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: $\left(\sin 25° =\frac{\sqrt{3}}{4}\right)$

$(5π/36)$

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}$