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A sphere with its back silvered can act as a retroreflector. When a fine beam of light is directed to the sphere as shown, it is refracted at the front surface, and focused on the rear interior surface. Then, it is reflected, and finally leaves the sphere in the incident direction. To achieve this, what is the approx. refractive index of the sphere?

I've tried thinking through this, but I don't quite understand why a precise refractive index is required since if a ray gets refracted into the sphere, if it gets reflected along the same ray as it exits. Regardless of refractive index, it should always leave with the same incident ray. Any thoughts?

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    $\begingroup$ Your last statement is manifestly untrue. Consider a sphere with refractive index one, in which case it just acts as a hemispherical mirror. Assuming the question means light rays parallel to the optical axis, these light rays will be reflected to pass through the focal point (in the paraxial approximation). So the angle of the reflected ray will depend on the distance from the optical axis. $\endgroup$ – John Rennie Apr 12 '16 at 16:19
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The only way a collimated beam will focus on the rear surface of a sphere is if the diameter of the sphere divided by the refractive index is equal to the focal length. Since the focal length is equal to the radius of the sphere divided by the change in refractive index (n-1), this can only occur when:

2R = nf = n (R/(n-1))

2 = n/(n-1).

Thus n=2. This would make a good homework question for an introductory optics course.

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