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 ————.
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Use the formula for refraction at a spherical surface: (n2/v) - (n1/u) = (n2 - n1)/R and then use the magnification formula m = (n1*v)/(n2*u).
The magnification produced by a spherical refracting surface is the ratio of image height to object height. For a point source on the axis, we look at the lateral magnification formula: $m = \frac{n_1 v}{n_2 u}$.
First, let's establish the image location using the lens maker's related formula for a single surface: $\frac{n_2}{v} - \frac{n_1}{u} = \frac{n_2 - n_1}{R}$. Here, $n_1 = 1$, $n_2 = 1.4$, $u = -4R$, and $R = R$.
Calculating image distance $v$: $$\frac{1.4}{v} + \frac{1}{4R} = \frac{0.4}{R}$$ Multiplying the whole equation by $4Rv$: $$1.4(4R) + v = 0.4(4v)$$ $$5.6R + v = 1.6v$$ $$5.6R = 0.6v$$ $$v = \frac{5.6}{0.6} R = \frac{56}{6} R = \frac{28}{3} R$$
Now, applying the magnification formula: $$m = \frac{n_1}{n_2} \cdot \frac{v}{u}$$ $$m = \frac{1}{1.4} \cdot \frac{28R/3}{-4R}$$ $$m = \frac{1}{1.4} \cdot \left( -\frac{28}{12} \right) = \frac{1}{1.4} \cdot \left( -\frac{7}{3} \right)$$ $$m = -\frac{7}{4.2} = -\frac{70}{42} = -\frac{5}{3} \approx -1.666$$ The magnitude of the magnification is $| -5/3 | = 1.66$.
