Question

In Young’s double slit experiment, the distance between the two slits is $0.1 \,mm$, the distance between the slits and the screen is $1\, m$ and the wave length of the light used is $600 \,nm$. 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 $I = 4I_{o} \cos^{2}\frac{\phi}{2},$ we get
$3I_{o}=4I_{o} \cos^{2}\frac{\phi}{2},$
$\Rightarrow \cos \frac{\phi}{2}=\frac{\sqrt{3}}{2}$
$\phi=\frac{\pi}{3}$
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
$\Delta\phi=\frac{\pi}{3}=\frac{yd}{3}=\frac{yd}{D}\times\frac{2\pi}{\lambda}$
$\Rightarrow y=\frac{\lambda D}{6d}=\frac{600\times10^{-9}\times1}{6\times0.1\times10^{-3}} $
$\Rightarrow y=1\times10^{-3}\, m$
$\Rightarrow y=1\,mm$