Q57 / 75 · JEE Main 2026 (21 January Shift 1) · Physics
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$5.0 \times 10^{-3}$
- B$6.0 \times 10^{-3}$
- Correct answerC$8 \times 10^{-4}$
- D$5.6 \times 10^{-4}$
Solution
Given: $d = 0.1$ cm, $D = 50$ cm, $n = 1.5$, Shift $= 0.2$ cm
Optical path difference by sheet: $\Delta = (n-1)t$
Shift of central fringe: Shift $= \frac{(n-1)t \times D}{d}$
$0.2 = \frac{(1.5-1) \times t \times 50}{0.1} = \frac{0.5 \times 50 \times t}{0.1} = 250t$
$t = \frac{0.2}{250} = 8 \times 10^{-4}$ cm
Optical path difference by sheet: $\Delta = (n-1)t$
Shift of central fringe: Shift $= \frac{(n-1)t \times D}{d}$
$0.2 = \frac{(1.5-1) \times t \times 50}{0.1} = \frac{0.5 \times 50 \times t}{0.1} = 250t$
$t = \frac{0.2}{250} = 8 \times 10^{-4}$ cm
Hence, the correct answer is option: 'C'.