Question

In a young’s double-slit experiment $\lambda =500nm,d=1m,D=1m$. The minimum distance from the central maximum for which the intensity is half of the maximum intensity is

• A. $2\times {10}^{-4}m$
• B. $1.25\times {10}^{-4}m$
• C. $4\times {10}^{-4}m$
• D. $2.5\times {10}^{-4}m$

Answer 1

The correct option is B $1.25\times {10}^{-4}m$

As relation of intensity with phase difference is
$I=4I_{0}co{s}^2\frac{\varphi }{2}$
Given that intensity is half the maximum
$I=\frac{4I_0}{2}=2I_0$

$\Rightarrow 4I_{0}co{s}^2\frac{\varphi }{2}=2I_0$
$\Rightarrow co{s}^2\frac{\varphi }{2}=\frac{1}{2}$
$\Rightarrow cos\frac{\varphi }{2}=\frac{1}{\sqrt{2}}$
$\Rightarrow \frac{\varphi }{2}=\frac{\pi }{4}$
$\Rightarrow \varphi =\frac{\pi }{2}$
Hence, path difference $=\frac{\lambda }{2\pi }\left(\frac{\pi }{2}\right)=\frac{\lambda }{4}$
So, distance $y=\frac{D\left(\lambda /4\right)}{d}$

Placing the values given, we will get $y=1.25\times {10}^{-4}m$