Question

In a young's double slit interference experiment the fringe pattern is observed on a screen palced at a distance D from the slits. The slits are separaated by a distance d and are illuminated by monochromatic light of wavelength $\lambda$ Find the distance from the central point where the intensity falls to (a) half the maximum, (b) one fourth of the maximum.

Answer 1

(i) When intensity is half the maximum,
$\Rightarrow \frac{I}{I_{max}}=\frac{1}{2}\Rightarrow \frac{4a^2{\cos }^2\left(\frac{\varphi }{2}\right)}{4a^2}\frac{1}{2}\Rightarrow {\cos }^2\left(\frac{\varphi }{2}\right)=\frac{1}{2}\Rightarrow \cos \left(\frac{\varphi }{2}\right)=\frac{1}{\sqrt{2}}\Rightarrow \frac{\varphi }{2}=\frac{\pi }{4}$
$\Rightarrow$ Path difference,
$x=\frac{1}{4}\Rightarrow y=\frac{+xD}{d}=\frac{\lambda D}{4d}$
(ii) When intensity is th the maximum,
$\frac{I}{I_{max}}=\frac{1}{4}\Rightarrow 4a^2{\cos }^2\left(\frac{\varphi }{2}\right)=\frac{1}{4}\Rightarrow {\cos }^2\left(\frac{\varphi }{2}\right)=\frac{1}{4}\Rightarrow \cos \left(\frac{\varphi }{2}\right)=\frac{1}{2}\Rightarrow \frac{\varphi }{2}=\frac{\pi }{3}$
Path difference, $x=\frac{\lambda }{3}\Rightarrow y=\frac{xD}{d}=\frac{\lambda }{3d}$