A point object kept at $P$ in front of a glass sphere of radius ' $R$ ' has its image formed at $Q$ such…

A point object kept at $P$ in front of a glass sphere of radius ' $R$ ' has its image formed at $Q$ such that $\mathrm{PO}=\mathrm{QO}$. The refractive index of material of glass sphere is 1.4. The distance PO is equal to
  1. $\frac{\mathrm{R}}{2}$
  2. 2R
  3. 5 R
  4. 6 R

Solution

Let $\mathrm{PO}=\mathrm{QO}=\mathrm{x}$ Given: $\mathrm{u}=-\mathrm{x}, \mathrm{v}={ }^{+}+\mathrm{x}$ We know, $\begin{aligned} & \frac{\mathrm{n}_2}{\mathrm{v}}-\frac{\mathrm{n}_1}{\mathrm{u}}=\frac{\mathrm{n}_2-\mathrm{n}_1}{\mathrm{R}} \\ & \Rightarrow \frac{1.4}{\mathrm{x}}+\frac{1}{\mathrm{x}}=\frac{0.4}{\mathrm{R}} \\ & \frac{2.4}{\mathrm{x}}=\frac{0.4}{\mathrm{R}} \\ \therefore \quad & x=6 \mathrm{R} \end{aligned}$

Asked in: MHT CET 2024 (09 May Shift 1)

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