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 :

For refraction at surface B:

Light travels from air ($$\mu_1 = 1$$) to glass ($$\mu_2 = 1.5$$).

$$u = -\frac{R}{2}, \quad R_{\text{curvature}} = -R$$

$$\frac{1.5}{V_B} - \frac{1}{-R/2} = \frac{1.5 - 1}{-R}$$

$$\frac{1.5}{V_B} + \frac{2}{R} = -\frac{0.5}{R}$$

$$\frac{1.5}{V_B} = -\frac{2.5}{R} \implies V_B = -0.6R$$

For refraction at surface A:

Light travels from air ($$\mu_1 = 1$$) to glass ($$\mu_2 = 1.5$$).

$$u = -\left(R + \frac{R}{2}\right) = -\frac{3R}{2}, \quad R_{\text{curvature}} = -R$$

$$\frac{1.5}{V_A} - \frac{1}{-3R/2} = \frac{1.5 - 1}{-R}$$

$$\frac{1.5}{V_A} + \frac{2}{3R} = -\frac{0.5}{R}$$

$$\frac{1.5}{V_A} = -\frac{7}{6R} \implies V_A = -\frac{9}{7}R \approx -1.286R$$

Distance between the two images measured from a common reference point:

$$\text{Separation} = \text{Total Distance } (AB) - |V_A| - |V_B|$$

$$\text{Separation} = 2R - \frac{9}{7}R - 0.6R$$

$$\text{Separation} = 2R - 1.286R - 0.6R = 0.114R$$