Question 11: Ray Optics

Question

A glass beaker has a solid, plano-convex base of refractive index 1.60, as shown in the figure. The radius of curvature of the convex surface (SPU) is $9 \mathrm{~cm}$, while the planar surface (STU) acts as a mirror. This beaker is filled with a liquid of refractive index $n$ up to the level QPR. If the image of a point object $\mathrm{O}$ at a height of $h$ (OT in the figure) is formed onto itself, then, which of the following option(s) is(are) correct?

Accepted Answer

Since STU is a plane mirror, we can take mirror image of the whole situation about it and final image can be assumed to be at a distance h below the base.
Since object and image are at same distance from equivalent lens, hence $$\mathrm{h}=2 \mathrm{~F}_{\mathrm{qq}}$$
$$\begin{aligned}
& \frac{1}{\mathrm{~F}_{\mathrm{eq}}}=\left(\frac{1.6-1}{1}ight)\left(\frac{2}{9}ight)+\frac{(\mathrm{n}-1)}{1}\left(\frac{-2}{9}ight) \
& \frac{1}{\frac{\mathrm{h}}{2}}=\frac{1.2}{9}+\frac{2(1-\mathrm{n})}{9} \
& \frac{2}{\mathrm{~h}}=\frac{3.2-2 \mathrm{n}}{9} \
& \mathrm{~h}=\frac{9}{1.6-\mathrm{n}} \mathrm{cm}
\end{aligned}$$
(A) for $$\mathrm{n}=1.42, \mathrm{~h}=50 \mathrm{~cm}$$
(B) for $$\mathrm{n}=1.35, \mathrm{~h}=36 \mathrm{~cm}$$
(C) for $$\mathrm{n}=1.45, \mathrm{~h}=60 \mathrm{~cm}$$
(D) for $$\mathrm{n}=1.48, \mathrm{~h}=75 \mathrm{~cm}$$

Answer

For $n=1.42, h=50 \mathrm{~cm}$.

Answer

For $n=1.35, h=36 \mathrm{~cm}$.

Answer

For $n=1.45, h=65 \mathrm{~cm}$.

Answer

For $n=1.48, h=85 \mathrm{~cm}$.

Ray Optics

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