Question

Intensity at point is ${\frac{1}{2}}^{th}$ of maximum intensity. If wavelength of light is $\lambda$ & separation between slits is d then, find angular position of point P is

• A. $si{n}^{-1}\left\{\frac{\lambda }{4d}\right\}$
• B. $si{n}^{-1}\left\{\frac{\lambda }{6d}\right\}$
• C. $si{n}^{-1}\left\{\frac{\lambda }{d}\right\}$
• D. $si{n}^{-1}\left\{\frac{\sqrt{3}\lambda }{d}\right\}$

Answer 1

The correct option is A $si{n}^{-1}\left\{\frac{\lambda }{4d}\right\}$

$I_{net}=I_{0}co{s}^2\left\{\frac{\varphi }{2}\right\}$$T_{net}=\frac{1}{2}{I}_0=I_{0}co{s}^2\left(\frac{\varphi }{2}\right)$
$\cos \left(\frac{\varphi }{2}\right)=\frac{1}{\sqrt{2}}=\cos \pi /4$

$\varphi =\frac{\pi }{2}=\frac{2\pi }{\lambda }\left(dsin\theta \right)$

Hence, $\theta$ wil be
$\theta =si{n}^{-1}\left\{\frac{\lambda }{4d}\right\}$