Question

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

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

Answer 1

The correct option is A $0.3\text{mm}$.

For first minimum, $\text{(a = width of the slit)}$
$a\sin {\theta }_1=\lambda$
$\Rightarrow \sin {\theta }_1=\frac{\lambda }{a}\Rightarrow {\theta }_1=\frac{\lambda }{a}(\because \lambda <<a)$

Similarly, for first maximum,
$a\sin {\theta }_2=\frac{3\lambda }{2}\Rightarrow {\theta }_2=\frac{3\lambda }{2a}$

Angular separation between them is,
$\Delta \theta ={\theta }_2-{\theta }_1=\frac{3\lambda }{2a}-\frac{\lambda }{a}=\frac{\lambda }{2a}$

$\therefore$ Linear separation between them is,
$\Delta x=D\cdot \Delta \theta =\frac{D\lambda }{2a}$
$=\frac{0.5\times (600\times {10}^{-9})}{2\times (0.5\times {10}^{-3})}=0.3\text{mm}$.

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