Two concave refracting surfaces of equal radii of curvature $R$ and refractive index $1.5$ face each other in air, as shown in the figure. A point object O is placed midway between P and B. The separation between the images of O formed by each refracting surface is
Answer: (D) $0.114R$
From the figure, $AP = R$ and $PO = OB = R/2$, so O is $1.5R$ from A and $0.5R$ from B. Each surface is concave towards the air, with its centre of curvature on the air side.
Use $\dfrac{\mu_2}{v} - \dfrac{\mu_1}{u} = \dfrac{\mu_2 - \mu_1}{R}$ with distances measured along the direction of the light (into the glass), so $u$ and the radius are both negative.
**Surface A:** $u = -1.5R$, radius $-R$:
$$\frac{1.5}{v} = \frac{0.5}{-R} + \frac{1}{-1.5R} = -\frac{7}{6R} \Rightarrow v = -\frac{9R}{7} \approx -1.286R$$
A virtual image $1.286R$ from A, on the air side.
**Surface B:** $u = -0.5R$, radius $-R$:
$$\frac{1.5}{v} = \frac{0.5}{-R} + \frac{1}{-0.5R} = -\frac{2.5}{R} \Rightarrow v = -0.6R$$
A virtual image $0.6R$ from B, on the air side, i.e. $2R - 0.6R = 1.4R$ from A.
Separation: $1.4R - 1.286R \approx 0.114R$
Solution by Sreeraj P, M.Sc Physics