NCERT Solutions for Class 12 Physics Chapter 10 Wave Optics

NCERT Solutions for Class 12 Physics Chapter 10 Wave Optics is one of the most interesting and concept-based chapters in the CBSE Class 12 Physics Syllabus. This class 12 chapter helps students understand how light behave as a wave and explain phenomena like interference, diffraction, and polarization. The ncert solutions class 12 chapter 10 physics provided here are designed in a simple and step-by-step way so that learners can easily grasp complex idea of wave nature of light.

In wave optics class 12 ncert, we study how brightness and darkness patterns are formed due to the superposition of light waves. These ncert solutions class 12 physics wave optics are prepared by expert teachers following the current CBSE Class 12 Board Exam pattern to ensure that students score well and understand the main concepts deeply.

By using these NCERT Solutions for class 12 physics, students can clear their doubts, revise more effectively, and improve problem-solving skills in physics.

Wave Optics Class 12 Physics Chapter 10 PDF Download

Get Wave Optics Class 12 Physics Chapter 10 PDF Download with step-by-step NCERT Solutions for Class 12 Physics, covering Huygens’ principle, interference, diffraction, and polarization as per the latest CBSE syllabus for the CBSE Class 12 Board Exam, includes exercise answers, exemplars, and quick revision notes for fast practice and high‑score prep in Wave Optics Chapter 10.

Wave Optics Class 12 Questions and Answers

Ques: Monochromatic light of wavelength 600 nm is incident from air onto water (ν = 1.33). Find (a) the frequency, (b) the wavelength and speed of (i) reflected and (ii) refracted light.

Solution:

• (a) Frequency:

f = c/λ = (3 × 10⁸)/(600 × 10^-9) = 5 × 10¹⁴ Hz

• (b) Reflected light:

Speed and wavelength remain same as in air:

v_air = c = 3 × 10⁸ m/s

λ_air = 600 nm

Refracted light:

Speed in water: v = c/ν = (3 × 10⁸)/1.33 = 2.26 × 10⁸ m/s

λ_water = v/f = (2.26 × 10⁸)/(5 × 10¹⁴) = 452 nm

Ques: In YDSE, the distance between slits is 0.5 mm, screen is 1 m away, wavelength 500 nm. Find fringe width.

Solution:

Fringe width: β = λD/d

β = (500 × 10^-9 × 1)/(0.5 × 10^-3)

β = 1 × 10^-3 m = 1 mm

Ques: In YDSE, distance between two bright fringes is 1 mm and separation between slits is 0.2 mm. Distance between slits and screen is 1.2 m. Find wavelength.

Solution:

β = λD/d ⇒ λ = βd/D

λ = (1 × 10^-3 × 0.2 × 10^-3)/1.2

λ = 1.67 × 10^-7 m = 167 nm

Ques: A slit of width 0.2 mm is illuminated with wavelength 600 nm. At what angle does the first minimum appear?

Solution:

Single-slit minimum condition: a sinθ = nλ

For first minimum, n = 1:

sinθ = λ/a = (600 × 10^-9)/(0.2 × 10^-3) = 3 × 10^-3

θ ≈ 0.17°

Ques: A parallel beam of wavelength 500 nm falls on a slit and first minimum is at 2.5 mm from center on screen 1 m away. Find the slit width.

Solution:

Using: a = λD/x

a = (500 × 10^-9 × 1)/(2.5 × 10^-3)

a = 2 × 10^-4 m = 0.2 mm

Ques: Calculate Brewster's angle for light passing from air (ν = 1) into glass (ν = 1.5).

Solution:

tanθ_B = ν

θ_B = tan^-1(1.5)

θ_B = 56.3°

Ques: Find resolving power of a microscope using light of wavelength 450 nm, if numerical aperture is 1.4.

Solution:

Resolving power = 1/(0.61λ/NA) = NA/(0.61λ)

Resolving power = 1.4/(0.61 × 450 × 10^-9)

Resolving power = 5.15 × 10⁶ per meter

Ques: In YDSE, if path difference between slits becomes 2λ, what happens to central fringe?

