Question

In a Young's double slit experiment, $d=1\text{mm},\lambda =6000\mathring{\text{A}}$ and $D=1\text{m}$. The two slits have equal intensity. The minimum distance between two points on the screen, having $75\%$ intensity of the maximum intensity is,

• A. $0.45\text{mm}$
• B. $0.40\text{mm}$
• C. $0.30\text{mm}$
• D. $0.20\text{mm}$

Answer 1

The correct option is D $0.20\text{mm}$

Lets look at the screen.


For minimum distance between the regions of $75\%$ of maximum intensity, we have to consider the two such regions on the two sides of a bright fringe.

Let $I_0$ is the intensity of each slit. So, the intensity at a maxima will be $4I_0$

$\therefore 75\%$ intensity will correspond to a point where intensity is $3I_0$.

Now, resultant intensity at a point is given by,

$I_R=I_0+I_0+2\sqrt{I_0}\sqrt{I_0}\cos \left(\varphi \right)$

$\Rightarrow 3I_0=2I_0(1+\cos \varphi )$

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

$\therefore \varphi =\frac{\pi }{3},2\pi -\frac{\pi }{3},2\pi +\frac{\pi }{3},........$

And, $\Delta x=\frac{\lambda }{6},\lambda -\frac{\lambda }{6},\lambda +\frac{\lambda }{6},........$


Now, $\Delta x=\frac{yd}{D}$

$\Rightarrow y=(\frac{D}{d}\times \frac{\lambda }{6}),(\frac{\lambda D}{d}-\frac{D}{d}\times \frac{\lambda }{6}),........$

$y=\frac{\beta }{6},\beta -\frac{\beta }{6},\beta +\frac{\beta }{6},........$

$y_{min}=\frac{\beta }{6}+\frac{\beta }{6}=\frac{\beta }{3}$

$\therefore y_{min}=\frac{1}{3}\times \frac{\lambda D}{d}$

$=\frac{6000\times {10}^{-10}\times 1}{3\times {10}^{-3}}=0.2\text{mm}$

Hence, $\left(D\right)$ is the correct answer.