Search any question & find its solution
Question:
Answered & Verified by Expert
In Young's double slit experiment, point $A$ on the screen has a path difference of $\lambda$ and point $B$ on the screen has a path difference of $\frac{\lambda}{4}$. What is the ratio of the intensities at point $A$ to $B$ ?
Options:
Solution:
1653 Upvotes
Verified Answer
The correct answer is:
2 : 1
In YDSE, let intensities of two identical coherent sources $S_1$ and $S_2$ is $I_0$. Then, resultant intensity,
$$
I=2 I_0 \cos ^2 \frac{\phi}{2}
$$
where, $\phi=$ phase difference.
When path difference is $\lambda$, then
$$
\phi_1=\frac{2 \pi}{\lambda} \text { (path difference) and } \phi_1=\frac{2 \pi}{\lambda} \lambda=2 \pi
$$
When path difference is $\frac{\lambda}{4}$, then
$$
\begin{aligned}
\phi_2 & =\frac{2 \pi}{\lambda} \frac{\lambda}{4}=\frac{\pi}{2} \\
\therefore \quad \frac{I_1}{I_2} & =\frac{2 I_0 \cos ^2 \frac{\phi_1}{2}}{2 I_0 \cos ^2 \frac{\phi_2}{2}}=\frac{\cos ^2\left(\frac{2 \pi}{2}\right)}{\cos ^2\left(\frac{\pi}{4}\right)}=\frac{\cos ^2(\pi)}{\cos ^2 \frac{\pi}{4}} \\
& =\frac{1}{(1 / 2)}=\frac{2}{1}
\end{aligned}
$$
Hence, $I_1: I_2=2: 1$
$$
I=2 I_0 \cos ^2 \frac{\phi}{2}
$$
where, $\phi=$ phase difference.
When path difference is $\lambda$, then
$$
\phi_1=\frac{2 \pi}{\lambda} \text { (path difference) and } \phi_1=\frac{2 \pi}{\lambda} \lambda=2 \pi
$$
When path difference is $\frac{\lambda}{4}$, then
$$
\begin{aligned}
\phi_2 & =\frac{2 \pi}{\lambda} \frac{\lambda}{4}=\frac{\pi}{2} \\
\therefore \quad \frac{I_1}{I_2} & =\frac{2 I_0 \cos ^2 \frac{\phi_1}{2}}{2 I_0 \cos ^2 \frac{\phi_2}{2}}=\frac{\cos ^2\left(\frac{2 \pi}{2}\right)}{\cos ^2\left(\frac{\pi}{4}\right)}=\frac{\cos ^2(\pi)}{\cos ^2 \frac{\pi}{4}} \\
& =\frac{1}{(1 / 2)}=\frac{2}{1}
\end{aligned}
$$
Hence, $I_1: I_2=2: 1$
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.