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)

$6\times {10}^{-3}rad$

Given : $\lambda =600nm=6\times {10}^{-7}$ $m$ $b=0.2mm=0.2\times {10}^{-3}$

Angular width of first minima from central maxima in diffraction pattern $\theta =\frac{\lambda }{d}$

$\therefore$ Angular width of central maxima, ${\theta }_c=2\theta =\frac{2\lambda }{d}$

$⟹$ ${\theta }_c=\frac{2\times 6\times {10}^{-7}}{0.2\times {10}^{-3}}=6\times {10}^{-3}$ $rad$