Questions & explanations
1. When unpolarised light is reflected from a glass surface, the reflected light is completely plane polarised. This phenomenon is explained by which law?
- Snell's law
- Malus's law
- Brewster's law
- Huygens' principle
Answer: Brewster's law
Brewster's law states that when light is incident at a particular angle (Brewster angle) on a transparent surface, the reflected light is completely polarised. At this angle, the reflected and refracted rays are perpendicular. Snell's law relates angles of incidence and refraction, Malus's law deals with intensity transmitted through polarisers, and Huygens' principle explains wave propagation.
2. In Young's double-slit experiment, the condition for a bright fringe at a point on the screen is that the path difference between the two waves arriving at that point is:
- an odd multiple of half wavelength.
- an even multiple of half wavelength.
- an integral multiple of wavelength.
- an odd multiple of quarter wavelength.
Answer: an integral multiple of wavelength.
For constructive interference (bright fringe), the path difference must be nλ, where n = 0, 1, 2, ... (integral multiple). An odd multiple of half wavelength causes destructive interference (dark fringe). Even multiples of half wavelength are equivalent to integral multiples of whole wavelength and also give bright fringes, but the standard statement uses integral multiples of λ.
3. In Young's double-slit experiment, interference fringes are formed on a screen. Which of the following statements correctly describes the fringes?
- The fringes are equally spaced and of equal intensity.
- The fringes are equally spaced but intensity decreases away from the center.
- The fringes are unequally spaced with constant intensity.
- The fringes are circular and concentric.
Answer: The fringes are equally spaced and of equal intensity.
For monochromatic light and identical slits, the interference fringes are straight lines (or hyperbolas locally appearing straight) with equal spacing (fringe width constant) and the maxima have the same intensity (assuming no diffraction envelope effects). Intensity does not decrease away from center; it is modulated by a cos² pattern with equal maxima. Fringes are not circular.
4. Which of the following correctly distinguishes a diffraction pattern from an interference pattern?
- In diffraction, all bright fringes have equal intensity; in interference, intensity decreases away from the centre.
- In diffraction, secondary maxima are of equal width; in interference, the central maximum is twice as wide.
- In diffraction, the central maximum is the brightest and widest; in interference, all bright fringes have nearly equal width and intensity if the slits are narrow.
- In diffraction, fringes are equally spaced; in interference, fringes become closer as the slit width increases.
Answer: In diffraction, the central maximum is the brightest and widest; in interference, all bright fringes have nearly equal width and intensity if the slits are narrow.
Diffraction from a single slit produces a central maximum that is both the brightest and the widest, with secondary maxima decreasing rapidly in intensity. In double-slit interference (with narrow slits), the bright fringes are of nearly equal width and intensity (modulated by the diffraction envelope if slits are finite in width). The other options contain incorrect comparisons.
5. Huygens' principle states that each point on a wavefront acts as a source of secondary wavelets. Which of the following is a correct application of Huygens' principle?
- Reflection of light can be explained by considering the envelope of secondary wavelets.
- Refraction cannot be explained by Huygens' principle.
- Huygens' principle only applies to light waves, not sound waves.
- The secondary wavelets travel only in the forward direction.
Answer: Reflection of light can be explained by considering the envelope of secondary wavelets.
Huygens' principle successfully explains both reflection and refraction by constructing the envelope of secondary wavelets. Refraction (Snell's law) is derived from the change in speed of wavelets. The principle applies to all waves, including sound. Secondary wavelets are spherical and travel in all directions, but the envelope determines the forward wavefront.
6. Which of the following is an essential condition for obtaining sustained interference of light?
- The two sources must be of different wavelengths.
- The two sources must have a constant phase difference.
- The two sources must have amplitudes that are very different.
- The two sources must be placed very far apart.
Answer: The two sources must have a constant phase difference.
Sustained interference requires coherent sources, which maintain a constant phase difference over time. Different wavelengths produce varying fringe patterns that are not stable. Amplitude difference affects contrast but not necessary for sustained interference. Distance between sources does not ensure coherence.
