In a single slit diffraction experiment, first minima for \( \lambda_1 = 600\,\mathrm{nm} \) coincides with first maxima for wavelength \( \lambda_2 \). The value of \( \lambda_2 \) is
Answer & explanation
Correct answer: option 1
The correct answer is Option (1) → 400 nm
For a single slit of width \( a \), the minima lie at \( a\sin\theta = n\lambda \), and the secondary maxima lie approximately midway, at \( a\sin\theta = (n + 1/2)\lambda \). The “first maxima” here is the first secondary maximum (\( n = 1 \)), since the central maximum is at \( \theta = 0 \).
\( \text{first minimum of } \lambda_1:\ a\sin\theta = \lambda_1 \)
\( \displaystyle \text{first secondary maximum of } \lambda_2:\ a\sin\theta = \frac{3}{2}\lambda_2 \)
The two coincide, so the angles are the same:
\( \displaystyle \frac{3}{2}\lambda_2 = \lambda_1 \;\Rightarrow\; \lambda_2 = \frac{2}{3} \times 600\,\mathrm{nm} = 400\,\mathrm{nm} \)