Question

Two cohorent monochromatic light beams of intensities 4I and 9I interfere in a Young's double slit experimental setup to produce a fringe pattern on the screen. The phase difference between the beams at two points P and Q on the screen are $\pi /2$ and $\pi /3$ respectively. Then the ratio of the two intensities $I_P/I_Q$ is

• A. 0
• B. $\frac{6}{19}$
• C. $\frac{13}{19}$
• D. $\frac{6}{13}$

Answer 1

The correct option is C $\frac{13}{19}$

We know that $I\propto A^2$ where $I$ is the intensity and $A$ is the amplitude.

Given that $I_1=4I$ and $I_2=9I$

Hence,

$\frac{A_1}{A_2}=\sqrt{\frac{I_1}{I_2}}$

$⟹\frac{A_1}{A_2}=\sqrt{\frac{4I}{9I}}$

$⟹\frac{A_1}{A_2}=\frac{2}{3}$

Let, $A_1=2A$ and $A_2=3A$

When, phase difference$\left(\varphi \right)$ is $\frac{\pi }{2}$ :-

Amplitude $A_P=\sqrt{A_1^2+A_2^2+2A_1{A}_2\cos \varphi }$

$⟹A_P=\sqrt{4A^2+9A^2+0}=\sqrt{13}A$

When, phase difference$\left(\varphi \right)$ is $\frac{\pi }{3}$ :-

Amplitude $A_Q=\sqrt{A_1^2+A_2^2+2A_1{A}_2\cos \varphi }$

$⟹A_Q=\sqrt{4A^2+9A^2+2.2A.3A.\frac{1}{2}}=\sqrt{19}A$

Now, $\frac{I_P}{I_Q}=\frac{A_P^2}{A_Q^2}$

$=\frac{13}{19}$

Hence, answer is option-(C).