Class 11 Physics Unit 9: Physical Optics – Important Questions & Answers with Solved Numericals.

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This post covers all important topics of Physical Optics, including light waves in the electromagnetic spectrum, wavefronts, Huygens’ principle, interference of light, Young’s double slit experiment, thin film interference, Michelson interferometer, diffraction (single slit, diffraction grating, X-ray diffraction), and polarization.

Perfect for exam preparation, quick revision, board exams, and competitive test practice, this comprehensive set of questions helps students strengthen conceptual understanding and problem-solving skills.

Learn and master all key concepts with step-by-step numerical solutions, clear explanations, and concise answers, making it ideal for students aiming for high marks and full mastery of Unit-9: Physical Optics.

SHORT QUESTIONS WITH BRIEF ANSWERS (Unit-7 Oscillation)

Part A: Conceptual Questions

Q1. Describe light waves as part of the electromagnetic spectrum.
Answer: Light waves are electromagnetic waves with wavelengths in the range of ~400–700 nm. They propagate as transverse waves, consisting of oscillating electric and magnetic fields perpendicular to each other and to the direction of propagation. Light is a part of the EM spectrum, lying between infrared and ultraviolet.

Q2. Explain the concept of wavefront.
Answer: A wavefront is an imaginary surface connecting all points of a wave that have the same phase. Types include:

Plane wavefront – far from point source.
Spherical wavefront – from point source.
Cylindrical wavefront – from line source.

Wavefronts help in understanding propagation, reflection, refraction, and diffraction.

Q3. State Huygens’ principle and explain its application.
Answer: Huygens’ Principle: Every point on a wavefront acts as a source of secondary spherical wavelets, and the new wavefront is the tangent to these secondary wavelets after a time interval.

Application: Explains reflection, refraction, and diffraction of waves.

Q4. State necessary conditions for interference of light.
Answer: For visible interference:

Sources must be coherent (constant phase difference).
Light must be monochromatic (same wavelength).
Amplitudes of interfering waves should be comparable.

Q5. Describe Young’s double slit experiment and its significance.
Answer: In YDSE, coherent light falls on two slits producing overlapping waves on a screen. Bright and dark fringes form due to constructive and destructive interference.

Evidence: Shows light behaves as a wave, as particles cannot produce stable fringes.

Q6. Explain the color pattern due to interference in thin films.
Answer: When light reflects from a thin film (like oil), partial reflection at the top surface and partial transmission/reflection at the bottom surface create two waves. Their path difference causes constructive/destructive interference. Variation with wavelength gives colorful patterns.

Q7. Describe the parts and working of Michelson interferometer.
Answer:
Parts: Source, beam splitter, two mirrors, movable mirror, screen.
Working: Light is split, reflected, recombined. Path difference produces interference fringes.
Uses: Measuring wavelength, small distances, refractive index of gases, precision metrology.

Q8. Explain diffraction and how it provides evidence for wave nature of light.
Answer: Diffraction is the bending of waves around edges or through narrow slits. Only waves exhibit diffraction; particle-like light cannot spread into geometrical shadows, proving light’s wave nature.

Q9. Explain polarization of light.
Answer: Polarization restricts the vibrations of light to a single plane. It is produced by Polaroids, reflection at Brewster’s angle, or double refraction. Malus’ law describes intensity variation: $I = I_0 \cos^2 \theta$ .

LONG / COMPREHENSIVE QUESTIONS WITH BRIEF EXAM ANSWERS

Q1a. Explain Huygens’ Principle and illustrate how it is used to construct a new wavefront.

Answer:

Huygens’ Principle states that every point on a wavefront acts as a source of secondary spherical wavelets. The new wavefront at any later time is the surface tangent to these wavelets. By marking points on the existing wavefront, drawing secondary wavelets from each point, and connecting the tangents of these wavelets, the new position of the wavefront is obtained. This principle helps to explain reflection, refraction, and diffraction of light, showing how waves propagate in space.

Q1. Explain Huygens’ Principle and construct a new wavefront after a time interval.

Answer:
Huygens’ Principle: Every point on a wavefront acts as a source of secondary spherical wavelets. The new wavefront at a later time is the tangent to these wavelets.

