Optics: JEE Main PYQ, 2020–2025

Answer: (A)

Principle: for a Galilean telescope in normal adjustment, the eyepiece's diverging focal point coincides with the image formed by the objective, so L = f_o − |f_e|. Substitute f_o = 60 cm and |f_e| = 10 cm to get L = 50 cm. The subtraction is the key step; the sum belongs to the astronomical telescope.

Answer: (A)

For thin lenses in contact, the powers add, so 1/f = 1/f1 + 1/f2. With the sign convention, the convex lens has f1 = +20 cm and the concave lens has f2 = −30 cm. Then 1/f = 1/20 − 1/30 = (3 − 2)/60 = 1/60, so f = +60 cm. The net power is positive, so the combination is converging, but weaker than the convex lens alone.

Answer: (B)

Interference maxima occur at d sinθ = nλ and single-slit minima at a sinθ = mλ. The ratio d/a = 6 means that n = 6 coincides with m = 1, so the sixth-order maximum is missing. Inside the central diffraction maximum (|sinθ| < λ/a) the visible maxima are n = 0, ±1, ±2, ±3, ±4, ±5, which is 11 fringes.

Answer: (C)

Two sources produce a stable interference pattern only if they are coherent, meaning the phase difference between their waves stays constant in time. Two independent light sources, such as two bulbs, emit light in short random bursts from many atoms, so their phase difference changes randomly. The interference pattern then averages out to uniform illumination. Coherence, not intensity or wavelength, is the condition that matters here.

Answer: (A)

Brewster's law for light going from a medium of index n into air is tan i_B = n₂/n₁ = 1/n. Here i_B = 30°, so 1/n = tan 30° = 1/√3 and n = √3. The critical angle at the same boundary satisfies sin C = 1/n = 1/√3 (C ≈ 35.3°). The key point is that the same ratio 1/n appears in both laws, as a tangent for the Brewster angle and as a sine for the critical angle.

Answer: (A)

The lens maker's formula is 1/f = (n_lens/n_medium - 1)(1/R1 - 1/R2). Here the lens material is air and the medium is glass, so n_rel = 1/1.5 = 2/3. For a biconvex shape, R1 = +10 cm and R2 = -10 cm, so (1/R1 - 1/R2) = 0.2 cm⁻¹. Then 1/f = (2/3 - 1)(0.2) = -1/15 cm⁻¹, so f = -15 cm. The lens is diverging, because a lens that is optically rarer than its surroundings behaves opposite to its shape.

Answer: (A)

Principle: the focal point of a spherical mirror lies midway between the pole and the centre of curvature. So f = R/2. For R = 40 cm, f = 20 cm. The focal length of a mirror is always half its radius of curvature, with the same sign convention for both concave and convex mirrors.

Answer: (C)

A sheet of thickness t and index μ adds an optical path of (μ − 1)t compared with air. The central fringe moves to the point where this is compensated. Number of fringes shifted = (μ − 1)t/λ = (0.5 × 6000 nm)/500 nm = 6. The shift is toward the slit covered by the sheet.

Answer: (A)

Work backwards from the final image. For the eyepiece, v = -25 cm and f = +5 cm, so 1/u = 1/v - 1/f = -1/25 - 1/5 = -6/25 and u = -25/6 cm. The image formed by the objective must lie 25/6 cm from the eyepiece, so its distance from the objective is v_o = 10 - 25/6 = 35/6 cm. The objective's magnification is m_o = v_o/f_o - 1 = 35/6 - 1 = 29/6, taken in magnitude. The eyepiece magnification for an image at the near point is m_e = 1 + D/f_e = 1 + 25/5 = 6. The total is m = m_o × m_e = (29/6) × 6 = 29.

Answer: (A)

Diffraction, the spreading of waves at an obstacle or opening, is prominent when the wavelength is comparable to the size of the obstacle or opening. Sound has wavelengths of about a metre, so it bends around the corner. Light has a wavelength near 500 nm, which is far smaller than any everyday corner, so the shadow is sharp.