CBSE 2025 · Region 5 · Set 1 · Q18 · 2 marks
A point object is placed in air at a distance $\displaystyle \mathrm{R} / 3$ in front of a convex surface of radius of curvature R, separating air from a medium of refractive index $\displaystyle \mathrm{n}(<4)$. Find the nature and position of the image formed.In Young's double slit experimental set-up, the intensity of the central maximum is $\displaystyle \mathrm{I}_{0}$. Calculate the intensity at a point where the path difference between two interfering waves is $\displaystyle \lambda / 3$.
A point object is placed in air at a distance $\displaystyle \mathrm{R} / 3$ in front of a convex surface of radius of curvature R, separating air from a medium of refractive index $\displaystyle \mathrm{n}(<4)$. Find the nature and position of the image formed.
In Young's double slit experimental set-up, the intensity of the central maximum is $\displaystyle \mathrm{I}_{0}$. Calculate the intensity at a point where the path difference between two interfering waves is $\displaystyle \lambda / 3$.
Marking-scheme solution
For refraction at convex surface:
$\displaystyle \frac{n_{1}}{-u}+\frac{n_{2}}{v}=\frac{n_{2}-n_{1}}{R}$
$\displaystyle \frac{n}{v}=\frac{n-1-3}{R}$
$\displaystyle v=\frac{n R}{n-4}$
For all values of $\displaystyle n<4$, the value of v is negative and greater than R.
Therefore the nature of the image is virtual and it is formed in front of the convex surface.
$\displaystyle \phi=\frac{2 \pi}{\lambda} \times \Delta x$
$\displaystyle =\frac{2 \pi}{\lambda} \times \frac{\lambda}{3}$
$\displaystyle =\frac{2 \pi}{3}$
$\displaystyle I^{\prime}=4 I \cos ^{2} \frac{\phi}{2}$ (Given $\displaystyle 4 I=I_{0}$)
$\displaystyle =I_{0} \cos ^{2} \frac{2 \pi}{6}$
$\displaystyle =\frac{I_{0}}{4}$
Ray Optics and Optical InstrumentsRefraction at Spherical Surfaces and by LensesApplyvery_short_answermedium
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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.