Q162 / 180 · KCET 2020 · Physics
In Young's double slit experiment, the distance between the slits and the screen is $1.2 \mathrm{~m}$ and the distance between the two slits is $2.4 \mathrm{~mm}$. If a thin transparent mica sheet of thickness $1 \mu \mathrm{m}$ and RI $1.5$ is introduced between one of the interfering beams, the shift in the position of central bright fringe is
- A$2 \mathrm{~mm}$
- B$0.5 \mathrm{~mm}$
- C$0.125 \mathrm{~mm}$
- Correct answerD$0.25 \mathrm{~mm}$
Solution
Given, $D=1.2 \mathrm{~m}, d=2.4 \mathrm{~mm}=2.4 \times 10^{-3} \mathrm{~m}$ $t=1 \mu \mathrm{m}=1 \times 10^{-6} \mathrm{~m}$ and $\mu=1.5$
When a transparent sheet is introduced in the path of one of the interfering beam, the shift in the position of central bright fringe is
$\begin{aligned}
y &=(\mu-1) t \frac{D}{d} \\
&=(1.5-1) 1 \times 10^{-6} \times \frac{1.2}{2.4 \times 10^{-3}} \\
&=\frac{0.5 \times 10^{-6} \times 1.2}{2.4 \times 10^{-3}} \\
&=0.25 \times 10^{-3} \text { or } 0.25 \mathrm{~mm}
\end{aligned}$
Hence, the correct answer is option: 'D'.