Light of wavelength \(\lambda_1\) and then light of wavelength \(\lambda_2\) is used with the same single slit and screen. The first minimum for \(\lambda_1\) is 3 mm from the centre. The second minimum for \(\lambda_2\) is 4 mm from the centre. The ratio \(\lambda_1 : \lambda_2\) is
Answer & explanation
Correct answer: option 1
The correct answer is Option (1) → \(3 : 2\)
For a given slit and screen, a minimum of order \( n \) lies at \( y_n = n\lambda D/a \). Both lights use the same slit and screen. So the wavelength is proportional to \( y_n/n \).
\( y_1 = 3\ \mathrm{mm}\ \ (n=1) \)
\( y_2 = 4\ \mathrm{mm}\ \ (n=2) \)
\( \begin{aligned} \frac{\lambda_1}{\lambda_2} &= \frac{3/1}{4/2} \\ &= \frac{3}{2} \end{aligned} \)
Therefore, \( \lambda_1 : \lambda_2 = 3 : 2 \).
NCERT (2026-27):
The single slit – “… The minima (zero intensity) are at”
\( \displaystyle \theta \approx \frac{n\lambda}{a} \)\( ,\quad n=\pm 1,\pm 2,\pm 3,\ldots \)