7. Which property distinguishes plane polarised light from unpolarised light?
- Plane polarised light has vibrations confined to a single plane perpendicular to the direction of propagation, while unpolarised light has vibrations in all directions perpendicular to propagation.
- Plane polarised light has a constant phase difference between its components, while unpolarised light does not.
- Plane polarised light undergoes refraction but not reflection, while unpolarised light undergoes reflection but not refraction.
- Plane polarised light always has a longer wavelength than unpolarised light.
Answer: Plane polarised light has vibrations confined to a single plane perpendicular to the direction of propagation, while unpolarised light has vibrations in all directions perpendicular to propagation.
In plane polarised light, the electric field vector oscillates along a fixed line (one plane) perpendicular to the direction of travel. In unpolarised light, the electric field vector vibrates randomly in all directions perpendicular to propagation. The other options do not correctly describe the difference.
8. In a single-slit diffraction experiment, the condition for a dark fringe (minimum) is given by (a is slit width, λ is wavelength, θ is angular position, and m is an integer):
- a sin θ = mλ, where m = ±1, ±2, ±3, ...
- a sin θ = (m + 1/2)λ, where m = 0, ±1, ±2, ...
- a cos θ = mλ, where m = 0, ±1, ±2, ...
- a sin θ = mλ, where m = 0, ±1, ±2, ...
Answer: a sin θ = mλ, where m = ±1, ±2, ±3, ...
For single-slit diffraction, minima occur at angles θ satisfying a sin θ = mλ, with m = ±1, ±2, ±3, ... (m = 0 gives the central maximum). The path difference between waves from the two edges of the slit is a sin θ, and destructive interference occurs when this equals an integer multiple of λ.
9. In a single-slit diffraction pattern, the angular width of the central maximum is compared to the angular width of a secondary maximum. Which statement is correct?
- The central maximum is twice as wide as a secondary maximum.
- The central maximum has the same width as a secondary maximum.
- The central maximum is half as wide as a secondary maximum.
- The central maximum is three times as wide as a secondary maximum.
Answer: The central maximum is twice as wide as a secondary maximum.
The central maximum is formed between the first minima at θ = ±λ/a, so its angular width is 2λ/a. Each secondary maximum lies between two consecutive minima (e.g., between λ/a and 2λ/a), giving an angular width of λ/a. Hence the central maximum is twice as wide.
10. In a Young's double-slit experiment, the slits are 0.5 mm apart and the screen is 1 m away. The wavelength of light is 500 nm. The distance of the third bright fringe from the central maximum is:
- 3 mm
- 1.5 mm
- 6 mm
- 0.5 mm
Answer: 3 mm
Fringe width β = λD/d = (500 × 10⁻⁹ m × 1 m) / (0.5 × 10⁻³ m) = 1 × 10⁻³ m = 1 mm. The third bright fringe (n=3) is at y = nβ = 3 mm. Option B (1.5 mm) would be for n=1.5 (not integer), C (6 mm) would be for n=6, D (0.5 mm) is for half fringe width.
11. Unpolarised light of intensity I₀ is incident on a polariser. The transmitted light then passes through an analyser whose polarising axis is at an angle of 60° to that of the polariser. What is the intensity of light emerging from the analyser?
- I₀/2
- I₀/4
- I₀/8
- I₀/16
Answer: I₀/8
After the polariser, the intensity becomes I₀/2 (since half the intensity of unpolarised light passes). Then Malus's law gives I = (I₀/2) cos²θ, with θ = 60°. cos 60° = 1/2, so cos²60° = 1/4. Thus final intensity = (I₀/2)(1/4) = I₀/8.
12. In Young's double-slit experiment, the intensity on the screen varies with position. Which of the following correctly describes the intensity pattern?
- The intensity varies linearly with distance from the central maximum.
- The intensity follows a cos² pattern, with maxima of equal intensity.
- The intensity is constant across the screen.
- The intensity decreases exponentially away from the central maximum.
Answer: The intensity follows a cos² pattern, with maxima of equal intensity.
For two coherent slits of equal width, the intensity distribution is I = I₀ cos²(π d sinθ/λ), which is a cos² function with equally spaced maxima of equal intensity I₀. It is not linear, constant, or exponential.