Class 12 Physics Notes – Chapter 10: Wave Optics
1. Introduction & Huygens Principle
In 1678, the Dutch physicist Christiaan Huygens put forward the wave theory of light. According to this theory, light travels in the form of waves.
Wavefront
A wavefront is defined as the locus of all points in a medium which oscillate in the same phase. The speed with which the wavefront moves outwards from the source is called the speed of the wave.
- Spherical Wavefront: Formed by a point source emitting waves uniformly in all directions.
- Plane Wavefront: At a large distance from a source, a small portion of a spherical wavefront appears plane.
Huygens Principle
It is essentially a geometrical construction to determine the shape of a wavefront at any later time:
- Each point of a wavefront is the source of a secondary disturbance, and the wavelets emanating from these points spread out in all directions with the speed of the wave.
- The new wavefront at any later time is given by the forward envelope (common tangent) of these secondary wavelets.
[Place Image Here: Diagram showing spherical and plane wavefronts and Huygens geometrical construction]
2. Refraction and Reflection of Plane Waves
Using Huygens principle, we can derive the laws of reflection and refraction.
Refraction (Snell’s Law)
When a plane wavefront is incident on a surface separating two media, it bends. If v1 and v2 are the speeds of light in medium 1 and 2 respectively, then:
Formula: (sin i) / (sin r) = v1 / v2 = n2 / n1
This proves Snell’s law of refraction. The wave theory correctly predicts that if light bends towards the normal (r < i), the speed of light in the second medium must be less than that in the first medium (v2 < v1).
Reflection
For reflection at a plane surface, the angle of incidence is equal to the angle of reflection: i = r.
3. Interference of Light Waves & Young’s Experiment
When two light waves superimpose, the resultant intensity is not simply the sum of individual intensities. This phenomenon is called Interference.
Coherent Sources
Two sources are said to be coherent if they emit light waves of the same frequency and have a constant phase difference. Interference patterns can only be stably observed with coherent sources.
Conditions for Interference
- Constructive Interference (Bright Fringes): Path difference = nλ (where n = 0, 1, 2, …)
- Destructive Interference (Dark Fringes): Path difference = (n + 1/2)λ
Young’s Double Slit Experiment (YDSE)
Thomas Young demonstrated interference using two narrow slits S1 and S2 illuminated by a single monochromatic source. The resulting pattern on the screen consists of alternate bright and dark fringes.
Fringe width (width of a dark or bright fringe) is given by: β = λ D / d
Where λ is wavelength, D is the distance to the screen, and d is the distance between the two slits.
[Place Image Here: Diagram of Young’s Double Slit Experiment showing interference fringes on a screen]
4. Diffraction
The phenomenon of bending of light around the corners of an obstacle or aperture and its encroachment into the region of geometrical shadow is called diffraction.
Single Slit Diffraction
When a monochromatic plane wave falls on a single slit of width ‘a’, the diffracted light produces a central bright maximum surrounded by alternate dark and bright secondary fringes.
- Condition for Minima (Dark fringes): a sinθ = nλ (n = ±1, ±2, …)
- The central maximum is the brightest and twice as wide as the other secondary maxima.
5. Polarisation
Light is a transverse electromagnetic wave. In ordinary (unpolarised) light, the electric field vector oscillates randomly in all directions perpendicular to the direction of propagation. If the electric field vector oscillates only in one specific plane, the light is called linearly polarised.
Malus’ Law
When completely plane-polarised light is incident on an analyzer (polaroid), the intensity I of the transmitted light varies directly as the square of the cosine of the angle θ between the pass-axis of the polarizer and the analyzer.
Formula: I = I0 cos2θ
Multiple Choice Questions (MCQs)
1. According to Huygens’ principle, light is a:
a) Particle
b) Wave
c) Both particle and wave
d) None of the above
Answer: (b) Wave
2. The locus of all particles of a medium vibrating in the same phase is called:
a) Wavelet
b) Ray
c) Wavefront
d) Pulse
Answer: (c) Wavefront
3. Which phenomenon proved that light waves are transverse in nature?
a) Interference
b) Diffraction
c) Polarisation
d) Refraction
Answer: (c) Polarisation
4. For constructive interference, the path difference between two waves must be:
a) nλ
b) (n + 1/2)λ
c) (2n + 1)λ
d) λ / 2
Answer: (a) nλ
5. In Young’s double slit experiment, if the distance between the slits (d) is halved, the fringe width will:
a) Be halved
b) Remain same
c) Be doubled
d) Become four times
Answer: (c) Be doubled
6. The central fringe in a single slit diffraction pattern is:
a) Dark
b) Bright and narrow
c) Bright and twice as wide as secondary maxima
d) Coloured
Answer: (c) Bright and twice as wide as secondary maxima
7. Two sources of light are said to be coherent if they produce waves of:
a) Equal intensity
b) Equal speed
c) Same frequency and constant phase difference
d) Same wavelength and varying phase difference
Answer: (c) Same frequency and constant phase difference
8. According to Malus’ Law, the transmitted intensity I through a polaroid is proportional to:
a) cos θ
b) cos2 θ
c) sin θ
d) sin2 θ
Answer: (b) cos2 θ
9. When unpolarised light of intensity I0 passes through a polaroid, the intensity of transmitted light is:
a) I0
b) I0 / 2
c) I0 / 4
d) Zero
Answer: (b) I0 / 2
10. The wave theory of light could NOT explain:
a) Refraction
b) Interference
c) Diffraction
d) Photoelectric effect
Answer: (d) Photoelectric effect
Textbook Exercise Solutions
Q 10.1: Monochromatic light of wavelength 589 nm is incident from air on a water surface. What are the wavelength, frequency and speed of (a) reflected, and (b) refracted light? Refractive index of water is 1.33.
