Physics - Wave Optics Question with Solution | TestHub

The position of the $n^{\text{th}}$ minimum in a single slit diffraction pattern is given by $y_n=\frac{n\lambda D}{a}$, where $a$ is the slit width.

The separation between the $1^{\text{st}}$ and $3^{\text{rd}}$ minima is:

$\Delta y=y_3-y_1=\frac{3\lambda D}{a}-\frac{\lambda D}{a}=\frac{2\lambda D}{a}$

Given $\Delta y=3\text{ mm}=3\times 10^{-3}\text{ m}$, $D=50\text{ cm}=0.5\text{ m}$, and $\lambda =600\text{ nm}=6\times 10^{-7}\text{ m}$.

Substituting the values:

$3\times 10^{-3}=\frac{2\times 6\times 10^{-7}\times 0.5}{a}$

$a=\frac{6\times 10^{-7}}{3\times 10^{-3}}$

$a=2\times 10^{-4}\text{ m}$

$a=0.2\text{ mm}$

Answer: 0.2 mm