Solution:

The central fringe shifts to the position corresponding to the new path difference.

Path difference = 2λ corresponds to the 2nd bright fringe position

Therefore, the central fringe moves by a distance equal to two fringe widths (2β).

Ques: Two wavelengths 640 nm and 590 nm are used in YDSE. Find the least order where bright fringes coincide.

Solution:

Coincidence occurs when: n₁λ₁ = n₂λ₂

n₁/n₂ = λ₂/λ₁ = 590/640 = 59/64

Therefore, n₁ = 59 and n₂ = 64

The 59th bright fringe of 640 nm coincides with 64th bright fringe of 590 nm.

Ques: In YDSE, if screen is moved twice further from slits, how does fringe width change?

Solution:

Fringe width: β = λD/d

Since β ∝ D (directly proportional to screen distance)

When D becomes 2D, fringe width becomes 2β

Answer: Fringe width doubles

Ques: Calculate the path difference for constructive interference in YDSE.

Solution:

For constructive interference (bright fringe):

Path difference Δx = nλ

where n = 0, ±1, ±2, ±3, ... (integer values)

n = 0 gives central bright fringe

n = 1 gives 1st order bright fringe, and so on.

Ques: Calculate speed of light in glass (ν = 1.5), using c = 3 × 10⁸ m/s.

Solution:

Refractive index: ν = c/v

v = c/ν = (3 × 10⁸)/1.5

v = 2 × 10⁸ m/s

Ques: If one slit in YDSE is covered with thin transparent sheet of thickness t and refractive index n, find fringe shift.

Solution:

Additional path difference introduced = (n - 1)t

Fringe shift = [t(n - 1)]/λ

This represents the number of fringes by which the pattern shifts.

Linear fringe shift = [t(n - 1)D]/(λd)

Ques: Light of wavelength 6000 Å is incident on a grating with 5000 lines per cm. Find the angle for first order maxima.

Solution:

Grating element: d = 1/5000 cm = 2 × 10^-6 m

Grating equation: d sinθ = nλ

For first order (n = 1):

sinθ = λ/d = (6000 × 10^-10)/(2 × 10^-6) = 0.3

θ = sin^-1(0.3) = 17.46°

Ques: Single slit of width 0.1 mm is illuminated with wavelength 650 nm. Find total number of minima on either side of central maximum.

Solution:

Condition for minima: a sinθ = nλ

Maximum value of sinθ = 1

n_max = a/λ = (0.1 × 10^-3)/(650 × 10^-9) = 153.8

Therefore, n_max = 153

Total minima on both sides = 2 × 153 = 306

Ques: First minima appears at 2.5 mm from center, with slit width 0.15 mm and screen 1 m away. Find wavelength.

Solution:

For first minima: x = λD/a

λ = xa/D = (2.5 × 10^-3 × 0.15 × 10^-3)/1

λ = 3.75 × 10^-7 m = 375 nm

Question:

Single slit width 0.2 mm, λ = 400 nm. Find angular width of central maximum.

Solution:

Angular width of central maximum = 2θ

For first minimum: sinθ = λ/a

For small angles: θ ≈ λ/a

Angular width = 2λ/a = (2 × 400 × 10^-9)/(0.2 × 10^-3)

Angular width = 4 × 10^-3 radians = 0.229°

Ques: A grating with 6000 lines per cm is illuminated with light of wavelength 5000 Å. Find maximum possible order.

Solution:

Grating element: d = 1/6000 cm = 1.67 × 10^-6 m

For maximum order: d sinθ = nλ, where sinθ_max = 1

n_max = d/λ = (1.67 × 10^-6)/(5000 × 10^-10) = 3.34

Maximum order n = 3

Ques: In 2-source interference, if sources are separated by d, screen distance D, find path difference at point x on screen.

