- #1
theguyoo
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Moved from a technical forum, so homework template missing.
An object is within a glass sphere of radius R with a refractive index of 1.5 . I'm trying to calculate the displacement of the virtual object relative to the actual when viewed from the side, such that the refracted ray emanating from the object becomes horizontal. I would like to know S (the image shift) as a function of d, the distance of the object from the center of the sphere.
Diagram: http://imgur.com/dCaTLvq
So far I've tried starting with Snell's law, [itex]sin\alpha = 1.5sin\beta [/itex], and manipulated it to get:
[itex]\frac{\sqrt{R^{2}}-x^{2}}{R}=1.5\frac{x \cdot d}{R\sqrt{R^{2}-2d\sqrt{R^{2}-x^{2}+d^{2}}}}[/itex]where and x is the distance form origin so [itex](S+d)^{2}+x^{2}=R^{2}[/itex] The problem is it's too messy to rearrange to make S the subject, but I'm guessing there's a better way.
Thanks.
Diagram: http://imgur.com/dCaTLvq
So far I've tried starting with Snell's law, [itex]sin\alpha = 1.5sin\beta [/itex], and manipulated it to get:
[itex]\frac{\sqrt{R^{2}}-x^{2}}{R}=1.5\frac{x \cdot d}{R\sqrt{R^{2}-2d\sqrt{R^{2}-x^{2}+d^{2}}}}[/itex]where and x is the distance form origin so [itex](S+d)^{2}+x^{2}=R^{2}[/itex] The problem is it's too messy to rearrange to make S the subject, but I'm guessing there's a better way.
Thanks.