Question

In Young’s double slit experiment, the distance between the two slits is $0.1mm$, the distance between the slits and the screen is $1m$ and the wave length of the light used is $600nm$. The intensity at a point on the screen is $75\%$ of the maximum intensity . What is the smallest distance of this point from the central fringe?

• A. 1.0 mm
• B. 2.0 mm
• C. 0.5 mm
• D. 1.5 mm

Answer 1

Answer: (A)

Let intensity of source is $I_0$.
Then, intensity of maxima is $4I_0$.
So, intensity at given point on screen $=75\%$ of maximum intensity $=3I_0$
Now, using $I=4I_o{\cos }^2\frac{\varphi }{2},$ we get
$3I_o=4I_o{\cos }^2\frac{\varphi }{2},$
$\Rightarrow \cos \frac{\varphi }{2}=\frac{\sqrt{3}}{2}$
$\varphi =\frac{\pi }{3}$
So, phase difference between raysreaching given point from upper andlower slit is $\frac{\pi }{3}.$
Hence, by using formula

For path difference in Young’s experiment, we get
$\Delta \varphi =\frac{\pi }{3}=\frac{yd}{3}=\frac{yd}{D}\times \frac{2\pi }{\lambda }$
$\Rightarrow y=\frac{\lambda D}{6d}=\frac{600\times 10^{-9}\times 1}{6\times 0.1\times 10^{-3}}$
$\Rightarrow y=1\times 10^{-3}m$
$\Rightarrow y=1mm$