Unit-9 Physical Optics – Complete Exercise Solutions | Class 11 Physics | FBISE

Unit-9 Physical Optics – Complete Exercise Solutions | Class 11 Physics | KPK Text Board, Peshawar | FBISE

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Click on the link below for complete Class 11 Physics notes, MCQs, Q&A, and Numericals.

Class 11 Physics – Unit-wise MCQs & Study Resources (Main-Page)

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This post provides a complete and fully solved exercise of Unit-9 Physical Optics for Class 11 Physics, strictly according to the KPK Textbook Board / FBISE syllabus. It includes all MCQs, short-answer questions, comprehensive (long) questions, and numerical problems, solved step-by-step using standard board-style notation, formulas, and clear explanations.

Whether you are preparing for annual board examinations, chapter tests, or entry tests, this guide is designed to be concept-clear, exam-oriented, and high-scoring, helping you master the chapter with confidence.

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1. Choose the best possible answer of the following MCQs.

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MCQ No. 1

The principle of Young’s double slit experiment is based on the division of:
a) Amplitude
b) Frequency
c) Velocity
d) Wavelength

Correct Answer is option a. Amplitude

Explanation: In Young’s double slit experiment, a single wave is divided into two coherent sources by dividing its amplitude, producing interference fringes.

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MCQ No. 2

Which one of the following properties proves the transverse wave nature of light?
a) Interference
b) Refraction
c) Polarization
d) Diffraction

Correct Answer is option c. Polarization

Explanation: Only transverse waves can be polarized, so polarization proves that light is a transverse wave.

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MCQ No. 3

Coloured fringes observed in soap bubbles are examples of:
a) Diffraction
b) Interference
c) Polarization
d) Reflection

Correct Answer is option b. Interference

Explanation: Thin film interference in soap bubbles produces constructive and destructive interference for different wavelengths, creating colours.

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MCQ No. 4

During a sunny day, we see objects in a classroom even when all electric lights are off, due to:
a) Reflection of light
b) Refraction of light
c) Diffraction of light
d) Interference of light

Correct Answer is option a. Reflection of light

Explanation: Sunlight enters the room and reflects from objects, making them visible.

Why Not Polarization?

We see objects due to reflection of scattered sunlight entering the room; diffraction effects are negligible because the size of windows is much larger than the wavelength of light.

👉 Illumination of rooms = Reflection + Scattering
👉 Diffraction = narrow slits, edges, gratings

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MCQ No. 5

The principle of Michelson interferometer is based on the division of:
a) Wavefront
b) Amplitude
c) Frequency
d) Speed of light

Correct Answer is option b. Amplitude

Explanation: Michelson interferometer divides the amplitude of a single light beam into two beams using a beam splitter.

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MCQ No. 6

In Young’s double slit experiment, the separation between the slits is halved and the distance between the slits and the screen is doubled. The fringe width is:
a) Halved
b) Unchanged
c) Doubled
d) Quadrupled

Correct Answer is option d. Quadrupled

Explanation:

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

If D is doubled and d is halved, then fringe width becomes four times.

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MCQ No. 7

Signal from a remote control to the device operated by it travels with the speed of:
a) Sound
b) Light
c) Ultrasonic
d) Supersonics

Correct Answer is option b. Light

Explanation: Remote controls use infrared electromagnetic waves, which travel at the speed of light.

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MCQ No. 8

Light of wavelength λ is incident normally on a diffraction grating for which the slit spacing is equal to 3λ/2. What is the sine of the angle between the second order maximum and the normal?

a) 1/6
b) 1/3
c) 2/3
d) 1

Correct Answer is option d. 1

Explanation:

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

$$\frac{3\lambda}{2} \sin \theta = 2\lambda$$$$\sin \theta = \frac{2\lambda}{3\lambda/2} = 1$$

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MCQ No. 9

Which of the following gives three regions of the electromagnetic spectrum in order of increasing wavelength, visible radiation?
a) Gamma rays, microwaves, visible radiation
b) Radio waves, ultraviolet, X-rays
c) Ultraviolet, infra-red, microwaves
d) Visible radiation, gamma rays, radio waves

Correct Answer is option c. Ultraviolet, infra-red, microwaves

Explanation: Wavelength increases in the order:
Gamma → X-rays → UV → Visible → Infrared → Microwaves → Radio waves.

