Q66 / 75 · JEE Main 2025 (29 Jan Shift 2) · Physics

Figure
Two concave refracting surfaces of equal radii of curvature and refractive index 1.5 face each other in air as shown in 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 :

  • A
    0.214 R
  • B
    0.411 R
  • C
    0.124 R
  • Correct answer
    D
    0.114 R

Solution

For B $\begin{aligned} & \frac{\mu_2}{\mathrm{~V}}-\frac{\mu_1}{\mathrm{u}}=\frac{\mu_2-\mu_1}{\mathrm{R}} \\ & \frac{1.5}{\mathrm{~V}}+\frac{1}{\frac{R}{2}}=\frac{0.5}{-\mathrm{R}} \\ & \frac{1.5}{\mathrm{~V}}=-\frac{1}{2 \mathrm{R}}-\frac{2}{\mathrm{R}} \\ & \frac{1.5}{\mathrm{~V}}=\frac{-5}{2 \mathrm{R}} \Rightarrow V_B=-0.6 \mathrm{R} \end{aligned}$ For A $\begin{aligned} & \frac{1.5}{V}+\frac{2}{3 R}=\frac{0.5}{-R} \\ & \frac{1.5}{V}=-\frac{1}{2 R}-\frac{2}{3 R} \\ & \frac{1.5}{V}=-\frac{7}{6 R} \\ & V_A=-\frac{9}{7} R \end{aligned}$ Distance between images $=2 \mathrm{R}-\left(0.6 \mathrm{R}+\frac{9}{7} \mathrm{R}\right)=0.114 \mathrm{R}$

Hence, the correct answer is option: 'D'.

Relevant Questions