Question

A diver shines a light up to the surface of a flat glass-bottomed boat at an angle of 37◦ relative to the normal. If the indices of refraction of air, water, and glass are 1.0, 1.33, and 1.4 respectively, at what angle does the light leave the glass (relative to its normal)? Answer in units of ◦ .

Answer 1

Approximately
`53^(\circ)`, assuming that the upper and lower surfaces of the glass on this boat are parallel.

Step-by-step explanation:

Assume that the upper and lower surfaces of the glass at the bottom of this ship are parallel. Refer to the diagram attached. The two normals would also be parallel to each other.

The following angles would be alternate interior angles between the two normals:

• The angle at which the light enters the glass, and
• The angle at which the light leaves the glass.

Since the two normals are parallel to each other, these two angles would have the same value. Let
`\theta_{\text{glass}}` denote the value of both of these angles.

Let
`\theta_{\text{src}}` denotes the angle at which a beam of light leaves the original medium (angle of incidence.) Let
`\theta_{\text{dst}}` denote the angle at which this beam of light enters the new medium.

Let
`n_\text{src}` and
`n_\text{dst}` denote the refractive indices of the original and the new medium, respectively. By Snell's Law:

`\begin{aligned}\frac{\sin(\theta_{\text{dst}})}{\sin(\theta_{\text{src}})} = \frac{n_{\text{src}}}{n_{\text{dst}}}\end{aligned}`.

Let
`\theta_{\text{water}}` denote the angle at which the beam of light in this question leaves the water. Let
`\theta_{\text{air}}` denote the angle at which this beam of light enters the air. It is given that
`\theta_{\text{water}} = 37^(\circ)`, while
`\theta_{\text{air}}` is the value that needs to be found.

Let
`n_{\text{air}}`,
`n_{\text{water}}`, and
`n_{\text{glass}}` denote the refractive index of air, water, and glass, respectively. By Snell's Law:

`\begin{aligned}\frac{\sin(\theta_{\text{glass}})}{\sin(\theta_{\text{water}})} = \frac{n_{\text{water}}}{n_{\text{glass}}}\end{aligned}`.

`\begin{aligned}\frac{\sin(\theta_{\text{air}})}{\sin(\theta_{\text{glass}})} = \frac{n_{\text{glass}}}{n_{\text{air}}}\end{aligned}`.

Thus:

`\begin{aligned} & \frac{\sin(\theta_{\text{air}})}{\sin(\theta_{\text{water}})} \\ =\; & \frac{\sin(\theta_{\text{glass}})}{\sin(\theta_{\text{water}})}* \frac{\sin(\theta_{\text{air}})}{\sin(\theta_{\text{glass}})} \\ =\; & \frac{n_{\text{water}}}{n_{\text{glass}}}* \frac{n_{\text{glass}}}{n_{\text{air}}} \\ =\; & \frac{n_{\text{water}}}{n_{\text{air}}}\end{aligned}`.

Since
`\theta_{\text{water}} = 37^(\circ)`:

`\begin{aligned} & \sin(\theta_{\text{air}})\\ =\; & \sin(\theta_{\text{water}}) * \frac{\sin(\theta_{\text{air}})}{\sin(\theta_{\text{water}})} \\ =\; & \sin(\theta_{\text{water}})* \frac{n_{\text{water}}}{n_{\text{air}}} \\ =\; & \sin(37^(\circ)) * (1.33)/(1.0) \\ \approx \; & 0.800 \end{aligned}`.

Therefore:

`\begin{aligned}\theta_{\text{air}} &= \arcsin(\sin(\theta_{\text{air}})) \\ & \approx \arcsin(0.800) \\ &\approx 53^(\circ) \end{aligned}`.

In other words, this beam of light would leave the glass at approximately
`53^(\cic)` from the normal.