A spherical interface lens of radius $$R$$ separates two media of refractive indices 1 and 1.4 respectively as shown in the figure below. A point source is placed at a distance of $$4R$$ in front of spherical interface. The magnitude of the magnification of point source image is _______.

$$\frac{n_2}{v} - \frac{n_1}{u} = \frac{n_2 - n_1}{R}$$

$$m = \frac{n_1 v}{n_2 u}$$

$$\frac{1.4}{v} - \frac{1}{-4R} = \frac{1.4 - 1}{R}$$

$$\frac{1.4}{v} + \frac{1}{4R} = \frac{0.4}{R}$$

$$v = \frac{1.4 \times 4R}{0.6} = \frac{5.6R}{0.6} = \frac{28}{3}R$$

$$|m| = \left| \frac{n_1 v}{n_2 u} \right|$$

$$|m| = \left| \frac{1 \times \left(\frac{28}{3}R\right)}{1.4 \times (-4R)} \right|$$

$$|m| = \frac{\frac{28}{3}R}{5.6R} = \frac{28}{3 \times 5.6} = \frac{28}{16.8} = \frac{5}{3} \approx 1.66$$