# Geometrical optics question.

## Homework Statement

A container with a layer of water (n=1.33) of 5 cm thick is over a block of acrylic (n=1.5) of 3 cm thick. An observer watches (perpendicularly from above) the lowest surface of the acrylic. What distance does it (the bottom surface) seems to be from the top of the water?

## Homework Equations

(n1/p) - (n2/q) = (n1-n2)/r

## The Attempt at a Solution

(n1/p) - (n2/q) = 0

q = n2p/n1

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## Homework Statement

A container with a layer of water (n=1.33) of 5 cm thick is over a block of acrylic (n=1.5) of 3 cm thick. An observer watches (perpendicularly from above) the lowest surface of the acrylic. What distance does it (the bottom surface) seems to be from the top of the water?

## Homework Equations

(n1/p) - (n2/q) = (n1-n2)/r

## The Attempt at a Solution

(n1/p) - (n2/q) = 0

q = n2p/n1
i think you can use the relation ship between real depth of an abject and apparent depth of the same object related to ratio of the refractive indices of the medium.
as the bottom is being viewed vertically so one can take paraxial rays close to the vertical and extend the emergent ray below to find the apparent position of the image.
what will be depth of the acrylic slab when viewed from water?
and what will be the depth of this image when viewed from above -the water surface?

andrevdh
Homework Helper
I would rather suggest to use Snell's law and a ray diagram Draw such a diagram with rays from a point on the bottom surface of the block.
These rays are refracted when they pass from the block into the water (ignore the glass)
and again as they exit the water surface. These rays will then indicate where the virtual
image of the point is located if you extend the rays backwards.