = 1

Explanation:

The first bright diffraction maximum is called the first-order spectrum.

MCQs No. 167

If the wavelength is doubled, the diffraction angle for the first order will:

a. Decrease

b. Increase

c. Remain unchanged

d. Become zero

The Correct Answer is option b. Increase

Explanation:

Larger wavelengths require larger diffraction angles to satisfy the grating equation.

MCQs No. 168

The resolving power of a diffraction grating increases with:

a. Fewer slits

b. Greater number of illuminated slits

c. Smaller wavelength only

d. Smaller screen distance

The Correct Answer is option b. Greater number of illuminated slits

Explanation:

More illuminated slits produce sharper principal maxima and better wavelength separation.

MCQs No. 169

Which instrument commonly uses a diffraction grating?

a. Spectrometer

b. Barometer

c. Voltmeter

d. Calorimeter

The Correct Answer is option a. Spectrometer

Explanation:

Modern spectrometers use diffraction gratings to analyze spectral lines accurately.

MCQs No. 170

If the slit width becomes extremely large compared to the wavelength, diffraction effects become:

a. More prominent

b. Negligible

c. Infinite

d. Unpredictable

The Correct Answer is option b. Negligible

Explanation:

Diffraction is noticeable only when the slit width is comparable to the wavelength.

MCQs No. 171

The central maximum in single-slit diffraction has the:

a. Least intensity

b. Greatest intensity

c. Same intensity as side maxima

d. Zero intensity

The Correct Answer is option b. Greatest intensity

Explanation:

The central maximum is the brightest part of the diffraction pattern.

MCQs No. 172

The side maxima in a diffraction pattern are:

a. Brighter than the central maximum

b. Equal in intensity to the central maximum

c. Less intense than the central maximum

d. Invisible

The Correct Answer is option c. Less intense than the central maximum

Explanation:

The intensity decreases rapidly for successive side maxima.

MCQs No. 173

Bragg's Law is based on the phenomenon of:

a. Reflection and interference

b. Polarization

c. Refraction

d. Dispersion

The Correct Answer is option a. Reflection and interference

Explanation:

Constructive interference occurs between X-rays reflected from successive crystal planes.

MCQs No. 174

Which branch of science extensively uses X-ray diffraction?

a. Crystallography

b. Meteorology

c. Astronomy

d. Oceanography

The Correct Answer is option a. Crystallography

Explanation:

X-ray diffraction is the fundamental technique for studying the internal structure of crystalline materials.

MCQs No. 175

A diffraction grating is preferred over a prism because it provides:

a. Lower cost only

b. Higher resolving power and greater accuracy

c. Less diffraction

d. Lower wavelength

The Correct Answer is option b. Higher resolving power and greater accuracy

Explanation:

A diffraction grating produces sharper spectral lines and separates closely spaced wavelengths more effectively than a prism.

MCQs No. 176

Light is incident on a glass surface at Brewster's angle. The reflected light is:

a. Unpolarized

b. Partially polarized

c. Completely plane polarized

d. Circularly polarized

The Correct Answer is option c. Completely plane polarized

Explanation:

At Brewster's angle, the reflected light becomes completely plane polarized, while the refracted light remains partially polarized.

MCQs No. 177

According to Brewster's Law, the refractive index of a medium is given by:

a. $n=\sin i_p$

b. $n=\cos i_p$

c. $n=\tan i_p$

d. $n=\cot i_p$

The Correct Answer is option c. $n=\tan i_p$

Explanation:

Brewster's Law states that the refractive index of a medium is equal to the tangent of the polarizing angle.

MCQs No. 178

The Brewster angle for a medium of refractive index √3 is:

a. 30°

b. 45°

c. 60°

d. 75°

The Correct Answer is option c. 60°

Explanation:

Using Brewster's Law,

tan ip = √3

Therefore,

ip = 60°.

MCQs No. 179

A transparent material has a polarizing angle of 45°. Its refractive index is:

a. 0.5

b. 1

c. 1.5

d. √2

The Correct Answer is option b. 1

Explanation:

According to Brewster's Law,

n = tan45°

= 1.

MCQs No. 180

If the refractive index of a medium increases, its Brewster angle will:

a. Increase

b. Decrease

c. Remain constant

d. Become zero

The Correct Answer is option a. Increase

Explanation:

Since tan ip = n, a larger refractive index corresponds to a larger Brewster angle.

MCQs No. 181

The reflected and refracted rays at Brewster's angle are inclined at:

a. 45°

b. 60°

c. 90°

d. 180°

The Correct Answer is option c. 90°

Explanation:

At Brewster's angle, the reflected and refracted rays are perpendicular to each other.

MCQs No. 182

A Polaroid sheet is primarily used to:

a. Increase light intensity

b. Produce plane-polarized light

c. Produce diffraction

d. Measure wavelength

The Correct Answer is option b. Produce plane-polarized light

Explanation:

A Polaroid transmits vibrations in only one direction, thereby producing plane-polarized light.

