# Homework Help: Image distance from the hemisphere lens

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1. Dec 22, 2017

### PeppaPig

1. The problem statement, all variables and given/known data

As shown in the figure, the distance between the object and hemisphere lens is R0. The hemisphere lens radius is R0. Find the distance between the image and the flat side of the lens if observing from the curved side of the lens. Refractive index of glass and air are n and 1 respectively.

2. Relevant equations
n1/s + n2/s' = (n2-n1)/R

3. The attempt at a solution
Refracting on the flat side
s is -R0

-1/R0 + n/s' = (n-1)/∞
s' = R0n

Did I do the first step correctly?
And what should I do next?

2. Dec 23, 2017

### TSny

Why the -1 in the first term?

Hint: Image of first step becomes the object for the next step.

3. Dec 23, 2017

### PeppaPig

The object distance is negative because the observer is on the curved side of the lens. Is that correct? If the first term image is at -R0n then the second term object distance is -R0n + R0 and second term n1 is 1 and n2 is n.

1/(-R0n+R0) + n/s' = (n-1)/-R0

Is that correct?

4. Dec 23, 2017

### TSny

The sign of the object distance does not depend on the presence of an observer. There are different choices of sign conventions that people use. So, it is important to consult your notes and/or textbook to review the sign conventions that you are using in your course. I suspect that you are using sign conventions such that the object distance is positive when the object is on the same side of the surface that the light strikes the surface. For the flat surface in your problem, the light is striking the left side of the flat surface. Since the object is also on the left side of the flat surface, the object distance is positive. But, again, you should verify that this corresponds to your sign convention.

In your original post, you got that s' = +Ron. Now you are saying s' = -Ron. After reviewing your sign conventions to get make sure you are using the correct sign for the object distance s for the flat surface, see what you get for the sign of s' for the first (flat) surface. Then, think again about what you should use for the object distance for the second (curved) surface.

Are you using the convention where n1 is the index for the medium through which the light is traveling just before it strikes the surface and n2 is the medium through which the light is traveling after it passes through the surface? If so, then I don't think you have the correct values for n1 and n2 for the second (curved) surface.

5. Dec 23, 2017

### PeppaPig

In the first term the s' is -R0n which mean that the image is on the left side of the flat surface.

Then in the second term, the distance of the object should be R0n + R0 and since the light strike from the left side, n1 should be n and n2 should be 1. Is that correct?

n/(R0n + R0) + 1/s' = (1 - n)/-R0

Last edited: Dec 23, 2017
6. Dec 23, 2017

### TSny

Yes, good.

You have a typographical error here
Yes.

I think that's right.

7. Dec 23, 2017

### PeppaPig

Thank you very much for advising.

I think I can solve the rest myself.

The result is (R0n+R0)/(n2-n-1) + R0 = R0n2/(n2-n-1) (Distance from the flat surface)

8. Dec 24, 2017

### TSny

I think that's right. As a bit of a check, you could let n = 1 and see if the final image comes out where it should.

9. Dec 24, 2017

### PeppaPig

If the sign refers from the observer then the result should be -R0n2/(n2-n-1). Is that correct?

10. Dec 24, 2017

### TSny

I'm not sure what "refers from the observer" means. The answer s' = n2/(n2 - n - 1) gives the distance of the final image as measured from the flat surface. For a negative value of s' (when n = 1.4, say) the image will be to the left of the flat surface. For a positive value of s' (when n = 1.7, say), the image is to the right of the flat surface. If you know where the observer is located, then of course you could figure out the location of the image relative to the observer.