Q. 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 $PO=PI$. The distance $PO$ equals to
A) 5 R
B) 3 R
C) 1.5 R
D) 2 R
Explanation
[Solution by Dr. Abubacker] With $u=-x$ and $v=+x$: $\frac{\mu_2}v-\frac{\mu_1}u=\frac{\mu_2-\mu_1}R\Rightarrow\frac{1.5}x+\frac1x=\frac{0.5}R$, so $x=5R$. Answer (1).
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.