Question

A parallel beam of monochromatic light of wavelength $600\text{nm}$ falls normally on a slit of width $0.5\text{mm}$. The resulting diffraction pattern is observed on a screen placed of $50\text{cm}$ from the slit. The linear separation between the first minimum and the first maximum on the same side of the central maximum is

• A. $3\text{mm}$
• B. $0.3\text{mm}$
• C. $0.003\text{mm}$
• D. $0.03\text{mm}$

Answer 1

The correct option is B $0.3\text{mm}$

Given:
$\lambda =600\text{nm};b=0.5\text{mm};D=50\text{cm}$

The angular position of the first minimum from the central maximum is given by,

$b\sin {\theta }_1=\lambda \Rightarrow \sin {\theta }_1=\frac{\lambda }{b}$

For small $\theta \to \sin \theta \approx \theta$

$\therefore {\theta }_1=\frac{\lambda }{b}$

The angular position of the first maximum from the central maximum is given by,

$b\sin {\theta }_2=\frac{3\lambda }{2}$

$\therefore {\theta }_2=\frac{3\lambda }{2b}$

Therefore, the angular separation between the first minimum and first maximum on the same side of the central maximum is given by,

$\Delta \theta ={\theta }_2-{\theta }_1$

$\Rightarrow \Delta \theta =\frac{3\lambda }{2b}-\frac{\lambda }{b}=\frac{\lambda }{2b}$

Therefore, linear separation will be,

$\Delta x=D\Delta \theta =\frac{D\lambda }{2b}$

$\Rightarrow \Delta x=\frac{0.5\times 600\times {10}^{-9}}{2\times 0.5\times {10}^{-3}}$

$\Rightarrow \Delta x=3\times {10}^{-4}\text{m}=0.3\text{mm}$

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