Construction Steps:

Draw the initial wavefront (plane or spherical).
Mark points on the wavefront.
Draw circles with radius = v × Δt around each point (v = wave speed, Δt = time interval).
Draw a tangent to all circles → this is the new wavefront.

Significance: Explains reflection, refraction, and diffraction.Q1b. Explain Huygens’ Principle and construct a new wavefront after a time interval.

Q2a. State the necessary conditions for interference of light. Explain Young’s Double Slit Experiment (YDSE) and its significance.

Answer:

Conditions for interference:

The light sources must be coherent (same frequency and constant phase difference).
The light must be monochromatic.
The amplitudes of interfering waves should be comparable.

Young’s Double Slit Experiment: In YDSE, a monochromatic coherent light source illuminates two narrow slits. Light from these slits overlaps on a screen, forming bright and dark fringes due to constructive and destructive interference. Bright fringes occur where the path difference is an integer multiple of the wavelength, while dark fringes occur at half-integer multiples.

Significance: YDSE provides direct evidence of the wave nature of light, as only waves can produce stable interference fringes.

Q2b. State the necessary conditions for interference of light and explain Young’s Double Slit Experiment (YDSE). Include a numerical calculation.

Answer:
Conditions:

Sources must be coherent (constant phase difference).
Light must be monochromatic.
Amplitudes should be comparable.

YDSE Explanation:

Coherent light passes through two narrow slits.
Overlapping waves on the screen form bright and dark fringes.
Bright fringe: Δ = mλ
Dark fringe: Δ = (m + ½)λ
Fringe width: β = λD/d

Q3a. Describe the interference of light in thin films and explain the origin of coloured patterns.

Answer:

When light reflects from a thin film (like an oil layer on water), part of the light reflects from the top surface and part from the bottom surface. The two reflected waves interfere depending on their path difference and any phase change at reflection. Constructive interference produces bright fringes, and destructive interference produces dark fringes.

The colour patterns arise because different wavelengths interfere differently, producing a spectrum of colours. The effect is most pronounced when the film thickness is comparable to the wavelength of light, leading to vivid coloured patterns seen in soap bubbles, oil films, or anti-reflective coatings.

Q3b. Explain colour patterns in thin film interference and calculate the minimum thickness for constructive reflection.

Answer:

Light reflects from the top and bottom surfaces of a thin film.
Path difference + phase difference → constructive or destructive interference.
Different wavelengths interfere differently → color patterns.

Q4. Describe the Michelson Interferometer and explain its working and applications.

Answer:

A Michelson Interferometer consists of a monochromatic light source, a beam splitter, two mirrors (one movable), and a screen. Light from the source is split into two beams by the beam splitter. Each beam travels a different path, reflects from the mirrors, and recombines at the beam splitter. The recombined light produces interference fringes, which shift when one mirror is moved.

Applications:

Measuring the wavelength of light.
Determining small distances precisely.
Measuring the refractive index of gases.
High-precision metrology and experiments in physics.

The device demonstrates wave interference and precision measurement.

Q5. Explain diffraction of light at a single narrow slit and discuss its significance.

Answer:

Diffraction is the bending of light waves when they pass through a narrow slit or around an obstacle. In single-slit diffraction, a central bright maximum appears on the screen, flanked by dimmer maxima separated by dark minima. The angular width of the central maximum depends on the wavelength of light and the width of the slit.

Significance:

Diffraction provides direct evidence of the wave nature of light.
It limits the resolving power of optical instruments like microscopes and telescopes.
Explains phenomena such as spreading of waves around edges and the formation of diffraction patterns.

Q6. Explain the use of a diffraction grating to determine the wavelength of light.

Answer:

A diffraction grating has many equally spaced parallel lines. When monochromatic light is incident on a grating, light diffracts at specific angles where constructive interference occurs. The grating equation,

d sin θ = nλ

, relates the grating spacing

d

, diffraction angle

θ

, diffraction order

n

, and wavelength

λ

. By measuring the angle of maxima, the wavelength of light can be accurately determined.

Diffraction gratings are widely used in spectroscopy, wavelength measurement, and optical instruments.

Q7. Explain polarization of light and Malus’ Law. Describe methods to produce and detect plane-polarized light.

