Find the distance of the image from the right edge of the sphere

[stunner5000pt]

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 $$\frac{1}{f} = (n-1) \frac{2}{R}$$
and $$\frac{1}{f} = \frac{1}{i} + \frac{1}{o}$$

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!

 

[einstone]

Do consider another reflection at the other surface of the sphere & use the formula f=nR/n-1 (i.e., n/ i -1/o =(n-1)/R.). ( The 'hint' is incomprehensible to me).
Regards,
Einstone.

 

[stunner5000pt]

what do you mena consider anotber reflection??

 

[kevinalm]

I imagine he meant refraction. And the hint makes no sense at all. A sphere is the worst case scenario of failure of the thin lense approximation, iirc. Pretty sure you have to solve with ray tracing. (Which is essentially what einstone suggests.) I know for a fact that spheres form a real image. And the image is _really_ bad, the aberations are abismal.

 

[James R]

What you need is the lensmakers' equation, I think.

 

[stunner5000pt]

the lens makers equation is

$$(n-1) (\frac{1}{r_{1}} - \frac{1}{r_{2}}) = \frac{1}{o} + \frac{1}{i}$$

but what is the object distance?/ Since the beams are parallel to each other is o infinity??

 

[stunner5000pt]

i got the answer

for those who were baffled here is the solution

firs find the image formed by the left side of the the sphere. the disatnce o here is infinity so 1/o approaches zero

the posotion of the object from the right (the image formed by the left) is p = 2r - i1. Then solve the image distance from the right hand side. And the answer comes out to be 0.5r (2-n) / (n-1)
its a pity i couldn't do this on my own, i had to resort to a hint suggested by my textbook's website

 

[stunner5000pt]

i do still have one more question though...

what can be said about the orientation of the image if glass n = 1.52 is used nad diamond n = 2.42 is used?/

with glass i get i =6/13 R and with diamond i get i = -21/142 R <0

what can be said about he orientation, Upright for diamond and inverted for glass??

 

[kevinalm]

First let me say that your work is probably what your textbook is looking for. However, in reality those equations are _not_ valid in the stated case of a thick lens (sphere). What will happen is that at a sufficiently high refractive index a real image forms within the body of the lens and at a sufficiently higher index that will become the object for another (yes, right side up) image farther along the optical path.

The problem with all the stated equations is that the intercept points of a single ray on both optical surfaces are assumed to be equidistant from the optical axis. This is approximately true for a "thin" lense but dead wrong for a sphere.

Your textbook uses a lousey example for an introductory optics discourse.