Optics 2 Question 14
14. A spherical surface of radius of curvature $R$, separates air (refractive index 1.0) from glass (refractive index 1.5). The centre of curvature is in the glass. A point object $P$ placed in air is found to have a real image $Q$ in the glass. The line $P Q$ cuts the surface at a point $O$ and $P O=O Q$. The distance $P O$ is equal to
$(1998,2 M)$
(a) $5 R$
(b) $3 R$
(c) $2 R$
(d) $1.5 R$
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Answer:
Correct Answer: 14. (a)
Solution:
- Let us say $P O=O Q=X$
Applying
$$ \frac{\mu _2}{v}-\frac{\mu _1}{u}=\frac{\mu _2-\mu _1}{R} $$
Substituting the values with sign
$$ \frac{1.5}{+X}-\frac{1.0}{-X}=\frac{1.5-1.0}{+R} $$
(Distances are measured from $O$ and are taken as positive in the direction of ray of light)
$$ \begin{aligned} & \therefore \quad \frac{2.5}{X}=\frac{0.5}{R} \\ & \therefore \quad X=5 R \end{aligned} $$
