CBSE 2025 · Region 5 · Set 1 · Q24 · 3 marks
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.
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)
$\displaystyle n \lambda_{1} \frac{D}{d}=(n+1) \lambda_{2} \frac{D}{d}$
$\displaystyle n \times 600 \times 10^{-9}=(n+1) \times 500 \times 10^{-9}$
$\displaystyle n=5$
$\displaystyle x=5 \times \frac{\lambda_{1} D}{d}$
$\displaystyle =\frac{5 \times 600 \times 10^{-9} \times 1}{10^{-3}}=3 \mathrm{~mm}$
$\displaystyle n_{1} \lambda_{1}=n_{2} \lambda_{2}$
$\displaystyle \frac{n_{1}}{n_{2}}=\frac{5}{6}$
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 2025board exam, with the answer as CBSE’s own marking scheme gives it. Where our answers come from.