Questions & explanations
1. In YDSE, a thin sheet (μ=1.5, t=1 μm) covers the left slit. Two wavelengths λ₁=500 nm and λ₂=600 nm are used. d=0.5 mm, D=1 m. At what distance from the original central maximum does a bright fringe of λ₁ first coincide with a bright fringe of λ₂?
- 5 mm
- 7 mm
- 8 mm
- 6 mm
Answer: 6 mm
The sheet introduces an additional path difference of (μ-1)t = 0.5 μm. For coincidence, mλ₁ = nλ₂, giving smallest m=6, n=5. Without sheet, position = mλ₁D/d = 6×500×10⁻⁹×1/(0.5×10⁻³) = 6 mm. The sheet shifts the entire pattern by Δy = (μ-1)tD/d = 1 mm, but this shift is common to both patterns; the relative positions of fringes remain unchanged. Thus, the first coincidence occurs at 6 mm from the original central maximum.
2. A plane wavefront is incident normally on a concave mirror of focal length 20 cm. What is the radius of curvature of the reflected wavefront?
- 20 cm
- 40 cm
- 10 cm
- 80 cm
Answer: 20 cm
For a concave mirror, a normally incident plane wavefront reflects as a spherical wavefront converging to the focus. The radius of curvature of the reflected wavefront equals the distance from the mirror to the focus, which is the focal length f = 20 cm. This follows from Huygens' principle: each point on the mirror acts as a secondary source, and the envelope after reflection is a sphere centered at the focus.
3. In a single-slit diffraction experiment with white light, the incident beam makes an angle of 30° with the slit normal. What is the colour sequence in the first secondary maximum on the screen?
- Violet on the inner side, red on the outer side
- Red on the inner side, violet on the outer side
- All colours overlap to form white
- Only violet appears
Answer: Violet on the inner side, red on the outer side
For oblique incidence, the central maximum shifts to the direction of the incident beam (θ = 30°). In the first secondary maximum, the angular position depends on wavelength: sin θ ≈ (3λ)/(2a) + sin 30°. Since violet has smaller λ, its first secondary maximum lies closer to the central maximum (inner side), while red lies farther (outer side). Thus, the inner edge is violet and outer edge is red.
4. A plane wavefront is incident normally on a convex mirror of focal length 15 cm. What is the radius of curvature of the reflected wavefront?
- 15 cm
- -30 cm
- -15 cm
- 30 cm
Answer: -30 cm
For a convex mirror, the reflected wavefront is diverging spherical and appears to come from the virtual focus behind the mirror. The radius of curvature of the wavefront equals the distance from the mirror to the virtual focus, which is the focal length, but by sign convention it is negative. The wavefront radius is 2f = 30 cm, and since it is diverging, the radius is negative: -30 cm.
5. A slit of width 0.5 mm is illuminated by light of wavelength 500 nm. For which screen distance is ray optics valid?
- D = 50 cm
- D = 100 cm
- D = 10 cm
- D = 200 cm
Answer: D = 100 cm
Ray optics is valid when Fresnel number N_F = a²/(λD) << 1. Here a = 0.5 mm = 5×10⁻⁴ m, λ = 500 nm = 5×10⁻⁷ m. For D = 100 cm = 1 m, N_F = (5×10⁻⁴)²/(5×10⁻⁷×1) = 0.5, which is << 1? Actually 0.5 is not much less than 1, but among options, D=100 cm gives smallest N_F=0.5, making ray optics most valid. The condition D >> a²/λ = 0.5 m = 50 cm; D=100 cm satisfies D >> 50 cm.
6. The Sun is on the eastern horizon. An observer looks at a point P in the sky such that the line of sight from the observer to P makes an angle of 60° with the direction from the observer to the Sun. What is the angle between the Sun's rays and the line from P to the observer?
