With light of wavelength 600 nm, the first minimum of a single-slit pattern is at an angle of \(0.15^\circ\). The same slit is now used with light of wavelength 400 nm. The angle of the second minimum is
Answer & explanation
Correct answer: option 1
The correct answer is Option (1) → \(0.20^\circ\)
For small angles the minima are at \( \theta \approx n\lambda/a \). The slit width does not change. So the angle of a minimum is proportional to the product \( n\lambda \).
\( \lambda = 600\ \mathrm{nm},\ \ \theta_1 = 0.15^\circ \)
\( n = 2,\ \ \lambda^{\prime} = 400\ \mathrm{nm} \)
\( \begin{aligned} \frac{\theta_2^{\prime}}{\theta_1} &= \frac{2\times 400}{1\times 600} \\ &= \frac{4}{3} \\ \theta_2^{\prime} &= \frac{4}{3}\times 0.15^\circ \\ &= 0.20^\circ \end{aligned} \)
Therefore, the second minimum for the shorter wavelength is at \( 0.20^\circ \).
NCERT (2026-27):
The single slit – “… The minima (zero intensity) are at”
\( \displaystyle \theta \approx \frac{n\lambda}{a} \)\( ,\quad n=\pm 1,\pm 2,\pm 3,\ldots \)