MCQs No. 183

Two Polaroids have their transmission axes parallel. The intensity of transmitted light is:

a. Zero

b. Maximum

c. Half

d. One-fourth

The Correct Answer is option b. Maximum

Explanation:

When the transmission axes are parallel, the maximum amount of polarized light passes through the second Polaroid.

MCQs No. 184

Two Polaroids have their transmission axes at 90° to each other. The transmitted intensity is:

a. Maximum

b. Half

c. Zero

d. Double

The Correct Answer is option c. Zero

Explanation:

Crossed Polaroids block all transmitted light because their transmission axes are perpendicular.

MCQs No. 185

According to Malus' Law, the transmitted intensity is proportional to:

a. sin²θ

b. cosθ

c. cos²θ

d. tan²θ

The Correct Answer is option c. cos²θ

Explanation:

Malus' Law states that

I = I₀ cos²θ,

where θ is the angle between the transmission axes.

MCQs No. 186

The transmission axes of two Polaroids make an angle of 60°. The transmitted intensity is:

a. I₀

b. I₀/2

c. I₀/4

d. Zero

The Correct Answer is option c. I₀/4

Explanation:

Using Malus' Law,

I = I₀ cos²60°

= I₀ × (1/2)²

= I₀/4.

MCQs No. 187

If the angle between two Polaroids is 45°, the transmitted intensity is:

a. I₀

b. I₀/2

c. I₀/4

d. Zero

The Correct Answer is option b. I₀/2

Explanation:

I = I₀ cos²45°

= I₀ × (1/√2)²

= I₀/2.

MCQs No. 188

Polarized sunglasses reduce glare because they absorb mainly:

a. Vertically polarized light

b. Horizontally polarized light

c. Circularly polarized light

d. Unpolarized light

The Correct Answer is option b. Horizontally polarized light

Explanation:

Glare from roads and water surfaces is largely horizontally polarized. Polarized sunglasses block this component.

MCQs No. 189

The phenomenon of polarization proves that light is:

a. Longitudinal

b. Transverse

c. Stationary

d. Mechanical

The Correct Answer is option b. Transverse

Explanation:

Only transverse waves can be polarized; therefore, polarization confirms the transverse nature of light.

MCQs No. 190

If unpolarized light passes through a single Polaroid, the transmitted intensity becomes:

a. I₀

b. I₀/2

c. I₀/4

d. Zero

The Correct Answer is option b. I₀/2

Explanation:

A Polaroid transmits only one component of unpolarized light, reducing its intensity to half.

MCQs No. 191

Which instrument is commonly used as an analyzer?

a. Convex lens

b. Plane mirror

c. Polaroid sheet

d. Prism

The Correct Answer is option c. Polaroid sheet

Explanation:

An analyser is a Polaroid used to examine the state of polarization of light.

MCQs No. 192

If the transmission axis of the analyzer is rotated continuously through 360°, the transmitted intensity becomes maximum:

a. Once

b. Twice

c. Four times

d. Never

The Correct Answer is option b. Twice

Explanation:

The intensity follows Malus' Law and reaches maximum whenever the analyzer is parallel to the plane of polarization, which occurs twice during one full rotation.

MCQs No. 193

The refractive index of glass is 1.5. The Brewster angle is approximately:

a. 42°

b. 56°

c. 68°

d. 75°

The Correct Answer is option b. 56°

Explanation:

tan ip = 1.5

Therefore,

ip ≈ 56°.

MCQs No. 194

A Polaroid is rotated from 0° to 90° with respect to another Polaroid. The transmitted intensity:

a. Continuously increases

b. Continuously decreases

c. Remains constant

d. Becomes infinite

The Correct Answer is option b. Continuously decreases

Explanation:

According to Malus' Law, the intensity decreases from maximum at 0° to zero at 90°.

MCQs No. 195

Which phenomenon is used in Liquid Crystal Display (LCD) technology?

a. Diffraction

b. Polarization

c. Reflection

d. Dispersion

The Correct Answer is option b. Polarization

Explanation:

LCDs operate by controlling polarized light with liquid crystal molecules.

MCQs No. 196

The transmission axis of a Polaroid is the direction along which it:

a. Absorbs all light

b. Transmits light vibrations

c. Reflects all light

d. Refracts light

The Correct Answer is option b. Transmits light vibrations

Explanation:

Only the component of light vibrating parallel to the transmission axis passes through the Polaroid.

MCQs No. 197

The intensity of polarized light becomes zero when the angle between the polarizer and analyser is:

a. 30°

b. 45°

c. 60°

d. 90°

The Correct Answer is option d. 90°

Explanation:

At 90°, the transmission axes are perpendicular (crossed Polaroids), so no light is transmitted.