Answer:
Polarization restricts light vibrations to a single plane. It is characteristic of transverse waves. Plane-polarized light can be produced using:

Polaroid filters, which allow vibrations in one direction.
Reflection at Brewster’s angle, where reflected light is polarized perpendicular to the plane of incidence.
Double refraction in certain crystals like calcite.

Malus’ Law: When plane-polarized light passes through an analyser at an angle $\theta$ to the transmission axis, the transmitted intensity is $I = I_0 \cos^2 \theta$ , where $I_0$ is the initial intensity.

Detection: Rotating a Polaroid analyser and observing the variation in intensity allows detection and analysis of plane-polarized light.

Q8. Differentiate interference and diffraction and explain how both phenomena demonstrate the wave nature of light.

Answer:

Interference occurs when two or more coherent waves overlap, producing regions of constructive and destructive interference (e.g., YDSE).
Diffraction is the bending of waves around obstacles or through narrow openings, producing a spreading pattern (e.g., single slit diffraction).

Both phenomena rely on superposition of waves. The appearance of fringes or patterns cannot be explained by particle theories of light, thus providing direct evidence for the wave nature of light.

Q9. Explain the colour patterns observed in Newton’s rings and their significance.

Answer:

Newton’s rings are formed by interference between light reflected from a convex lens placed on a flat glass surface. A thin air film forms between the surfaces, and the path difference between reflected rays causes circular interference fringes.

Bright rings: constructive interference.
Dark rings: destructive interference.

Significance:

Provides experimental evidence for wave interference.
Can be used to measure the radius of curvature of lenses.
Demonstrates thin film interference in circular geometry.

Q10. Discuss the applications of Michelson interferometer in precision measurements and the determination of small distances.

Answer:

The Michelson interferometer splits light into two beams that traverse different paths and recombine, producing interference fringes. By carefully moving a mirror and counting the number of fringe shifts:

Wavelength of light can be measured accurately.
Small displacements or changes in length can be determined with high precision.
It is used in refractive index measurements, detection of gravitational waves, and metrology.

The interferometer demonstrates how wave interference can be used for precise quantitative measurements.

Miscellaneous Numerical Problems with Proper Solutions

Numerical 1: Young’s Double Slit Experiment

Question: Calculate the fringe width on the screen if monochromatic light of wavelength 600 nm passes through two slits separated by 0.25 mm, and the screen is placed at a distance of 2 m.

Given:
λ = 600 nm = 600 × 10⁻⁹ m
d = 0.25 mm = 0.25 × 10⁻³ m
D = 2 m

To Find: Fringe width β

Solution:

\beta = \frac{\lambda D}{d} = \frac{600 \times 10^{-9} \times 2}{0.25 \times 10^{-3}} = 4.8 \text{ mm}

Answer: β = 4.8 mm

Numerical 2: Thin Film Interference

Question: Find the minimum thickness of a thin film of refractive index 1.5 for which reflected light of wavelength 600 nm gives a bright fringe.

Given:
λ = 600 nm
n = 1.5
First bright fringe (m = 0)

To Find: Thickness t

Solution:
Constructive interference for reflected light (with phase change at top surface):

2nt = (m + \frac{1}{2})\lambda

t = \frac{\lambda}{4n} = \frac{600}{4 \times 1.5} = 100 \text{ nm}

Answer: t = 100 nm

Numerical 3: Michelson Interferometer

Question: A movable mirror of a Michelson interferometer is shifted by 0.2 mm, producing 800 fringe shifts. Determine the wavelength of light used.

Given:
ΔL = 0.2 mm = 0.2 × 10⁻³ m
N = 800

To Find: Wavelength λ

Solution:

\lambda = \frac{2 \Delta L}{N} = \frac{2 \times 0.2 \times 10^{-3}}{800} = 5 \times 10^{-7} \text{ m}

Answer: λ = 500 nm

Numerical 4: Single Slit Diffraction

Question: A slit of width 0.2 mm is illuminated by light of wavelength 500 nm. Find the angular width of the central diffraction maximum.

