JEE PYQ: Wave Optics Question 19
Question 19 - 2020 (07 Jan Shift 1)
Visible light of wavelength $6000 \times 10^{-8}$ cm falls normally on a single slit and produces a diffraction pattern. It is found that the second diffraction minimum is at 60$°$ from the central maximum. If the first minimum is produced at $\theta_1$, then $\theta_1$ is close to:
(1) 20$°$
(2) 30$°$
(3) 25$°$
(4) 45$°$
Show Answer
Answer: (3)
Solution
Given, $\lambda = 6000 \times 10^{-8}$ cm
Second diffraction minimum at 60$°$ i.e., $\theta_2 = 60°$
Using, $d\sin\theta = n\lambda$
$d\sin\theta_2 = 2\lambda$ (for 2nd minima)
$\Rightarrow d \times \frac{\sqrt{3}}{2} = 2\lambda$ …(i)
$\Rightarrow \frac{\lambda}{d} = \frac{\sqrt{3}}{4}$
For first minima, $d\sin\theta_1 = \lambda$
$\Rightarrow \sin\theta_1 = \frac{\lambda}{d} = \frac{\sqrt{3}}{4} = 0.43 \Rightarrow \theta_1 < 30°$
Hence closest option, $\theta_1 = 25°$
BETA
