Question

In young's double slit experiment,d = 1 mm.$\lambda =6000$ A & D = 1 m.The slits produe same intensity o the screen.The minimum distance between two points on the screen having 75% intensity of the maximum intensity is:

• A. 0.45 mm
• B. 0.40 mm
• C. 0.30 mm
• D. 0.20 mm

Answer 1

The correct option is C 0.30 mm

Given that,

$d=1mm=0.001m$

$\lambda =6000\dot{A}=6\times {10}^{-7}m$

$D=1m$

Intensity in YDSE is given by

$I=I_{max}{\cos }^2\left(\frac{\pi d}{\lambda D}y\right)$

According to question,

$I_{max}{\cos }^2\left(\frac{\pi d}{\lambda D}y\right)=75$

$I_{max}{\cos }^2\left(\frac{\pi d}{\lambda D}y\right)=\frac{3}{4}{I}_{max}$

$\left(\frac{\pi d}{\lambda D}y\right)=\frac{\pi }{3}$

$y=\frac{\lambda D}{3d}$

$y=\frac{6\times {10}^{-7}\times 1}{3\times 0.001}$

$y=3\times {10}^{-4}m$

$y=0.3mm$