A glass rod 96 cm long with an index of refraction of 1.6 has its ends ground to

Question

A glass rod 96 cm long with an index of refraction of 1.6 has its ends ground to convex spherical surfaces of radii 8 cm and 16 cm. A point object is in air on the axis of the rod 19.1 cm from the end with the 8 cm radius.
(a) Find the image distance due to refraction at the first surface. (Include the sign.)
. 70.65895954 cm <----- part is answered correctly already.

(b) Find the location of the final image distance due to refraction at both surfaces.
. ______________ cm (from the 8 cm diameter surface)

I just need help with part (b) . Will rate lifesaver if answer is correct, thanks.

Explanation / Answer

I will have: 1/20 + 1.6/s = (1.6-1)/8. So then I get 64 cm. This means that the image of the object is 64 cm inside the glass tube. I am not sure about the second one. I know that the image from the 8 cm mirror will serve as the object for the 16 cm mirror. So then do I have that the image from the 8 cm mirror is (96-64) from the end of the 16 cm mirror? I am not sure about the signs. I know that distance is positive if it is on the light incident side. So would distance from the image of 8 cm to the radius of the 16 cm mirror be negative because the light incident side would come from the outside of the glass tube? And wouldn't the indices of refraction be the same? Since the equation is n1/s1+n2/s2 = (n2-n1)/r. Would I have 1.6/32 + 1/s2 = (-.6)/16? I think the image of the 16cm will be on the outside of the glass tube. Thanks

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