Question

In a $YDSE$ experiment, $d=1\text{mm}$, $\lambda =6000A^{\circ }$ and $D=1\text{m}$. The minimum distance between two points on screen having $75\%$ intensity of the maximum intensity will be

• A. $0.50\text{mm}$
• B. $0.40\text{mm}$
• C. $0.30\text{mm}$
• D. $0.20\text{mm}$

Answer 1

The correct option is B $0.40\text{mm}$

Intensity at a given point on the screen is given by,

$I=I_{max}{\cos }^2\left(\frac{\varphi }{2}\right)$

According to the question,

$I=75\%\text{of}{I}_{max}=\frac{3}{4}{I}_{max}$

$\Rightarrow \frac{3}{4}{I}_{max}=I_{max}{\cos }^2\left(\frac{\varphi }{2}\right)$

$\Rightarrow \frac{\varphi }{2}=\frac{\pi }{6}\text{and}\frac{5\pi }{6}$

$\Rightarrow \varphi =\frac{\pi }{3}\text{and}\frac{5\pi }{3}$


Using $\varphi =\left(\frac{2\pi }{\lambda }\right)\Delta x\text{and}\Delta x=\frac{yd}{D}$

$\Rightarrow \varphi =\left(\frac{2\pi }{\lambda }\right)\left(\frac{yd}{D}\right)$

For, $\varphi =\frac{\pi }{3}$

$y_1=\frac{\lambda D}{6d}=\frac{(6000\times {10}^{-10})\left(1\right)}{\left(6\right)\left({10}^{-3}\right)}=0.1\times {10}^{-3}$

$\therefore y_1=0.10\text{mm}$

Similarly, for $\varphi =\frac{5\pi }{3}$

$y_2=5\left(\frac{\lambda D}{6d}\right)=5\times 0.1\times {10}^{-3}\text{m}=0.50\text{mm}$

The distance between these two points,
$\Delta y=y_2-y_1=0.40\text{mm}$

Hence, option $\left(\text{B}\right)$ is correct.