CBSE 2026 · Region 1 · Set 1 · Q18 · 2 marks
A beam of light consisting of two wavelengths $\displaystyle 400$ nm and $\displaystyle 600$ nm is used to illuminate a single slit of width $\displaystyle 1$ mm. Find the least distance of the point from the central maximum where the dark fringes due to both wavelengths coincide on the screen placed $\displaystyle 1.5$ m from the slit.In a Young's double-slit experimental set-up with slit separation $\displaystyle 0.6$ mm a beam of light consisting of two wavelengths $\displaystyle 440$ nm and $\displaystyle 660$ nm is used to obtain interference pattern on a screen kept $\displaystyle 1.5$ m in front of the slits. Find the least distance of the point from the central maximum where the bright fringes due to both the wavelengths coincide.
A beam of light consisting of two wavelengths $\displaystyle 400$ nm and $\displaystyle 600$ nm is used to illuminate a single slit of width $\displaystyle 1$ mm. Find the least distance of the point from the central maximum where the dark fringes due to both wavelengths coincide on the screen placed $\displaystyle 1.5$ m from the slit.
In a Young's double-slit experimental set-up with slit separation $\displaystyle 0.6$ mm a beam of light consisting of two wavelengths $\displaystyle 440$ nm and $\displaystyle 660$ nm is used to obtain interference pattern on a screen kept $\displaystyle 1.5$ m in front of the slits. Find the least distance of the point from the central maximum where the bright fringes due to both the wavelengths coincide.
Marking-scheme solution
At the point of coincidence for dark fringes.
$\displaystyle d \sin \theta=n \lambda_{2}=(n+1) \lambda_{1}$
$\displaystyle n \times 600=(n+1) \times 400$
$\displaystyle n=2$
Position of dark fringe from the central maxima
$\displaystyle y=\dfrac{(n+1) \lambda_{1} D}{d}$
$\displaystyle y=\dfrac{3 \times 400 \times 10^{-9} \times 1.5}{1 \times 10^{-3}}$
$\displaystyle y=1.8 \mathrm{~mm}$
Alternatively:
$\displaystyle y=\dfrac{n \lambda_{2} D}{d}$
$\displaystyle y=\dfrac{2 \times 600 \times 10^{-9} \times 1.5}{1 \times 10^{-3}}$
$\displaystyle y=1.8 \mathrm{~mm}$
At the point of coincidence for bright fringes
$\displaystyle \dfrac{n \lambda_{2} D}{d}=\dfrac{(n+1) \lambda_{1} D}{d}$
$\displaystyle n \times 660=(n+1) \times 440$
$\displaystyle n=2$
Position of bright fringe from the central maxima
$\displaystyle y=\dfrac{n \lambda_{2} D}{d}$
$\displaystyle y=\dfrac{2 \times 660 \times 10^{-9} \times 1.5}{0.6 \times 10^{-3}}$
$\displaystyle y=3.3 \mathrm{~mm}$
Alternatively:
$\displaystyle y=\dfrac{(n+1) \lambda_{1} D}{d}$
$\displaystyle y=3.3 \mathrm{~mm}$
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CBSE Class 12 Physics past-paper question from the 2026board exam, with the answer as CBSE’s own marking scheme gives it. Where our answers come from.