Question

In YDSE The intensity of central bright fringe is $8mW/m^2$. What will be the intensity at $\lambda /6$ path difference?

• A. $8mW/m^2$
• B. $6mW/m^2$
• C. $4mW/m^2$
• D. $2mW/m^2$

Answer 1

Answer: (B)

$6mW/m^2$

Given that,

Intensity of central bright fringe $I_0=8mW/m^2$

We know that,

$I=I_0{\cos }^2\left(\frac{\varphi }{2}\right)$

$I=8{\cos }^2\left(\frac{\varphi }{2}\right)$

Now,

Phase difference = $\frac{2\pi }{\lambda }$ x path difference

$\varphi =\frac{2\pi }{\lambda }\times \frac{\lambda }{6}$

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

Now, the intensity is

$I=8\times {\cos }^2\left(\frac{\pi }{6}\right)$

$I=8\times \frac{\sqrt{3}}{2}\times \frac{\sqrt{3}}{2}$

$I=6mW/m^2$

Hence, the intensity is $6mW/m^2$