Question

In a young's double slit experiment $\lambda =500nm,d=$ $1.0mm$ and $D=1m$. Find the minimum distance from the central maximum for which the intensity is half of the maximum intensity:

• A. $1.25\times 10^{-3}m$
• B. $1.25\times 10^{-4}m$
• C. $2.25\times 10^{-1}m$
• D. $125m$

Answer 1

Answer: (B)

$I=4I_0{\cos }^2\varphi /2$
$\frac{\left(4I_o\right)}{2}=4I_o{\cos }^2\varphi /2$ [Intensity is half of the maximum]
$\therefore \frac{1}{\sqrt{2}}=\cos \varphi /2$
$90^{\circ }=\varphi$
Minimum distance from central maximum corresponding to $\varphi =90^{\circ }$ is
$x=\frac{\varphi }{2\pi }\times \beta$
$\Rightarrow \frac{\pi /2}{2\pi }\times \frac{\lambda D}{d}$
$x=\frac{1}{4}\times \frac{500\times 10^{-9}\times 1}{1\times 10^{-3}}$
$x=1.25\times 10^{-4}m$