CBSE 2026 · Region 2 · Set 1 · Q21 · 2 marks
A $\displaystyle 5$ cm long pencil is placed along the principal axis of a concave mirror of focal length $\displaystyle 20$ cm such that its nearest end is at a distance of $\displaystyle 25$ cm from the mirror. Calculate the length of the image of the pencil.
OR In a Young's double-slit experiment, a beam of light consisting of two wavelengths $\displaystyle 500$ nm and $\displaystyle 600$ nm is used. The interference fringes are observed at a screen placed $\displaystyle 1.8$ m away from the plane of slits (slit separation $\displaystyle 0.3$ mm). Calculate the least distance from the central maximum where the bright fringes due to both the wavelengths coincide.
Marking-scheme solution
$\displaystyle \dfrac{1}{\mathrm{f}}=\dfrac{1}{\mathrm{v}}+\dfrac{1}{u}$
$\displaystyle \dfrac{1}{v_{1}}=\dfrac{1}{\mathrm{f}}-\dfrac{1}{u_{1}}=\dfrac{1}{(-20)}-\dfrac{1}{(-25)}$ (For near end of the pencil)
$\displaystyle v_{1}=-100 \mathrm{~cm}$
$\displaystyle \dfrac{1}{v_{2}}=\dfrac{1}{(-20)}-\dfrac{1}{(-30)}$ (For far end of the pencil)
$\displaystyle v_{2}=-60 \mathrm{~cm}$
Length of the image $\displaystyle \mathrm{L}=\left|v_{1}\right|-\left|v_{2}\right|=40 \mathrm{~cm}$
OR
* $\displaystyle n_{1} \lambda_{1}=n_{2} \lambda_{2}$$\displaystyle \dfrac{n_{1}}{n_{2}}=\dfrac{\lambda_{2}}{\lambda_{1}}=\dfrac{600}{500}=\dfrac{6}{5}$
$\displaystyle y_{6}^{\max }=n_{1} \dfrac{\lambda_{1} D}{d}=\dfrac{6 \times 500 \times 10^{-9} \times 1.8}{0.3 \times 10^{-3}}$
$\displaystyle =18 \mathrm{~mm}$
Alternatively: $\displaystyle n \lambda_{1}=(n+1) \lambda_{2}$, we get $\displaystyle n=5$$\displaystyle y_{5}^{\max }=\dfrac{n_{2} \lambda_{2} \mathrm{D}}{d}=\dfrac{5 \times 600 \times 10^{-9} \times 1.8}{0.3 \times 10^{-3}}=18 \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.