- #1

- 166

- 0

## Homework Statement

The length of each side of an opaque cubic container is 5 cm and a coin is placed at the center of the base as shown. At this instant, the observer is unable to see the coin. However, the coin can just be seen when the container is fully filled with liquid A. Find the refractive index of liquid A. (You may regard the coin as a point)

## Homework Equations

Snell's Law

[tex]n_1\sin\theta_1=n_2\sin\theta_2[/tex]

## The Attempt at a Solution

I assumed that the apparent position of the coin is on the diagonal line, which means when the ray is refracted at the top-right corner, the refracted angle will be 45 degrees. Hence

[tex]n\sin\tan^{-1}\frac{1}{2}=\sin\frac{\pi}{2}[/tex]

and

[tex]n=1.58[/tex]

However, the solution give is 2.24 (it would be a rather unusual refractive index for a liquid), which is (I think) calculated by assuming that the apparent position of the coin is on the surface of the liquid. Which assumption is correct and why?

Thanks in advance

P.S. why aren't we allowed to attach svg images?

#### Attachments

Last edited: