CBSE 2025 · Region 4 · Set 1 · Q18 · 2 marks
Find the angle of diffraction (in degrees) for first secondary maximum of the pattern due to diffraction at a single slit. The width of the slit and wavelength of light used are $\displaystyle 0.55$ mm and $\displaystyle 550$ nm , respectively.
Marking-scheme solution
$\displaystyle \theta=(2 n+1) \frac{\lambda}{2 a}$
For first secondary maximum $\displaystyle n=1$:
$\displaystyle \theta=\frac{3 \lambda}{2 a}$
$\displaystyle \theta=\frac{3 \times 550 \times 10^{-9}}{2 \times 0.55 \times 10^{-3}}$
$\displaystyle \theta=1.5 \times 10^{-3}$ radian $\displaystyle =0.086$ degree
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CBSE Class 12 Physics past-paper question from the 2025board exam, with the answer as CBSE’s own marking scheme gives it. Where our answers come from.