Question

In a YDSE apparatus, two identical slits are separated by $1$ and distance between the slits and screen is $1m$.The wavelength of light used is $6000A^{\circ }$. The minimum distance between two points on screen having $75\%$ intensity of the maximum intensity is :

• A. $0.9$
• B. $0.40$
• C. $0.30$
• D. $0.20$

Answer 1

The correct option is A $0.9$

Given,

Distance between slit and screen is $D=1m$

Slit width is $d=1m$

Wavelength of light is $\lambda =6000A^0=6\times {10}^{-7}m$

$\frac{I}{I_0}=\frac{75}{100}=\frac{3}{4}$

Intensity of light in the fringe pattern is

$I=I_0{cos}^2\left(\frac{\theta }{2}\right)$

By equating

$\frac{3}{4}{I}_0=I_0{cos}^2\left(\frac{\theta }{2}\right)\frac{\theta }{2}=\frac{\pi }{12},\frac{5\pi }{6}\theta =\frac{\pi }{6},\frac{5\pi }{3}$

Let $△\theta$ is path difference

$\frac{\pi }{6}=\frac{\pi yd}{\lambda D}y=\frac{\lambda D}{6d}$

Substitute all value in above equation

$y=\frac{6\times {10}^{-7}\times 1}{6\times 1}=0.1\mu m$

And

$\frac{5\pi }{3}=\frac{\pi yd}{\lambda D}y=\frac{5\lambda D}{3d}$

Substitute all value in above equation

$y=\frac{5\times 6\times {10}^{-7}\times 1}{3\times 1}=1\mu m$

$△y=1-0.1=0.9\mu m$