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stunner5000pt
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A narrow parallel incident beam of light falls from the left on a solid glass sphere at normal incidence. The radius of the sphere is R and its index of refraction is n = 1.52. Find the distance of the image from the right edge of teh sphere
Hint was to treat the sphere like a thin lens
Since this beam is parallel rays the object disatnce approaches infinity, and thus 1/o approaches zero.
also [tex] \frac{1}{f} = (n-1) \frac{2}{R} [/tex]
and [tex] \frac{1}{f} = \frac{1}{i} + \frac{1}{o} [/tex]
and since 1/o appraoches zero then focus distance = image distance = R/2(0.52). So tis means that the image will occur at the focus of the right edge but it will be a virtual image occurring isnide the sphere?? So what can be said about the orientation of thius image?
please help with this! Help would be greatly appreciated!
Hint was to treat the sphere like a thin lens
Since this beam is parallel rays the object disatnce approaches infinity, and thus 1/o approaches zero.
also [tex] \frac{1}{f} = (n-1) \frac{2}{R} [/tex]
and [tex] \frac{1}{f} = \frac{1}{i} + \frac{1}{o} [/tex]
and since 1/o appraoches zero then focus distance = image distance = R/2(0.52). So tis means that the image will occur at the focus of the right edge but it will be a virtual image occurring isnide the sphere?? So what can be said about the orientation of thius image?
please help with this! Help would be greatly appreciated!
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