Question

In a Young’s double slit experiment, $\lambda =500\text{nm},d=1.0\text{mm}$ and $D=1.0\text{m}$. Find the minimum distance from the central maximum for which the intensity is half of the maximum intensity.

• A. $1.25\text{nm}$
• B. $0.125\text{nm}$
• C. $12.5\text{nm}$
• D. None

Answer 1

The correct option is B $0.125\text{nm}$

As we know that the intensity is given by
$I=I_{max}{\cos }^2\left(\frac{\varphi }{2}\right)$

$\frac{I_{max}}{2}=I_{max}{\cos }^2\left(\frac{\varphi }{2}\right)$

${\cos }^2\left(\frac{\varphi }{2}\right)=\frac{1}{2}$

$\cos \left(\frac{\varphi }{2}\right)=\frac{1}{\sqrt{2}}$

$\frac{\varphi }{2}=\frac{\pi }{4}$

$\varphi =\frac{\pi }{2}$
As path difference
$\Delta x=\frac{\lambda \varphi }{2\pi }=\frac{\lambda }{2\pi }\times \frac{\pi }{2}=\frac{\lambda }{4}$

$\Delta x=\frac{yd}{D}=\frac{\lambda }{4}$

$y=\frac{\lambda D}{4d}=\frac{500\times {10}^{-9}\times 1}{4\times {10}^{-3}}$

$y=0.125\text{mm}$