Visible light of wavelength 600 nm falls normally on a single slit and diffraction pattern is obtained on a screen. It is found that the second diffraction minima is at $(π/3)$. If the first minimum is obtained at $θ_1$, then $θ_1$ is
Given: $\left(\sin 25° =\frac{\sqrt{3}}{4}\right)$
Answer & explanation
Correct answer: option 3
The correct answer is Option (3) → $(5π/36)$
Given: Wavelength of light λ = 600 nm
For diffraction minima,
$a\sin\theta = m\lambda$
For second minimum,
$a\sin\theta_2 = 2\lambda$ and $\theta_2 = \frac{\pi}{3}$
For first minimum,
$a\sin\theta_1 = \lambda$
Upon dividing both equations:
$\frac{\sin\theta_1}{\sin\theta_2} = \frac{1}{2}$
$\Rightarrow \sin\theta_1 = \frac{1}{2}\sin\frac{\pi}{3} = \frac{1}{2}\times\frac{\sqrt{3}}{2} = \frac{\sqrt{3}}{4}$
Given, $\sin25^\circ = \frac{\sqrt{3}}{4}$
∴ $θ_1 = 25° = \frac{5\pi}{36}$
Final Answer: $ θ_1 = 25° = \frac{5\pi}{36}$