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 _______.

Given:
Refractive index of the first medium, \(n_1 = 1\)
Refractive index of the second medium, \(n_2 = 1.4 = \dfrac{7}{5}\)
Object distance, \(u = -4R\)
Radius of curvature of the spherical interface, \(R = +R\) (since the surface is convex towards the rarer medium)

Using the formula for refraction at a single spherical surface:
\(\dfrac{n_2}{v} - \dfrac{n_1}{u} = \dfrac{n_2 - n_1}{R}\)

Substituting the given values:
\(\dfrac{1.4}{v} - \dfrac{1}{-4R} = \dfrac{1.4 - 1}{R}\)

\(\dfrac{7}{5v} + \dfrac{1}{4R} = \dfrac{0.4}{R}\)

\(\dfrac{7}{5v} + \dfrac{1}{4R} = \dfrac{2}{5R}\)

\(\dfrac{7}{5v} = \dfrac{2}{5R} - \dfrac{1}{4R}\)

\(\dfrac{7}{5v} = \dfrac{8 - 5}{20R} = \dfrac{3}{20R}\)

\(v = \dfrac{7 \times 20R}{5 \times 3} = \dfrac{28R}{3}\)

The transverse magnification \(m\) for a spherical refracting surface is given by:
\(m = \dfrac{n_1 v}{n_2 u}\)

Substituting the values of \(n_1\), \(n_2\), \(u\), and \(v\):
\(m = \dfrac{1 \times \left(\dfrac{28R}{3}\right)}{1.4 \times (-4R)}\)

\(m = \dfrac{\dfrac{28R}{3}}{-\dfrac{28R}{5}}\)

\(m = -\dfrac{5}{3}\)

The magnitude of the magnification is:
\(|m| = \dfrac{5}{3} \approx 1.66\)

Answer: \(1.66\)