Question

In $\text{YDSE}$, the two slits are separated by $0.1\text{mm}$ and they are $0.5\text{m}$ from the screen. The wavelength of light used is $5000\dot{\text{A}}$. Find the distance between the $7^{th}$ maxima and ${11}^{th}$ minima on the upper side of the screen.

• A. $3.45\text{mm}$
• B. $7.60\text{mm}$
• C. $8.75\text{mm}$
• D. $10.2\text{mm}$

Answer 1

The correct option is C $8.75\text{mm}$

Given:
$d=0.1\text{mm}={10}^{-4}\text{m}$
$D=0.5\text{m}$
$\lambda =5000\dot{\text{A}}=5.0\times {10}^{-7}\text{m}$

The position of minima is given by,

$y_{min}=\frac{(2n-1)\lambda D}{2d}$

The position of maxima is given by,

$y_{max}=\frac{n\lambda D}{d}$

$\Delta y=y_{11min}-y_{7max}=\frac{(2\times 11-1)\lambda D}{2d}-\frac{7\lambda D}{d}$

$=\frac{7\lambda D}{2d}=\frac{7\times 5.0\times {10}^{-7}\times 0.5}{2\times {10}^{-4}}$

$=8.75\times {10}^{-3}\text{m}=8.75\text{mm}$

Hence, option $\left(C\right)$ is correct.