In Fraunhofer diffraction by a single slit, if the slit width is doubled, how does the angular width of the central maximum change?
Answer: B
Angular width of central maximum = 2λ/a. If slit width a doubles, angular width becomes 2λ/(2a) = λ/a, which is half the original.
Q.82Hard
Two slits of widths w₁ and w₂ produce a diffraction pattern with intensity ratio I₁:I₂ = 4:1. What is the ratio of their widths?
Answer: A
Intensity is proportional to (slit width)². If I₁:I₂ = 4:1, then w₁:w₂ = √4:√1 = 2:1.
Q.83Hard
A biconvex lens (n = 1.5) has both radii of curvature equal to 20 cm. What is its focal length?
Answer: B
Using lens maker's formula: 1/f = (n-1)[1/R₁ + 1/R₂] = (0.5)[201 + 201] = (0.5)(202) = 201. Therefore f = 20 cm.
Q.84Hard
In a double-slit experiment, if one slit is covered with a transparent film of thickness t and refractive index n, the central bright fringe shifts. What is the path difference introduced?
Answer: A
Optical path through film = nt. Geometric path = t. Extra optical path = nt - t = (n-1)t. This causes a phase shift equivalent to a path difference of (n-1)t.
Q.85Hard
A monochromatic light source of wavelength λ is incident on a diffraction grating with 500 lines/mm. The second-order maximum is at 30°. What is the wavelength?
Answer: A
Grating equation: d·sin(θ) = m·λ. Here d = 1/(500 × 10³) = 2 × 10⁻⁶ m. For m = 2: (2 × 10⁻⁶)·sin(30°) = 2·λ. (2 × 10⁻⁶)·(0.5) = 2·λ. λ = 500 nm.
Advertisement
Q.86Hard
A parallel beam of light undergoes diffraction through a circular aperture of diameter D. The radius of the first dark ring in Fraunhofer diffraction is proportional to:
Answer: A
For Fraunhofer diffraction by circular aperture (Airy disk), the radius of first dark ring = 1.22λf/D, which is proportional to λ/D.
Q.87Hard
In Fraunhofer diffraction through a circular aperture, the angular radius of the first dark ring (Airy disk) is θ = 1.22λ/D. For D = 1 mm and λ = 500 nm, what is this angle in radians?
In a compound microscope, the objective has focal length 0.5 cm and eyepiece has focal length 5 cm. If the least distance of distinct vision is 25 cm, the magnifying power is:
Answer: C
Magnifying power m = (v₁/u₁) × (D/fe) ≈ -(L/fo) × (D/fe) = (025.5) × (525) = 50 × 5 = 250. With proper calculation considering tube length, result is approximately 1250
Q.89Hard
In a fiber optics cable, light undergoes total internal reflection. The minimum refractive index of the core material needed to guide light from air is:
Answer: C
For total internal reflection at core-cladding boundary: n_core × sin(θc) = n_cladding × 1. For propagation at 45°, n_core ≥ √2 ≈ 1.414
Q.90Hard
When light passes through a glass slab of thickness t and refractive index n, the lateral displacement is maximum when the angle of incidence is:
Answer: D
Lateral displacement d = t × sin(i - r)/cos(r). Displacement is maximum when ∂d/∂i = 0, which occurs approximately at high incident angles, practically around 60°
Q.91Hard
In Brewster's angle incidence, the reflected and refracted rays are:
Answer: B
At Brewster's angle θB, tan(θB) = n₂/n₁. The reflected ray and refracted ray are perpendicular (θB + (90° - θr) = 90°)
Q.92Hard
A thin air gap of thickness d is created between two glass plates. For constructive interference in reflected light with wavelength λ, the condition is:
Answer: A
For air gap with phase change at one surface, constructive interference occurs when 2d = mλ (accounting for the phase change of π at reflection from denser medium)
Q.93Hard
In a diffraction grating with 5000 lines/cm, the second order spectrum is observed at angle 30°. The wavelength of light is:
Answer: B
d = 1/(5000×10^2) = 2×10^-6 m. Using d sin θ = mλ: 2×10^-6 × sin(30°) = 2 × λ. λ = 10^-56 = 200 nm
Q.94Hard
In a compound microscope, the objective has focal length 0.5 cm and eyepiece has focal length 5 cm. The distance between them is 20 cm. The magnifying power is approximately:
Answer: B
M = -(v₀/u₀) × (D/f_e) where v₀ ≈ L = 20 cm, u₀ ≈ f₀ for final image at infinity. M ≈ (020.5) × (525) = 40 × 5 = 200. For near point at lens: ~400.
Q.95Hard
A concave mirror forms a real image 4 times larger than the object. The object is placed at distance u from the mirror. The focal length is:
Answer: A
Magnification m = -4 (real, inverted). m = -v/u, so v = 4u. Using mirror formula: 1/f = 1/u + 41u = 45u, therefore f = 4u/5. Rechecking: f = u/3 when m=-3.
Q.96Hard
In a Newton's rings experiment, dark fringe appears when the air gap thickness is:
Answer: A
Dark fringes occur at the center (t=0) and when 2t = mλ (m = 1,2,3...), giving constructive interference with phase change at glass surface.
Q.97Hard
In photoelectric effect, the stopping potential V₀ for a metal is 2V. If the frequency of incident light is doubled, the new stopping potential will be:
Answer: D
Using hf = Φ + eV₀, when frequency doubles: h(2f) = Φ + eV'₀. Since V'₀ = (2hf - Φ)/e = 2(hf/e) - Φ/e = 2V₀ + hf/e - Φ/e, the increase is not exactly double due to work function dependency.
Q.98Hard
A 1 gram sample of a radioactive isotope with atomic mass 100 undergoes decay. Initial activity is 10¹⁵ Bq. The decay constant is approximately:
Answer: B
N = (1g/100g·mol⁻¹) × Nₐ = 6.02×10²¹. λ = A/N = 10¹⁵/(6.02×10²¹) ≈ 1.66×10⁻⁷ s⁻¹. (Re-check: should be 10¹⁵/6.02×10²¹ ≈ 1.66×10⁻⁷)
Q.99Hard
An electron and a proton have the same kinetic energy. The ratio of their de Broglie wavelengths (λₑ/λₚ) is:
Answer: A
λ = h/√(2mKE). For same KE: λₑ/λₚ = √(mₚ/mₑ) ≈ √1836 ≈ 42.8
Q.100Hard
A nucleus of mass number A and atomic number Z emits an alpha particle. The recoil kinetic energy of the daughter nucleus is 0.5 MeV. What is the kinetic energy of the alpha particle? (Assume non-relativistic motion)
Answer: D
By momentum conservation, Pα = Pdaughter. KEα/KEdaughter = mdaughter/mα = (A-4)/4. If KEdaughter = 0.5 MeV, then KEα = 0.5 × 4/(A-4). For typical nuclei (A~200), KEα ≈ 2 × 0.5/(0.8) ≈ 1.25 MeV. Closer approximation gives ~2 MeV.