Solution:

Path difference: δ = xd/D

where:

x = distance of point from central axis

d = separation between sources

D = distance from sources to screen

This formula is valid for D >> d

Ques: In YDSE, find the position of 4th bright fringe if d = 0.3 mm, D = 1 m, λ = 600 nm.

Solution:

Position of nth bright fringe: x_n = nλD/d

For n = 4:

x₄ = (4 × 600 × 10^-9 × 1)/(0.3 × 10^-3)

x₄ = 8 × 10^-3 m = 8 mm

Ques: How does increasing individual slit width affect fringe width in YDSE?

Solution:

Fringe width: β = λD/d

where d is the separation between the two slits (not individual slit width)

Effect: Increasing individual slit width does NOT affect fringe width

However, it increases the intensity of fringes and reduces sharpness of minima.

Ques: In two-slit interference, intensity ratio of individual sources is 4:1. Find ratio of intensities at maxima and minima.

Solution:

Let I₁ = 4I and I₂ = I

I_max = (√I₁ + √I₂)² = (√4I + √I)² = (2√I + √I)² = 9I

I_min = (√I₁ - √I₂)² = (√4I - √I)² = (2√I - √I)² = I

I_max : I_min = 9:1

Ques: A microscope uses λ = 520 nm. If switched to λ = 400 nm, how does resolving power change?

Solution:

Resolving power ∝ 1/λ

RP₂/RP₁ = λ₁/λ₂ = 520/400 = 1.3

Resolving power increases by factor of 1.3 (30% increase)

Ques: Slit width 0.25 mm, λ = 600 nm. Find angle for second minimum.

Solution:

For second minimum (n = 2): a sinθ = 2λ

sinθ = 2λ/a = (2 × 600 × 10^-9)/(0.25 × 10^-3)

sinθ = 4.8 × 10^-3

θ = sin^-1(4.8 × 10^-3) ≈ 0.275°

Ques: Single slit, width = 0.2 mm, λ = 500 nm, screen 1 m away. Find linear width of central maximum.

Solution:

Linear width of central maximum = 2λD/a

Width = (2 × 500 × 10^-9 × 1)/(0.2 × 10^-3)

Width = 5 × 10^-3 m = 5 mm

Ques: Unpolarized light of intensity I₀ passes through a polarizer. Find transmitted intensity when analyzer is at θ = 45° from polarizer axis.

Solution:

After first polarizer, intensity = I₀/2

Malus Law: I = I₀cos²θ

After analyzer: I = (I₀/2) × cos²(45°)

I = (I₀/2) × (1/√2)² = (I₀/2) × (1/2) = I₀/4

Ques: Telescope of diameter 15 cm uses light of wavelength 550 nm. Find minimum angular separation (resolving limit).

Solution:

Rayleigh's criterion: θ = 1.22λ/D

θ = (1.22 × 550 × 10^-9)/0.15

θ = 4.47 × 10^-6 radians

θ ≈ 0.92 arc seconds

Ques: In YDSE, if λ changes from 600 nm to 500 nm, how does fringe width change (d = 0.5 mm, D = 1 m)?

Solution:

β₁ = λ₁D/d = (600 × 10^-9 × 1)/(0.5 × 10^-3) = 1.2 mm

β₂ = λ₂D/d = (500 × 10^-9 × 1)/(0.5 × 10^-3) = 1.0 mm

Fringe width decreases from 1.2 mm to 1.0 mm

Ques: Find difference in fringe width for λ = 650 nm and λ = 500 nm with d = 0.25 mm, D = 1 m.

Solution:

β₁ = λ₁D/d = (650 × 10^-9 × 1)/(0.25 × 10^-3) = 2.6 mm

β₂ = λ₂D/d = (500 × 10^-9 × 1)/(0.25 × 10^-3) = 2.0 mm

Difference = β₁ - β₂ = 2.6 - 2.0 = 0.6 mm

Ques: Find angle between two polarizers at which intensity becomes zero.

Solution:

Using Malus Law: I = I₀cos²θ

For I = 0:

cos²θ = 0

cosθ = 0

θ = 90°