A solid glass sphere of refractive index 1.5 and radius 10 cm is…

Question

A solid glass sphere of refractive index 1.5 and radius 10 cm is placed in air. A point object lies on the axis 30 cm in front of the sphere's nearest surface. Light refracts at the front surface, travels through the glass, and refracts at the back surface. Find the distance (in cm) of the final image from the back surface of the sphere (take a positive value to mean the image is on the far side, i.e. real).

Accepted Answer

First surface (air->glass, R=+10): n2/v - n1/u = (n2-n1)/R.

With u=-30: 1.5/v = 0.5/10 + 1/(-30) = 0.05 - 0.0333 = 1/60, so v1 = 90 cm (measured from front surface).

This image lies 90-20 = 70 cm beyond the back surface, so it is a virtual object for the back surface at u2 = +70.

Second surface (glass->air, R=-10): n2/v - n1/u = (n2-n1)/R gives 1/v - 1.5/70 = (1-1.5)/(-10) = 0.05, so 1/v = 0.05 + 1.5/70 = 0.05 + 0.021428 = 0.071428, v2 = 14 cm.

Final answer14

Final answer: 14

A solid glass sphere of refractive index 1.5 and radius 10 cm is…

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