Question
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 4 R in front of spherical interface. The magnitude of the magnification of point source image is $\_\_\_\_$ .
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Solution
Sol. [Key Concept: Use the refraction-at-a-spherical-surface relation $\frac{n_{2}}{v} - \frac{n_{1}}{u} = \frac{n_{2} - n_{1}}{R}$, then magnification $m = \frac{n_{1}v}{n_{2}u}$.]
$${m = \frac{n_{1}}{n_{2}}\frac{v}{u} }{\frac{n_{2}}{v} - \frac{n_{1}}{u} = \frac{n_{2} - n_{1}}{R} }{\frac{1.4}{v} + \frac{1}{4R} = \frac{1.4 - 1}{R} }{\frac{1.4}{v} = \frac{4}{10R} - \frac{1}{4R} }{v = \frac{56R}{6} = + \frac{28}{3}R }{m = \frac{1}{1.4} \cdot \frac{28R/3}{( - 4R)} = - 1.67 }$$
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