JEE Main · 2026 · 21 Jan 2026 · Shift 1 · Wave Optics

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In a double slit experiment the distance between the slits is 0.1 cm and the screen is placed at 50 cm from the slits plane. When one slit is covered with a transparent sheet having thickness

t

and refractive index

n (= 1.5)

, the central fringe shifts by 0.2 cm . The value of t is ____ cm .

(A) $8 \times 1 0^{- 4}$ ✓ Correct
(B) $6.0 \times 1 0^{- 3}$
(C) $5.6 \times 1 0^{- 4}$
(D) $5.0 \times 1 0^{- 3}$

Answer: (A)

Solution

Covering one slit with the sheet shifts the central fringe by

Δ = \frac{D}{d} (n - 1) t

. Solving for the thickness,

t = \frac{Δ d}{D ( n - 1 )} = \frac{0.2 \times 0.1}{50 \times ( 1.5 - 1 )} = \frac{0.02}{25} = 8 \times 1 0^{- 4} cm

JEE Main · 2026 · 21 Jan 2026 · Shift 1 · Single correct option

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