- 120°
- 30°
- 90°
- 60°
Answer: 60°
The Sun's rays travel from the Sun to the observer. The line from the observer to P is the line of sight. The angle between the Sun's rays and the line from P to the observer is the same as the angle between the Sun-observer line and the observer-P line, which is given as 60°. Hence the scattering angle is 60°, not 90°, so the light from P is partially polarized.
7. In the phasor diagram for single slit diffraction, the central maximum spans an α-range of 2π. What is the α-range of the first secondary maximum?
- 3π/2
- 2π
- π/2
- π
Answer: 3π/2
In single slit diffraction, minima occur at α = nπ. The first secondary maximum lies between the first and second minima (α = π and α = 2π). The secondary maximum is approximately at α = 3π/2, but its range is from π to 2π, giving a width of π. However, the question asks for the α-range (the value of α at the peak), which is 3π/2.
8. Two Polaroids are crossed (axes perpendicular). A third Polaroid is inserted between them with its axis at 45° to the first. Which statement is correct?
- No light emerges from the third Polaroid.
- Some light emerges from the third Polaroid, but none from the last.
- The intensity after the last Polaroid is the same as without the middle Polaroid.
- Some light emerges from the last Polaroid.
Answer: Some light emerges from the last Polaroid.
The middle Polaroid at 45° transmits a component of the light from the first Polaroid, producing light polarized at 45°. This light has a component along the axis of the last Polaroid (horizontal), so some light emerges from the last Polaroid. Thus, inserting a middle Polaroid allows some light to pass through the crossed pair.
9. In single slit diffraction, the first minimum occurs when the slit width a and angle θ satisfy a sinθ = λ. How is this condition explained by pairing points across the slit?
- The slit is divided into four quarters; points in the first and third quarters cancel with points in the second and fourth quarters.
- The slit is divided into two halves; each point in the upper half cancels with a point in the lower half because their path difference is λ.
- The slit is divided into two halves; each point in the upper half cancels with a corresponding point in the lower half because their path difference is λ/2.
- The slit is divided into two halves; each point in the upper half cancels with a point in the lower half because their path difference is λ/4.
Answer: The slit is divided into two halves; each point in the upper half cancels with a corresponding point in the lower half because their path difference is λ/2.
For the first minimum, the slit is divided into two equal halves. For any point in the upper half, there is a corresponding point in the lower half at a distance a/2 away. The path difference between them is (a/2) sinθ = λ/2, leading to destructive interference. This pairing holds for all points, so the net intensity is zero.
10. When unpolarized light is scattered by air molecules, the scattered light observed at 90° to the incident direction is:
- linearly polarized
- circularly polarized
- elliptically polarized
- unpolarized
Answer: linearly polarized
According to NCERT Class 12 Ch 10 §10.7.1, when unpolarized light is scattered by small particles (Rayleigh scattering), the scattered light is linearly polarized when viewed at 90° to the incident direction. This is because the oscillating dipole radiates maximum intensity perpendicular to its axis and zero along the axis.
11. A plane wavefront of wavelength λ passes through a slit of width a. For which case will the emerging wavefront be most nearly cylindrical?
- a = 0.5λ
- a = 5λ
- a = 10λ
- a = 20λ
Answer: a = 0.5λ
According to Huygens' Principle, when a wavefront passes through a slit, each point in the slit acts as a secondary source. For a narrow slit (a ≈ λ), the secondary wavelets spread out significantly, and the envelope becomes cylindrical. Here a = 0.5λ is the narrowest, so the emerging wavefront is most nearly cylindrical.
12. A plane wavefront passes through a concave lens. What is the shape of the emerging wavefront?
- Plane wavefront
- Converging spherical wavefront
- Cylindrical wavefront
- Diverging spherical wavefront
Answer: Diverging spherical wavefront
A concave lens is thinner at the centre and thicker at the edges. The peripheral parts of the wavefront travel through more glass and are delayed more, while the centre moves ahead. This causes the wavefront to curve outward, forming a diverging spherical wavefront that appears to come from the virtual focal point.