Physics · 2026
JEE Main · 21 January 2026, Shift 1 · Q39
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…
In a double slit experiment the distance between the slits is $\displaystyle 0.1$ cm and the screen is placed at $\displaystyle 50$ cm from the slits plane. When one slit is covered with a transparent sheet having thickness $\displaystyle t$ and refractive index $\displaystyle n(=1.5)$, the central fringe shifts by $\displaystyle 0.2$ cm . The value of $\displaystyle t$ is
$\displaystyle \_\_\_\_$ cm.
Official answer
From NTA’s final answer key for this paper.
(3)
$\displaystyle 8 \times 10^{-4}$
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JEE Main 2026 Physics question, with the answer from NTA’s final answer key. Where our answers come from.