JEE Main 2026 — Jan 21, Shift 2 — Physics solution

JEE Main2026Jan 21, Shift 2Physics

In a Young's double slit experiment set up, the two slits are kept 0.4 mm apart and screen is placed at 1 m from slits. If a thin transparent sheet of thickness 20 $μ$ m is introduced in front of one of the slits then centre bring fringe shifts by 20 mm on the screen. The refractive index of transparent sheet is given by $\frac{α}{10}$ , where $α$ is _________.

Your answer

Solution

Introducing a sheet of refractive index $μ$ and thickness $t$ in front of one slit shifts the central fringe by $y_{s hi f t} = \frac{( μ - 1 ) t D}{d}$ , where $D$ is the screen distance and $d$ the slit separation.

Plug in: $20 \times 1 0^{- 3} = \frac{( μ - 1 ) \times 20 \times 1 0 ^{- 6} \times 1}{0.4 \times 1 0 ^{- 3}}$ .

Simplify the right side: $\frac{( μ - 1 ) \times 20 \times 1 0 ^{- 6}}{0.4 \times 1 0 ^{- 3}} = (μ - 1) \times 0.05$ .

So $0.02 = (μ - 1) \times 0.05 \Rightarrow μ - 1 = 0.4 \Rightarrow μ = 1.4$ .

Since $μ = \frac{α}{10}$ , we get $α = 14$ .

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