CBSE 2025 · Region 4 · Set 2 · Q18 · 2 marks
In a double slit experiment, the two slits are $\displaystyle 1.5$ mm apart. The slits are illuminated by a mixture of lights of wavelengths of $\displaystyle 600$ nm and $\displaystyle 400$ nm and the interference pattern is observed on a screen $\displaystyle 1.5$ m away from the slits. Find the minimum distance of the point from the central maximum at which bright fringes of the interference patterns of the two wavelengths coincide.
Marking-scheme solution
$\displaystyle n \lambda_{1}=(n+1) \lambda_{2}$
$\displaystyle n \times 600=(n+1) \times 400$
$\displaystyle \therefore n=2$
$\displaystyle \therefore x=\frac{2 \lambda_{1} D}{d}$
$\displaystyle =\frac{2 \times 600 \times 10^{-9} \times 1.5}{1.5 \times 10^{-3}}$
$\displaystyle x=1.2 \times 10^{-3} \mathrm{~m}$
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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.