# Thread: Completely confused by this question

1. ## Completely confused by this question

An opaque container is 15 cm deep. It contains only a single coin. When looking into the container at a viewing angle of 50o relative to the vertical side of the container, you see nothing on the bottom. When the container is filled with water, you see the coin (from the same viewing angle) on the bottom of, and just beyond, the side of the container. How far is the coin from the side of the container?

2. Originally Posted by chicagomike
An opaque container is 15 cm deep. It contains only a single coin. When looking into the container at a viewing angle of 50o relative to the vertical side of the container, you see nothing on the bottom. When the container is filled with water, you see the coin (from the same viewing angle) on the bottom of, and just beyond, the side of the container. How far is the coin from the side of the container?
With your problem the optical dense medium is water. The index of refraction of water is approximately

$\displaystyle n_{water}=1.333$

The light comes from the coin that means the light passes through the dense medium into the light optical medium. Use the law of refraction:

$\displaystyle \frac{\sin(\alpha)}{\sin(50^\circ)}=\frac1{1.333}$ .... which will yield .... $\displaystyle \alpha = 35.077^\circ$

To calculate the length of x (that's the distance of the coin from the right side of the container) use Tangens:

$\displaystyle \tan(\alpha)=\frac x{15}$ .... which yield .... $\displaystyle x = 10.53\ cm$