Question

Light of wavelength $600 \, nm$ 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,
$x=\left(2 \theta \right)D$
Where, $2\theta :$ Angular width of central maximum.
We know that, $x=2\frac{\lambda D}{d}$
Angular width $2\theta .D=2\frac{\lambda D}{d}$
$\Rightarrow \, 2\theta =\frac{2 \lambda }{d}=\frac{2 \times 600 \times 1 0^{- 9}}{0.2 \times 1 0^{- 3}}$
$=6\times 10^{- 7}\times 10^{4}$
$=6\times 10^{- 3} \, rad$