MHT CET · Physics · Ray Optics
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
- A \(\frac{\mathrm{R}}{2}\)
- B 2R
- C 5 R
- D 6 R
Answer & Solution
Correct Answer
(D) 6 R
Step-by-step Solution
Detailed explanation
Let \(\mathrm{PO}=\mathrm{QO}=\mathrm{x}\)
Given: \(\mathrm{u}=-\mathrm{x}, \mathrm{v}={ }^{+}+\mathrm{x}\)
We know,
\(\begin{aligned}
& \frac{\mathrm{n}_2}{\mathrm{v}}-\frac{\mathrm{n}_1}{\mathrm{u}}=\frac{\mathrm{n}_2-\mathrm{n}_1}{\mathrm{R}} \\
& \Rightarrow \frac{1.4}{\mathrm{x}}+\frac{1}{\mathrm{x}}=\frac{0.4}{\mathrm{R}} \\
& \frac{2.4}{\mathrm{x}}=\frac{0.4}{\mathrm{R}} \\
\therefore \quad & x=6 \mathrm{R}
\end{aligned}\)
Given: \(\mathrm{u}=-\mathrm{x}, \mathrm{v}={ }^{+}+\mathrm{x}\)
We know,
\(\begin{aligned}
& \frac{\mathrm{n}_2}{\mathrm{v}}-\frac{\mathrm{n}_1}{\mathrm{u}}=\frac{\mathrm{n}_2-\mathrm{n}_1}{\mathrm{R}} \\
& \Rightarrow \frac{1.4}{\mathrm{x}}+\frac{1}{\mathrm{x}}=\frac{0.4}{\mathrm{R}} \\
& \frac{2.4}{\mathrm{x}}=\frac{0.4}{\mathrm{R}} \\
\therefore \quad & x=6 \mathrm{R}
\end{aligned}\)
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