Exercises
(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.
Answer
Given: λ = 589 nm = 589 × 10⁻⁹ m, c = 3 × 10⁸ m s⁻¹, n = 1.33.
(a) Reflected light travels back in the same medium (air), so nothing changes:
λ = 589 nm; ν = c/λ
⇒ ν = 3 × 10⁸ / (589 × 10⁻⁹)
⇒ ν ≈ 5.09 × 10¹⁴ Hz; speed = 3 × 10⁸ m s⁻¹
(b) For refracted light the frequency stays the same, while speed and wavelength reduce by the factor n:
ν ≈ 5.09 × 10¹⁴ Hz
v = c/n
⇒ v = 3 × 10⁸ / 1.33
⇒ v ≈ 2.26 × 10⁸ m s⁻¹
λ′ = λ/n → λ′ = 589/1.33 → λ′ ≈ 443 nm
(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.
Answer
(a) Spherical (diverging spherical wavefront centred on the source).
(b) Plane: the lens makes the emerging rays parallel, so the wavefront is plane.
(c) Plane: the star is so far away that the small part of its huge spherical wavefront reaching the Earth is practically flat.
(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?
Answer
(a) v = c/n
⇒ v = 3.0 × 10⁸ / 1.5
⇒ v = 2.0 × 10⁸ m s⁻¹
(b) No. The refractive index of glass depends on wavelength (this is why a prism disperses white light). Violet light has a larger refractive index than red, so violet travels slower than red in a glass prism.
(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.
Answer
Given: d = 0.28 mm = 0.28 × 10⁻³ m, D = 1.4 m, n = 4, yₙ = 1.2 cm = 1.2 × 10⁻² m.
Position of the nth bright fringe yₙ = nλD/d
⇒ λ = yₙ d/(nD)
λ = (1.2 × 10⁻² × 0.28 × 10⁻³)/(4 × 1.4)
⇒ λ = 3.36 × 10⁻⁶ / 5.6
⇒ λ = 6.0 × 10⁻⁷ m = 600 nm
(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?
Answer
For two slits giving equal intensities I₀, the resultant intensity is I = 4I₀ cos²(φ/2), where the phase difference φ = (2π/λ) × path difference.
Path difference λ
⇒ φ = 2π
⇒ I = 4I₀ cos² π = 4I₀
⇒ K = 4I₀
Path difference λ/3
⇒ φ = 2π/3
⇒ I = 4I₀ cos²(π/3) = 4I₀ × (1/2)² = I₀
I = K/4
(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?
Answer
The slit separation d and screen distance D are not stated in the question, so the answers are first written in terms of D/d. The NCERT answer key uses D/d = 600 (for example D = 1.2 m and d = 2 mm); numerical values are given for that case.
(a) y₃ = 3λD/d
⇒ y₃ = 3 × 650 × 10⁻⁹ × (D/d) = 1.95 × 10⁻⁶ (D/d) m
With D/d = 600: y₃ = 1.95 × 10⁻⁶ × 600
⇒ y₃ ≈ 1.17 × 10⁻³ m = 1.17 mm
(b) Bright fringes coincide where n × 650 nm = m × 520 nm, i.e. n/m = 520/650 = 4/5. The smallest whole numbers are n = 4 (for 650 nm) and m = 5 (for 520 nm), giving a common path difference of 4 × 650 = 5 × 520 = 2600 nm.
y = 2600 × 10⁻⁹ × (D/d) = 2.6 × 10⁻⁶ (D/d) m
⇒ with D/d = 600: y = 1.56 × 10⁻³ m = 1.56 mm
Important Extra Questions
State Huygens principle.
Answer
Every point on a wavefront acts as a source of secondary wavelets that spread out in the forward direction with the speed of the wave. The common tangent (envelope) to all these secondary wavelets after a time t gives the position of the new wavefront.
What are coherent sources? Why can two independent sodium lamps not produce a sustained interference pattern?
Answer
Coherent sources emit waves of the same frequency with a constant (time-independent) phase difference. Two independent lamps emit light in random, rapidly changing phases, so the phase difference between them changes many times within the time our eye can respond; the interference pattern shifts so fast that only uniform average illumination is seen.
Write the expression for fringe width in Young's double-slit experiment. How does it change if the whole apparatus is immersed in water?
Answer
Fringe width β = λD/d. In water the wavelength becomes λ/n, so the fringe width decreases by the factor n (about 1.33), i.e. the fringes become narrower.
In single slit diffraction, where are the minima located? What is the angular width of the central maximum?
Answer
Minima occur at angles θ where a sin θ ≈ aθ = nλ (n = ±1, ±2, ...), a being the slit width. The central maximum extends from θ = −λ/a to +λ/a, so its angular width is 2λ/a.
Unpolarised light of intensity I₀ passes through two polaroids whose pass axes are inclined at 60°. Find the intensity of the emergent light.
Answer
After the first polaroid: I₁ = I₀/2. By Malus' law, after the second: I₂ = I₁ cos² 60°
⇒ I₂ = (I₀/2) × (1/4)
⇒ I₂ = I₀/8