Question

A monochromatic light source of wavelength $\lambda$ is placed at S. Three slits $S_1,S_2$ and $S_3$ are equidistance from the source S and the point P on the screen. $S_{1}P-S_{2}P=\lambda /6$ and $S_{1}P-S_{3}P=2\lambda /3$. If I be the intensity at P when only one slit is open, the intensity at P when all the three slits are open is?

• A. $3$I
• B. $5$I
• C. $8$I
• D. Zero

Answer 1

The correct option is A $3$I

Given, $S_{1}P-S_{2}P=\frac{\lambda }{6}$

Hence, phase difference between $S_1$ and $S_2,\varphi =\frac{\lambda }{6}\times \frac{2\pi }{\lambda }=\frac{\pi }{3}$

So intensity between $S_1$ and $S_2$ is,

$I^{\prime }=I{\cos }^2{\left(\frac{\pi }{6}\right)}^2=\frac{3I}{4}.........\left(i\right)$

Given, $S_{1}P-S_{3}P=\frac{2\lambda }{3}$

Hence, phase difference between $S_1$ and $S_3,\varphi =\frac{2\lambda }{3}\times \frac{2\pi }{\lambda }=\frac{4\pi }{3}$

So intensity between $S_1$ and $S_2$ is,

$I^{\prime \prime }=I{\cos }^2{\left(\frac{2\pi }{3}\right)}^2=\frac{I}{4}.........\left(ii\right)$

The intensity at P when all the three slits are open is, $I\frac{I^{\prime }}{I^{\prime \prime }}=3I$