Q. Given a thin convex lens (refractive index $\mu_2$), kept in a liquid (refractive index $\mu_1$, $\mu_1<\mu_2$) having radii of curvatures $|R_1|$ and $|R_2|$. Its second surface is silver polished. Where should an object be placed on the optic axis so that a real and inverted image is formed at the same place?
A) $\frac{\mu_1|R_1|\cdot|R_2|}{\mu_2\left(|R_1|+|R_2|\right)-\mu_1|R_2|}$
B) $\frac{\mu_1|R_1|\cdot|R_2|}{\mu_2\left(|R_1|+|R_2|\right)-\mu_1|R_1|}$
C) $\frac{(\mu_2+\mu_1)|R_1|}{(\mu_2-\mu_1)}$
D) $\frac{\mu_1|R_1|\cdot|R_2|}{\mu_2\left(2|R_1|+|R_2|\right)-\mu_1\sqrt{|R_1|\cdot|R_2|}}$
Explanation
[Solution by Dr. Abubacker] The image coincides with the object only if the rays hit the silvered surface normally, i.e. they diverge from its centre of curvature, a distance $|R_2|$ before the lens. Refraction at the first surface with $u=-x$, $v=-|R_2|$: $\frac{\mu_2}{-|R_2|}+\frac{\mu_1}x=\frac{\mu_2-\mu_1}{|R_1|}$. So $\frac{\mu_1}x=\frac{\mu_2-\mu_1}{|R_1|}+\frac{\mu_2}{|R_2|}$ and $x=\frac{\mu_1|R_1||R_2|}{\mu_2(|R_1|+|R_2|)-\mu_1|R_2|}$. Answer (1).
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.