# Weird Optics Question

1. May 6, 2013

### jaumzaum

I was attempting to solve the following problem from IIT-JEE 1991 (actually I first found it on the book Solved problems in Physics)

[Broken]

It asks for the image viewed by an observer at left. Ok, an observer at left would seen an image that passes through a spherical dioptre glass-air only. It doesn't matter the water at the other side, as the been rays going left will NEVER pass trough it after being irradiated by the object. In the same reasoning I would say an observer at right would see an image formed only by a plane dioptre. Happens that the solution in the book considered the image to be secondary (they found a primary image formed when the rays passed trough the spherical dioptre and then used an "apparent distance in glass" formula (that I translate as "plane dioptre"). I really don't know why they used it, and I want to know what is the right answer, as I didn't find anywhere else in internet the solution for the problem.

The book answer is 35cm right of P.

2/15 + 1/p' = (1-2)/(-10)
p' = -30cm = 30cm right of P

Thanks
John

Last edited by a moderator: May 6, 2017
2. May 6, 2013

### dlgoff

Wouldn't the refraction at the water surface make it appear further?

3. May 6, 2013

### jaumzaum

I would say the refraction in the water surface applies only for the beams irradiated to the right, this beams will form an image at 3,33cm left of the surface. But the beams will continue to move right! They will never turn left to be refracted again by the air surface. An observer at right will see one image only, 3,33 cm from CD, as well as an observer at left will see only one image at 30cm from P.
How can the beams be refracted by the water surface if they never passed trough it?

4. May 6, 2013

### dlgoff

I see what you mean. Looking in one of my old physics books, there is a similar situation where they say, "The image formed by the first surface of a lens serves as the object for the second surface."

edit: but I don't think that is the same as your problem.

Last edited: May 6, 2013
5. May 7, 2013

### jaumzaum

This would be completely valid if, for example, the object stands out of the lens. The beams irradiated by it would pass trough both surfaces, being refracted 2 times, so the image is secondary: the image formed by the first refraction would serve as an object to the second surface. But that's not the case, the beams do not pass trough both surfaces, they cannot form a 2-refractions image.

6. May 7, 2013

### jaumzaum

Is there anyone to tell me if I'm right or wrong?

7. May 7, 2013

### haruspex

Hi jaumzaum,
I'm no expert on optics, but I went back to first principles and came up with the same answer as you have, 30cm from P. More generally, if r is the radius, O the centre of curvature, d the distance of the object behind O and p the distance of the image behind O, I got 1/p + 1/r = (1/d+1/r)/n.
My guess is that the water is there to confuse you. Having calculated that the image is "in" the water, you may be tempted to think the water will have an affect.