JEE Main 2026 — Jan 21, Shift 1 — Physics solution

JEE Main2026Jan 21, Shift 1Physics

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
$8 \times 1 0^{- 4}$
B
$6.0 \times 1 0^{- 3}$
C
$5.6 \times 1 0^{- 4}$
D
$5.0 \times 1 0^{- 3}$

Solution

The central-fringe shift due to a thin sheet covering one slit: $Δ x = \frac{( μ - 1 ) t D}{d}$ .

Solving for $t$ : $t = \frac{Δ x \cdot d}{D ( μ - 1 )} = \frac{0.2 \times 0.1}{50 \times 0.5} = \frac{0.02}{25} = 8 \times 1 0^{- 4}$ cm.

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