Question

Light of wavelength $600\eta m$ is incident normally on a slit of width $0.2mm$. 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)

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

$\because$ 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