Question

A two-slit interference experiment uses coherent light of wavelength $5\times {10}^{-7}m$. Intensity in the interference pattern for the following points are $I_1,I_2,I_3$ and $I_4$, respectively.
(1) A point that is close to one slit than the other by $5\times {10}^{-7}m$.
(2) A point where the light waves received from the two slits are out of phase by $\frac{4\pi }{3}rad$
(3) A point that is closer to one slit than the other by $7.5\times {10}^{-7}m$
(4) A point where the light waves received by the two slits are out of phase by $\frac{\pi }{3}rad$
Then which of the following statements is/are correct?

• A. $I_1>I_4>I_2>I_3$
• B. $I_3>I_2>I_1>I_4$
• C. $3I_2=I_4$
• D. $I_3=0$

Answer 1

The correct option is A $I_1>I_4>I_2>I_3$

Let $I_0$ be the maximum intensity that can be on the screen.

For (1),

Phase difference at the point=$\delta =5\times {10}^{-7}\times \frac{2\pi }{\lambda }=2\pi$

Hence, $I_1=I_{0}co{s}^2\left(\frac{\delta }{2}\right)=I_0$

For (2),

$\delta =\frac{4\pi }{3}$

Hence, $I_2=I_{0}co{s}^2\left(\frac{\frac{4\pi }{3}}{2}\right)=\frac{1}{4}{I}_0$

For (3),

$\delta =7.5\times {10}^{-7}\times \frac{2\pi }{\lambda }=3\pi$

Hence, $I_3=I_{0}co{s}^2\left(\frac{3\pi }{2}\right)=0$

For (4),

$\delta =\frac{\pi }{3}$

Hence, $I_4=I_{0}co{s}^2\left(\frac{\pi }{6}\right)=\frac{3}{4}{I}_0$

Hence, $I_1>I_4>I_2>I_3$