Solution:
Given: λ = 589 nm = 589 × 10-9 m, c = 3 × 108 m/s, n = 1.33.
(a) For reflected light: The medium remains air, so speed and wavelength do not change.
Speed v = 3 × 108 m/s.
Wavelength λ = 589 nm.
Frequency f = c / λ = (3 × 108) / (589 × 10-9) = 5.09 × 1014 Hz.
(b) For refracted light: The frequency of light depends only on the source, so it remains unchanged.
Frequency f = 5.09 × 1014 Hz.
Speed in water v = c / n = (3 × 108) / 1.33 = 2.26 × 108 m/s.
Wavelength in water λ’ = λ / n = 589 / 1.33 = 443 nm.
Q 10.2: What is the shape of the wavefront in each of the following cases:
(a) Light diverging from a point source.
(b) Light emerging out of a convex lens when a point source is placed at its focus.
(c) The portion of the wavefront of light from a distant star intercepted by the Earth.
Solution:
(a) The wavefront is Spherical.
(b) The wavefront is Plane (since rays become parallel after passing through the lens).
(c) The wavefront is Plane (as the star is at a very large distance, a small portion of its spherical wavefront appears flat).
Q 10.3: (a) The refractive index of glass is 1.5. What is the speed of light in glass? (Speed of light in vacuum is 3.0 × 108 m s-1).
(b) Is the speed of light in glass independent of the colour of light? If not, which of the two colours red and violet travels slower in a glass prism?
Solution:
(a) Speed in glass v = c / n = (3.0 × 108) / 1.5 = 2.0 × 108 m/s.
(b) No, the speed of light in glass depends on the colour (wavelength) of light. The refractive index of glass is higher for violet light than for red light. Hence, violet light travels slower than red light in a glass prism.
Q 10.4: In a Young’s double-slit experiment, the slits are separated by 0.28 mm and the screen is placed 1.4 m away. The distance between the central bright fringe and the fourth bright fringe is measured to be 1.2 cm. Determine the wavelength of light used in the experiment.
Solution:
Given: d = 0.28 mm = 0.28 × 10-3 m, D = 1.4 m, n = 4, y4 = 1.2 cm = 1.2 × 10-2 m.
The distance of the nth bright fringe from the centre is yn = n λ D / d.
1.2 × 10-2 = 4 × λ × 1.4 / (0.28 × 10-3)
λ = (1.2 × 10-2 × 0.28 × 10-3) / (4 × 1.4)
λ = 3.36 × 10-6 / 5.6 = 0.6 × 10-6 m = 600 nm.
Q 10.5: In Young’s double-slit experiment using monochromatic light of wavelength λ, the intensity of light at a point on the screen where path difference is λ, is K units. What is the intensity of light at a point where path difference is λ/3?
Solution:
We know that phase difference φ = (2π / λ) × Path difference.
Intensity formula: I = 4 I0 cos2(φ / 2).
Case 1: Path difference = λ ⇒ Phase difference φ = 2π.
I = 4 I0 cos2(2π / 2) = 4 I0 cos2(π) = 4 I0 (1) = 4 I0.
Given, this intensity is K. So, K = 4 I0.
Case 2: Path difference = λ / 3 ⇒ Phase difference φ = (2π / λ) × (λ / 3) = 2π / 3.
New Intensity I’ = 4 I0 cos2[ (2π/3) / 2 ] = 4 I0 cos2(π / 3).
I’ = 4 I0 (1/2)2 = 4 I0 (1/4) = I0.
Since 4 I0 = K, we get I0 = K / 4.
So, the intensity will be K / 4 units.
Q 10.6: A beam of light consisting of two wavelengths, 650 nm and 520 nm, is used to obtain interference fringes in a Young’s double-slit experiment.
(a) Find the distance of the third bright fringe on the screen from the central maximum for wavelength 650 nm.
(b) What is the least distance from the central maximum where the bright fringes due to both the wavelengths coincide?
(Note: In general NCERT format, assume Distance to screen as ‘D’ and slit separation as ‘d’ if numerical values are not provided, or plug them in if given elsewhere in the question).
Solution:
Given: λ1 = 650 nm, λ2 = 520 nm.
(a) The distance of the 3rd bright fringe for λ1 is:
y3 = 3 λ1 D / d = 3 × (650 × 10-9) D / d = 1.95 × 10-6 (D/d) m.
(b) Let the nth bright fringe of λ1 coincide with the (n+1)th bright fringe of λ2.
n λ1 D / d = (n + 1) λ2 D / d
n λ1 = (n + 1) λ2
n × 650 = (n + 1) × 520
650 n = 520 n + 520 ⇒ 130 n = 520 ⇒ n = 4.
So, the 4th bright fringe of 650 nm coincides with the 5th bright fringe of 520 nm.
Least distance y = n λ1 D / d = 4 × (650 × 10-9) D / d = 2.6 × 10-6 (D/d) m.