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MCQ No. 10

Two monochromatic radiations X and Y are incident normally on a diffraction grating. The second order maximum for X coincides with the third order maximum for Y. What is the ratio (λₓ / λᵧ)?

a) 1/2

b) 2/3
c) 3/2
d) 2/1

Correct Answer is option c. 3/2

Explanation:

$$n_X \lambda_X = n_Y \lambda_Y$$

$$2 \lambda_X = 3 \lambda_Y$$$$\frac{\lambda_X}{\lambda_Y} = \frac{3}{2}$$

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MCQ No. 11

The tip of a needle does not give a sharp image. It is due to:
a) Polarization
b) Interference
c) Diffraction
d) Refraction

Correct Answer is option c. Diffraction

Explanation: Light bends around sharp edges due to diffraction, causing the image of very small objects to be blurred.

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CONCEPTUAL QUESTIONS

Give short response to the following questions

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1. A soap bubble looks black when it bursts, why?

Answer:

When the soap bubble bursts, the thin film of water disappears. Interference of light occurs only in thin films, so when the film vanishes, no interference occurs and the bubble appears black.

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2. What is the difference between interference and diffraction?

Answer:

Interference is produced by the superposition of light waves from two or more coherent sources, whereas diffraction is produced by the bending and spreading of light waves around obstacles or through narrow apertures.

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3. In a Michelson interferometer a second glass plate is also used, why?

Answer:

The second glass plate is used to compensate for the optical path difference introduced by the beam splitter so that both light beams travel through equal thickness of glass, ensuring accurate interference fringes.

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4. How can you explain Brewster’s law of polarization?

Answer:

Brewster’s law states that light reflected from a surface is completely polarized when the reflected and refracted rays are at right angles. It is given by:

$$\tan \theta_B = n$$

where $\theta_B$ is Brewster’s angle and $n$ is the refractive index.

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5. What is meant by path difference in interference of waves?

Answer:

Path difference is the difference in distances traveled by two interfering waves from their sources to a point of observation. It determines whether constructive or destructive interference occurs.

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6. Why is it not possible to see interference where light beams from car headlamps overlap?

Answer:
Because light from car headlamps is not coherent. Interference requires coherent sources with constant phase difference, which ordinary lamps cannot provide.

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7. A telephone pole casts a clear shadow in light from a distant headlamp, but no such effect is noticed for sound from a car horn. Why?

Answer:

Light has a very small wavelength, so diffraction around the pole is negligible and a sharp shadow is formed. Sound has a large wavelength, so it diffracts around the pole and no sharp shadow is produced.

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8. Why is it not possible to obtain diffraction of X-rays using Young’s double slit experiment?

Answer:

X-rays have extremely small wavelengths, so the slit width must be of atomic dimensions, which is not possible to achieve in Young’s double slit experiment.

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9. Can we apply Huygens’ principle to radar waves?

Answer:

Yes. Radar waves are electromagnetic waves, and Huygens’ principle applies to all types of wave motion, including electromagnetic waves.

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10. How would you justify that light waves are transverse?

Answer:

Light shows polarization, and only transverse waves can be polarized. Therefore, light waves are transverse in nature.

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COMPREHENSIVE QUESTIONS

Give extended response to the following question

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1. What is meant by the dual nature of light? Discuss the history about the nature of light in detail.

Answer:

The dual nature of light means that light exhibits both wave-like and particle-like properties. In some experiments light behaves as a wave, while in others it behaves as a stream of particles called photons.

Historically, Newton proposed the corpuscular theory of light, according to which light consists of tiny particles. This theory explained reflection and refraction but failed to explain interference and diffraction. Later, Huygens proposed the wave theory of light, which explained reflection, refraction, interference, and diffraction successfully. Young and Fresnel experimentally confirmed the wave nature of light through interference and diffraction experiments.

However, the wave theory could not explain the photoelectric effect. Einstein explained this phenomenon by proposing that light consists of discrete packets of energy called photons, thus confirming the particle nature of light. Therefore, light shows dual behavior depending on the experiment performed.

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2. Explain the diffraction of X-rays by crystals and derive Bragg’s law.

Answer:
X-rays have very small wavelengths comparable to interatomic distances in crystals. When X-rays fall on a crystal, the regularly spaced atoms act as a diffraction grating and reflect X-rays from different atomic planes. Constructive interference occurs when the path difference between rays reflected from successive planes is an integral multiple of the wavelength.

Consider two parallel crystal planes separated by distance d. X-rays strike the planes at an angle θ and are reflected. The path difference between the two reflected rays is equal to 2d sinθ. For constructive interference, this path difference must equal nλ.

Thus,

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

This equation is known as Bragg’s law, and it is used to determine the wavelength of X-rays or the spacing between crystal planes.

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3. Describe the experimental arrangement for Young’s double slit experiment and derive the expression for fringe spacing.

Answer:

In Young’s double slit experiment, a monochromatic light source illuminates a narrow slit S, which acts as a coherent source. Light from S falls on two closely spaced slits S₁ and S₂, which act as secondary coherent sources. A screen is placed at a distance D from the slits to observe interference fringes.

