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Physics · 2025 · 3 marks
CBSE 2025 · Region 5 · Set 1 · Q24
In Young's double slit experiment, the separation between the two slits is 1.0 mm and the screen is 1.0 m away from the slits. A beam of light…
In Young's double slit experiment, the separation between the two slits is $\displaystyle 1.0$ mm and the screen is $\displaystyle 1.0$ m away from the slits. A beam of light consisting of two wavelengths $\displaystyle 500$ nm and $\displaystyle 600$ nm is used to obtain interference fringes. Calculate :
(a)
the distance between the first maxima for the two wavelengths.
(b)
the least distance from the central maximum, where the bright fringes due to both the wavelengths coincide.
Marking-scheme solution
(a)
Distance $\displaystyle =\frac{n \lambda_{1} D}{d}-\frac{n \lambda_{2} D}{d}$
For $\displaystyle n=1$:
Distance $\displaystyle =\frac{(600-500) \times 10^{-9} \times 1}{10^{-3}}$
$\displaystyle =10^{-4} \mathrm{~m}$
(b)
$\displaystyle n \lambda_{1} \frac{D}{d}=(n+1) \lambda_{2} \frac{D}{d}$
$\displaystyle n \times 600 \times 10^{-9}=(n+1) \times 500 \times 10^{-9}$
$\displaystyle n=5$
$\displaystyle x=5 \times \frac{\lambda_{1} D}{d}$
$\displaystyle =\frac{5 \times 600 \times 10^{-9} \times 1}{10^{-3}}=3 \mathrm{~mm}$
OR
$\displaystyle n_{1} \lambda_{1}=n_{2} \lambda_{2}$
$\displaystyle \frac{n_{1}}{n_{2}}=\frac{5}{6}$
therefore $\displaystyle n=5$
Position of 5th bright fringe for $\displaystyle \lambda_{1}$ ($\displaystyle 600$ nm): $\displaystyle x=5 \times \frac{\lambda_{1} D}{d}=3 \mathrm{~mm}$
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CBSE Class 12 Physics past-paper question from the 2025 board exam, with the answer as CBSE’s own marking scheme gives it. Where our answers come from.