Magnification of Point Source Image by Spherical Refracting Surface

To find the magnitude of the magnification of the image formed by the spherical refracting surface, we use the refraction formula for a single spherical surface:

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

Given:

• Refractive index of the first medium, $n_1 = 1$
• Refractive index of the second medium, $n_2 = 1.4 = \frac{7}{5}$
• Object distance, $u = -4R$
• Radius of curvature of the spherical surface, $+R$ (center of curvature lies in the second medium)

Substituting the given values into the refraction formula:

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

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

$\frac{1.4}{v} = \frac{0.4}{R} - \frac{0.25}{R}$

$\frac{1.4}{v} = \frac{0.15}{R}$

$v = \frac{1.4}{0.15} R = \frac{28}{3} R$

The formula for the transverse magnification ($m$) for refraction at a spherical interface is given by:

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

Substituting the values of $n_1$, $n_2$, $u$, and $v$:

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

$m = \frac{\frac{28}{3} R}{-5.6 R} = -\frac{28}{16.8} = -\frac{5}{3} \approx -1.666$

The magnitude of the magnification is:

$|m| = \frac{5}{3} \approx 1.66$

Thus, the correct option is A.