The path difference between waves reaching a point on the screen depends on the slit separation d and the angle θ. Constructive interference occurs when the path difference equals nλ, and destructive interference occurs when it equals (2n+1)λ/2.

The distance between successive bright or dark fringes is called fringe spacing β and is given by:

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

This shows that fringe spacing depends on wavelength, distance to the screen, and slit separation.

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4. State and explain Huygens’ principle. What is the difference between spherical and plane wavefronts?

Answer:

Huygens’ principle states that every point on a wavefront acts as a source of secondary wavelets, and the new wavefront at a later time is the envelope of these wavelets.

A spherical wavefront is produced by a point source of light, and its shape is spherical. A plane wavefront is produced by a source at an infinite distance, such as sunlight, and its shape is a plane. Plane wavefronts are parallel surfaces, while spherical wavefronts are curved.

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5. Explain the interference effect produced by thin films.

Answer:

When light falls on a thin transparent film, part of the light is reflected from the top surface and part from the bottom surface. These reflected rays interfere with each other. Depending on the thickness of the film, wavelength of light, and angle of incidence, constructive or destructive interference occurs, producing coloured patterns.

Soap bubbles and oil films on water show colourful patterns due to thin film interference. The colours depend on the film thickness and the wavelength of light.

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6. What is the principle of interference of light? Discuss the necessary conditions for interference.

Answer:

The principle of interference states that when two or more light waves overlap, the resultant intensity depends on the phase difference between them. Constructive interference occurs when waves are in phase, and destructive interference occurs when they are out of phase.

The necessary conditions for interference are that the sources must be coherent, have the same frequency, maintain constant phase difference, have comparable intensities, and their waves must overlap in space.

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7. What is a diffraction grating? How can the wavelength of light be measured with it?

Answer:
A diffraction grating is an optical device consisting of a large number of equally spaced parallel slits. When monochromatic light falls on a grating, it produces principal maxima at certain angles.

The condition for maxima is given by:

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

where d is slit spacing, θ is the angle of diffraction, n is the order of maximum, and λ is the wavelength.

By measuring θ and knowing d, the wavelength of light can be calculated.

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8. Describe the construction and working of Michelson’s interferometer and determine the wavelength of light.

Answer:

Michelson’s interferometer consists of a beam splitter, two mirrors, a compensating plate, and a viewing telescope. The beam splitter divides the light into two beams, which travel different paths and are reflected back by mirrors. These beams recombine to produce interference fringes.

When one mirror is moved, the path difference changes, and the fringes shift. By counting the number of fringes moved and the distance moved by the mirror, the wavelength of light is determined using:

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

where d is the distance moved by the mirror and N is the number of fringes shifted.

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9. What is meant by plane polarized light? How does this phenomenon prove that light is transverse?

Answer:
Plane polarized light is light in which vibrations occur in only one plane perpendicular to the direction of propagation. Polarization can be produced using Polaroid filters or reflection at Brewster’s angle.

Only transverse waves can be polarized, while longitudinal waves cannot. Since light shows polarization, it proves that light waves are transverse in nature.

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4. Numerical Problems:

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Numerical Example 1

Problem Statement:
In a Young’s double slit experiment, the separation of the slits is 1 mm and red light of wavelength 620 nm falls on it. Determine the distance between the central bright band and the fifth bright fringe on a screen 3 m away from the slits.

Given:
Slit separation, d = 1 mm = 1 × 10⁻³ m
Wavelength, λ = 620 nm = 620 × 10⁻⁹ m
Distance of screen, D = 3 m
Order, n = 5

To Find:
Distance of 5th bright fringe from centre (y₅)

Formula:
Fringe width: β = λD / d
Position of nth bright fringe: yₙ = nβ

Solution:
β = (620 × 10⁻⁹ × 3) / (1 × 10⁻³)
β = 1.86 × 10⁻³ m

y₅ = 5 × 1.86 × 10⁻³
y₅ = 9.3 × 10⁻³ m

Answer:
Distance = 9.3 mm

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Numerical Example 2

Problem Statement:

Two parallel slits are illuminated by light of two wavelengths, one of which is 5.8 × 10⁻⁷ m. The fourth dark line of the known wavelength coincides with the fifth bright line of the unknown wavelength. Find the unknown wavelength.

Given:

λ₁ = 5.8 × 10⁻⁷ m
Fourth dark: (n − 0.5)λ₁ = 3.5λ₁
Fifth bright: 5λ₂

To Find:
Unknown wavelength λ₂

Formula:
(3.5)λ₁ = 5λ₂

Solution:
λ₂ = (3.5 / 5) λ₁
λ₂ = 0.7 × 5.8 × 10⁻⁷
λ₂ = 4.06 × 10⁻⁷ m

Answer:
λ₂ = 4.06 × 10⁻⁷ m

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Numerical Example 3

Problem Statement:
In a Michelson interferometer, the movable mirror is moved 0.1 mm. How many dark fringes pass through the reference point if light of wavelength 580 nm is used?