MCQs No. 198

Malus' Law is applicable to:

a. Unpolarized light only

b. Plane-polarized light passing through an analyser

c. Diffraction of light

d. Reflection from mirrors

The Correct Answer is option b. Plane-polarized light passing through an analyzer

Explanation:

Malus' Law describes how the intensity of plane-polarized light changes after passing through an analyzer.

MCQs No. 199

Which of the following confirms both the wave nature and transverse nature of light?

a. Reflection

b. Refraction

c. Polarization

d. Dispersion

The Correct Answer is option c. Polarization

Explanation:

Polarization is a wave phenomenon and can occur only in transverse waves, proving that light is a transverse electromagnetic wave.

MCQs No. 200

A researcher wants to determine the refractive index of a transparent material by measuring its polarizing angle. Which law should be applied?

a. Snell's Law

b. Bragg's Law

c. Brewster's Law

d. Malus' Law

The Correct Answer is option c. Brewster's Law

Explanation:

Brewster's Law relates the refractive index of a transparent medium to its polarizing angle through the equation n = tan ip.

📚 Continue Your Physics Preparation

If you found this quiz helpful, explore our complete collection of Physics learning resources.
🔬 Physics MCQs Hub

More Study Resources for Students and Teachers

Computer Science 11th Class Notes (Main Page)
Computer Science 12th Class Notes (Main Page)

************************************

1. English Poems Analysis Hub– A to Z Index (English Poetry)
(https://poemanalysishub.blogspot.com/p/english-poems-analysis-to-z-index.html)
2. Website for School and College Level Mathematics
(https://askraremaths.blogspot.com)
3. ASK Rare Class Notes – Study Resources for Students & Teachers
(https://askrareclassnotes.blogspot.com/)

© 2026 & onwards Academic Skills and Knowledge (ASK)

Note:  Write me in the comments box below for any query and also Share this information with your classmates and friends.

Best Tools and Supplies for Students and Teachers:

📚 Grammar Books – Improve Writing, Score Higher
🪑 Study Chairs – Sit Comfortably, Study Better
🛋️ Study Desks – Organized Space, Better Learning
💡 Desk Lamps – Bright Light, Better Focus
⌚ Digital Watches– Manage Time, Improve Results
📱 Phone Stands – Hands-Free Learning Made Easy
🎧 Headphones – Focus Better, Learn Without Distractions
⌨️ Wireless Keyboards – Type Faster, Work Smarter
🖱️ Wireless Mouse – Smooth Control, Better Workflow
🖨️ Printers – Print Smart, Save Time
Disclosure: The above links on this page are affiliate links. This means that if you make a purchase through them, I may earn a small commission at no additional cost to you. Your support helps me maintain, improve, and continue providing valuable free content on this website.
With Best Regards
ASKRare Team

🌍📖

"Knowledge grows when it is shared."

At ASK Rare Class Notes, we believe quality education should be freely accessible to everyone. Learn, share, inspire, and help others succeed.

Physical Optics MCQs (Level 2) – 100 Advanced & Numerical MCQs with Answers

100 Advanced & Numerical MCQs (Level -2) on Physical Optics, Physics (Unit-Wise MCQs Practice):

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

This chapter-wise MCQ collection includes:

This MCQ collection covers:

This question bank helps students to:

Level-II – 100 Advanced & Numerical MCQs (MCQs 101–200)

MCQs No. 101

MCQs No. 102

MCQs No. 103

MCQs No. 104

MCQs No. 105

MCQs No. 106

MCQs No. 107

MCQs No. 108

MCQs No. 109

MCQs No. 110

MCQs No. 111

MCQs No. 112

MCQs No. 113

MCQs No. 114

MCQs No. 115

MCQs No. 116

MCQs No. 117

MCQs No. 118

MCQs No. 119

MCQs No. 120

MCQs No. 121

MCQs No. 122

MCQs No. 123

MCQs No. 124

MCQs No. 125

MCQs No. 126

MCQs No. 127

MCQs No. 128

MCQs No. 129

MCQs No. 130

MCQs No. 131

MCQs No. 132

MCQs No. 133

MCQs No. 134

MCQs No. 135

MCQs No. 136

MCQs No. 137

MCQs No. 138

MCQs No. 139

MCQs No. 140

MCQs No. 141

MCQs No. 142

MCQs No. 143

MCQs No. 144

MCQs No. 145

MCQs No. 146

MCQs No. 147

MCQs No. 148

MCQs No. 149

MCQs No. 150

MCQs No. 151

MCQs No. 152

MCQs No. 153

MCQs No. 154

MCQs No. 155

MCQs No. 156

MCQs No. 157

MCQs No. 158

MCQs No. 159

MCQs No. 160

MCQs No. 161

MCQs No. 162

MCQs No. 163

MCQs No. 164

MCQs No. 165

MCQs No. 166

MCQs No. 167

MCQs No. 168

MCQs No. 169

MCQs No. 170

MCQs No. 171

MCQs No. 172

MCQs No. 173

If you found this quiz helpful, explore our complete collection of Physics learning resources.
🔬 Physics MCQs Hub