Search any question & find its solution
Question:
Answered & Verified by Expert
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
Options:
Solution:
2086 Upvotes
Verified Answer
The correct answer is:
$0.25 \mathrm{~mm}$
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
$\begin{aligned}
y &=(\mu-1) t \frac{D}{d} \\
&=(1.5-1) 1 \times 10^{-6} \times \frac{1.2}{2.4 \times 10^{-3}} \\
&=\frac{0.5 \times 10^{-6} \times 1.2}{2.4 \times 10^{-3}} \\
&=0.25 \times 10^{-3} \text { or } 0.25 \mathrm{~mm}
\end{aligned}$
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
$\begin{aligned}
y &=(\mu-1) t \frac{D}{d} \\
&=(1.5-1) 1 \times 10^{-6} \times \frac{1.2}{2.4 \times 10^{-3}} \\
&=\frac{0.5 \times 10^{-6} \times 1.2}{2.4 \times 10^{-3}} \\
&=0.25 \times 10^{-3} \text { or } 0.25 \mathrm{~mm}
\end{aligned}$
Looking for more such questions to practice?
Download the MARKS App - The ultimate prep app for IIT JEE & NEET with chapter-wise PYQs, revision notes, formula sheets, custom tests & much more.