Question

Light of wavelength $600\, \eta m$ is incident normally on a slit of width $0.2 \,mm$. The angular width of central maxima in the diffraction pattern is (measured from minimum to minimum)

• A. $6 \times 10^{-3}$ rad
• B. $4 \times 10^{-3}$ rad
• C. $2.4 \times 10^{-3}$ rad
• D. $4.5 \times 10^{-3}$ rad

Answer 1

Answer: (A)

$\lambda=600\, nm =600 \times 10^{-9} m$
$d=0.2 \,mm =0.2 \times 10^{-3} m$

$\because$ Linear width of central maximum,
$x=(2 \theta) \cdot D$
where, $2 \theta$ : Angular width of central maximum.
We know that, $x=2 \frac{\lambda D}{d}$
Angular width $2 \theta . D=\frac{2 \lambda D}{d}$
$\Rightarrow 2\theta = \frac{2\lambda}{d} = \frac{2\times 600\times 10^{-9}}{0.2 \times 10^{-3}}$
$ = 6\times 10^{-7} \times 10^4$
$= 6\times 10^{-3} $ rad