Given:
Mirror movement, d = 0.1 mm = 1 × 10⁻⁴ m
Wavelength, λ = 580 nm = 580 × 10⁻⁹ m

To Find:
Number of fringes, N

Formula:
N = 2d / λ

Solution:
N = (2 × 1 × 10⁻⁴) / (580 × 10⁻⁹)
N = 344.8 ≈ 345

Answer:
Number of fringes = 345

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Numerical Example 4

Problem Statement:
A soap film has refractive index 1.40. How thick must the film be if it appears black for mercury light of wavelength 546.1 nm incident normally?

Given:
μ = 1.40
λ = 546.1 nm = 546.1 × 10⁻⁹ m

To Find:
Thickness t

Formula:

For thin film in reflected light, the condition for dark (black) film at normal incidence is:

$$2 \mu t = \frac{\lambda}{2}$$

So the minimum thickness is:

$$t = \frac{\lambda}{4\mu}$$

Solution:

$$t = \frac{546.1 \times 10^{-9}}{4 \times 1.40}$$

$$\mathrm{<br>}$$$$t = \frac{546.1 \times 10^{-9}}{5.6}$$

$$t = 97.52 \times 10^{-9} \text{ m}$$$$t = 9.75 \times 10^{-8} \text{ m}$$

Answer:
Thickness = 9.75 × 10⁻⁸ m

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Numerical Example 5

Problem Statement:
A diffraction grating has 5000 lines per cm. At what angle does the second order spectrum of sodium light (λ = 589 nm) occur?

Given:
Lines per cm = 5000
d = 1 / 5000 cm = 2 × 10⁻⁶ m
λ = 589 × 10⁻⁹ m
n = 2

To Find:
Angle θ

Formula:
d sinθ = nλ

Solution:
sinθ = (2 × 589 × 10⁻⁹) / (2 × 10⁻⁶)
sinθ = 0.589
θ = 36.1°

Answer:
θ = 36.1°

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Numerical Example 6

Problem Statement:
Light is incident normally on a grating with 250 lines/mm. Find the wavelength for which the second order deviation is 12°.

Given:
Lines/mm = 250
d = 1/250 mm = 4 × 10⁻⁶ m
n = 2
θ = 12°

To Find:
λ

Formula:
d sinθ = nλ

Solution:
λ = d sinθ / n
λ = (4 × 10⁻⁶ × sin12°) / 2
λ = 4.16 × 10⁻⁷ m

Answer:
λ = 4.16 × 10⁻⁷ m

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Numerical Example 7

Problem Statement:
In an X-ray diffraction experiment, first order image is observed at 5° for crystal spacing 2.8 × 10⁻¹⁰ m. Find the wavelength.

Given:
d = 2.8 × 10⁻¹⁰ m
θ = 5°
n = 1

To Find:
λ

Formula:
nλ = 2d sinθ

Solution:
λ = 2 × 2.8 × 10⁻¹⁰ × sin5°
λ = 4.88 × 10⁻¹¹ m

Answer:
λ = 4.88 × 10⁻¹¹ m

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Numerical Example 8

Problem Statement:
An X-ray beam of wavelength 0.48 × 10⁻¹⁰ m gives Bragg reflection at 20° for first order. Find the crystal plane spacing.

Given:
λ = 0.48 × 10⁻¹⁰ m
θ = 20°
n = 1

To Find:
d

Formula:
d = nλ / (2 sinθ)

Solution:
d = (0.48 × 10⁻¹⁰) / (2 × sin20°)
d = 7.02 × 10⁻¹¹ m

Answer:
d = 7.02 × 10⁻¹¹ m

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Numerical Example 9

Problem Statement:
Crystal plane spacing in NaCl is 0.282 nm. First order Bragg maximum occurs at 7°. Find the wavelength.

Given:
d = 0.282 nm = 0.282 × 10⁻⁹ m
θ = 7°
n = 1

To Find:
λ

Formula:
λ = 2d sinθ

Solution:
λ = 2 × 0.282 × 10⁻⁹ × sin7°
λ = 6.87 × 10⁻¹¹ m

Answer:

λ = 6.87 × 10⁻¹¹ m 

λ = 0.0687 nm 

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Click on the link below for complete Class 11 Physics notes, MCQs, Q&A, and Numericals.

Class 11 Physics – Unit-wise MCQs & Study Resources (Main-Page)

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