Wave Optics — Previous-Year Questions

Q1. According to Huygens' principle, each point on a wavefront acts as a source of:

Q2. The shape of the wavefront from a distant star reaching Earth is essentially:

Q3. Huygens' wave theory of light could NOT explain:

Q4. When a plane wavefront strikes a concave mirror at its focus, the reflected wavefront is:

Q1. In Young's experiment, fringe width is β. If the entire apparatus is immersed in water (μ = 4/3), the new fringe width is:

Q2. Two coherent sources have intensities in the ratio 1 : 4. The ratio of I_max to I_min on the screen is:

Q3. In a YDSE, slit separation d = 1 mm, screen distance D = 1 m. Light of λ = 600 nm. Fringe width is:

Q4. If white light is used in YDSE, the central fringe is:

Q5. A glass plate (μ = 1.5, t = 4 μm) is placed before one slit. For λ = 600 nm, the number of fringes by which the pattern shifts is:

Q6. In YDSE, the path difference at a point on the screen is 11λ/2. The point is:

Q1. In a single-slit experiment, the angular width of the central maximum doubles if:

Q2. Light of wavelength 500 nm passes through a slit of width 0.5 mm. Width of central maximum on a screen 2 m away:

Q3. First-minimum diffraction angle for a single slit of width 0.1 mm illuminated by light of λ = 600 nm:

Q4. In a single-slit diffraction pattern, the ratio of intensities of the central maximum to the first secondary maximum is approximately:

Q5. If the slit width in a single-slit experiment is reduced to half, the angular width of the central maximum:

Q1. Two sources of light will produce sustained interference if they:

Q2. Two coherent sources of intensities I and 4I produce an interference pattern. Maximum and minimum intensities are:

Q3. Two coherent light sources S₁ and S₂ have intensities I₀ each. The intensity at a point where the phase difference is π/2 is:

Q4. Why are two independent sources of light never coherent?

Q1. Unpolarized light of intensity I₀ passes through two polarizers with their axes at 30°. The transmitted intensity is:

Q2. Two polarizers are crossed (90° apart). A third polarizer is inserted between them with its axis at 45° to both. Final intensity (initial unpolarized I₀):

Q3. Light intensity through an analyzer is I₀/4 of the incident. Angle between polarizer and analyzer is:

Q4. Polarized light of intensity I₀ falls on an analyzer; transmitted intensity is found to be I₀/2. The angle is:

Q5. If the angle between the transmission axes of two polarizers is increased from 0° to 60°, the intensity of transmitted light through the pair (for unpolarized incident):

Q1. The Brewster angle for light going from air (n = 1) into a medium of refractive index √3 is:

Q2. At Brewster's angle, the angle between reflected and refracted rays is:

Q3. Light is incident on a glass surface from air. If the reflected ray is fully polarized, the angle of incidence is approximately (n_glass = 1.5):

Q4. Brewster angle for an interface depends on:

Q5. At what angle of incidence on a glass surface (n = 1.5) is the reflected light completely polarized?

Q1. A diffraction grating has 5000 lines/cm. Find the angle of the first-order maximum for λ = 500 nm.

Q2. Resolving power of a grating with N = 10000 lines at first order is:

Q3. In a diffraction grating, the maximum order possible for λ = 600 nm and d = 1 μm is:

Q4. In a grating with d = 4a (slit spacing 4× slit width), the missing order is at:

Q1. A telescope has aperture 25 cm. The minimum angular separation it can resolve for λ = 500 nm is:

Q2. Resolving power of a microscope can be increased by:

Q3. Why is electron microscopy able to resolve atoms while optical microscopy cannot?

Q4. Two stars are just resolved by a telescope of aperture D. For finer resolution, you should:

Q5. The Rayleigh criterion gives θ_min = 1.22 λ/D. The factor 1.22 arises from:

Q1. An oil film on water (n_oil = 1.4) is viewed in reflected light. The minimum thickness for bright reflection at λ = 560 nm (normal incidence) is:

Q2. A soap film in air (n = 1.33). Minimum thickness for the film to appear BRIGHT in reflected light at λ = 532 nm (normal incidence):

Q3. An anti-reflection coating on a lens reduces reflection by destructive interference. For wavelength 550 nm and coating index 1.38, minimum thickness:

Q4. In Newton's rings, the radius of the m-th dark ring (in reflection) varies as:

Q5. Why does a soap film appear black just before bursting?