Wave Optics Question 2
Question 2 - 2024 (01 Feb Shift 2)
A microwave of wavelength $2.0 \mathrm{~cm}$ falls normally on a slit of width $4.0 \mathrm{~cm}$. The angular spread of the central maxima of the diffraction pattern obtained on a screen $1.5 \mathrm{~m}$ away from the slit, will be:
(1) $30^{\circ}$
(2) $15^{\circ}$
(3) $60^{\circ}$
(4) $45^{\circ}$
Show Answer
Answer: (3)
Solution:
For first minima a $\sin \theta=\lambda$
$\sin \theta=\frac{\lambda}{\mathrm{a}}=\frac{1}{2}$
$\theta=30^{\circ}$
Angular spread $=60^{\circ}$
