Optics: JEE Advanced PYQ, 2007–2026

Answer: (B)

The principle: accommodation is the change in the power of the eye lens, achieved by changing its curvature with the ciliary muscles. A nearer object forms its image behind the lens at a greater distance, so the lens must converge rays more strongly. Greater power means shorter focal length. So the focal length of the eye lens decreases when the eye views a near object.

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: (A)

The slab appears 6/1.5 = 4 cm thick, so the mark seems raised by 6 − 4 = 2 cm: shift = t(1 − 1/μ).

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: (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: (C)

The sheet replaces a length t of air by a length t of glass. The extra optical path is (μ−1)t = (1.5−1) × 6 μm = 3 μm. Each wavelength of extra path moves the pattern by one fringe width, so the shift is 3 μm / 600 nm = 5 fringe widths. The pattern moves towards the side of the slit covered by the sheet.

Answer: (D)

The central diffraction maximum extends to sin θ = λ/a on either side. A bright fringe of order m lies at sin θ = mλ/d, and is inside if m/d < 1/a, that is m < d/a = 6. The orders 0, ±1, ±2, ±3, ±4, ±5 qualify, 11 fringes in all. The fringes of order ±6 would fall exactly on the diffraction minima and are absent.