Optics Ques 37
- A slab of material of refractive index 2 shown in figure has a curved surface $A P B$ of radius of curvature $10 cm$ and a plane surface $C D$. On the left of $A P B$ is air and on the right of $C D$ is water with refractive indices as given in the figure. An object $O$ is placed at a distance of $15 cm$ from the pole $P$ as shown. The distance of the final image of $O$ from $P$, as viewed from the left is ……
$(1991,2 M)$
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Answer:
Correct Answer: 37.$30 cm$ to the right of $P$. Image will be virtual
Solution:
Formula:
Refraction at Spherical Surfaces:
- Rays starting from $O$ will suffer single refraction from spherical surface $A P B$. Therefore, applying
$$ \begin{aligned} \frac{\mu _2}{v}-\frac{\mu _1}{u} & =\frac{\mu _2-\mu _1}{R} \\ \frac{1.0}{v}-\frac{2.0}{-15} & =\frac{1.0-2.0}{-10} \\ \frac{1}{v} & =\frac{1}{10}-\frac{1}{7.5} \\ v & =-30 cm \end{aligned} $$
$$ \text { or } $$
Therefore, image of $O$ will be formed at $30 cm$ to the right of $P$. Note that image will be virtual. There will be no effect of CED.
BETA
