A spherical interface lens of radius R separates two media of...

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: $1.66$