Given:
a = 0.2 mm = 0.2 × 10⁻³ m
λ = 500 nm = 500 × 10⁻⁹ m

To Find: Angular width θ

Solution:

\theta \approx \frac{\lambda}{a} = \frac{500 \times 10^{-9}}{0.2 \times 10^{-3}} = 0.0025 \text{ rad}

\theta \approx 0.14^\circ

#Answer: θ ≈ 0.14°

Numerical 5: Diffraction Grating

Question: A diffraction grating has 5000 lines/cm. First-order maximum of monochromatic light is observed at an angle of 17.5°. Determine the wavelength of light.

Given:
N = 5000 lines/cm = 5 × 10⁵ lines/m
θ = 17.5°
n = 1

To Find: Wavelength λ

Solution:

d = \frac{1}{N} = \frac{1}{5 \times 10^5} = 2 \times 10^{-6} \text{ m}

\lambda = d \sin \theta = 2 \times 10^{-6} \times \sin 17.5^\circ

\lambda \approx 2 \times 10^{-6} \times 0.300 = 6 \times 10^{-7} \text{ m}

Answer: λ ≈ 600 nm

Numerical 6: Plane Polarized Light through Polaroid

Question: Plane-polarized light of intensity 200 W/m² passes through a Polaroid analyzer at 45°. Find the transmitted intensity.

Given:
I₀ = 200 W/m²
θ = 45°

To Find: Transmitted intensity I

Solution:

I = I_0 \cos^2 \theta = 200 \times \cos^2 45^\circ = 200 \times 0.5 = 100 \text{ W/m²}

Answer: I = 100 W/m²

Numerical 7: YDSE Path Difference

Question: In Young’s double slit experiment, the distance between the slits is 0.3 mm, the screen is 1.5 m away, and the wavelength of light is 500 nm. Calculate the path difference for the 3rd bright fringe from the central maximum.

Given:
d = 0.3 mm = 0.3 × 10⁻³ m
D = 1.5 m
λ = 500 nm = 500 × 10⁻⁹ m
m = 3

To Find: Path difference Δ

Solution:

\Delta = m \lambda = 3 \times 500 \text{ nm} = 1500 \text{ nm} = 1.5 \times 10^{-6} \text{ m}

Answer: Δ = 1.5 × 10⁻⁶ m

Numerical 8: Newton’s Rings

Question: A plano-convex lens of radius 50 cm is placed on a flat glass plate and illuminated with light of wavelength 600 nm. Calculate the radius of the 5th dark ring.

Given:
R = 50 cm = 0.5 m
λ = 600 nm = 6 × 10⁻⁷ m
m = 5

To Find: Radius of dark ring r₅

Solution:
For dark rings: $r_m^2 = m \lambda R$

r_5^2 = 5 \times 6 \times 10^{-7} \times 0.5 = 1.5 \times 10^{-6}

r_5 = \sqrt{1.5 \times 10^{-6}} \approx 0.001225 \text{ m} \approx 1.225 \text{ mm}

Answer: r₅ ≈ 1.225 mm

Numerical 9: Minimum Thickness for Destructive Interference

Question: Find the minimum thickness of a thin film of refractive index 1.33 in air so that reflected light of wavelength 500 nm produces a dark fringe.

Given:
λ = 500 nm
n = 1.33

To Find: Minimum thickness t

Solution:
For destructive interference (with phase change at top surface):

2 n t = m \lambda, \text{ m = 0 for minimum}

t = \frac{\lambda}{2 n} = \frac{500}{2 \times 1.33} \approx 188 \text{ nm}

Answer: t ≈ 188 nm

Numerical 10: Multiple Order Diffraction

Question: A diffraction grating of 10,000 lines/cm is used. Calculate the angle for the second-order maximum of light of wavelength 500 nm.

Given:
N = 10,000 lines/cm = 1 × 10⁶ lines/m
n = 2
λ = 500 nm = 5 × 10⁻⁷ m

To Find: Angle θ

Solution:

d = \frac{1}{N} = \frac{1}{1 \times 10^6} = 1 \times 10^{-6} \text{ m}

\sin \theta = \frac{n \lambda}{d} = \frac{2 \times 5 \times 10^{-7}}{1 \times 10^{-6}} = 1

Since sin θ = 1, θ = 90° → maximum is at grazing angle; higher order not observed.

Answer: Second-order maximum not possible (θ = 90°).

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)