JEE Mains · Physics · STD 12 - Ray optics and optical instruments
- JEE Mains 2025
- Medium
- MCQ
A spherical surface of radius of curvature \(R\), separates air from glass (refractive index \(=1.5\) ). The centre of curvature is in the glass medium. A point object ' \(O\) ' placed in air on the optic axis of the surface, so that its real image is formed at ' \(I\) ' inside glass. The line OI intersects the spherical surface at P and \(\mathrm{PO}=\mathrm{PI}\). The distance PO equals to ________.
Pick an option to check your answer.
Answer & Solution
Correct Answer
(A) 5 R
Step-by-step Solution
Detailed explanation
\(\begin{aligned} & \mathrm{PO}=\mathrm{u}=-\mathrm{x} \\ & \mathrm{PI}=\mathrm{v}=\mathrm{x} \\ & \mathrm{PO}=\mathrm{PI} \\ & \frac{\mu_2}{\mathrm{v}}-\frac{\mu_1}{\mathrm{u}}=\frac{\mu_2-\mu_1}{\mathrm{R}} \\ & \frac{1.5}{\mathrm{x}}+\frac{1}{\mathrm{x}}=\frac{1}{2 \mathrm{R}} \\ & \frac{5}{2 \mathrm{x}}=\frac{1}{2 \mathrm{R}} \\ & \mathrm{X}=5 \mathrm{R}\end{aligned}\)
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