Q145 / 160 · TS EAMCET 2021 (06 Aug Shift 1) · Physics

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 $B$ ?

  • A
    1 : 1
  • Correct answer
    B
    2 : 1
  • C
    1 : 2
  • D
    4 : 1

Solution

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, $I_1: I_2=2: 1$

Hence, the correct answer is option: 'B'.

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