Question

Rays of the Sun are seen to make a 31.0° angle to the vertical beneath the water. At what angle above the horizon is the Sun?

Answer 1

Concepts and reason

The concept used to solve this problem is Snell's law. Use Snell's law to find the angle of the sun above the horizon.

Fundamentals

"Snell's law states that when light travels between two given media, then the ratio of the sines of the angles of incidence and refraction of wave are constant." The expression for Snell's law is as follows:

\(\frac{n_{1}}{n_{2}}=\frac{\sin \theta_{r}}{\sin \theta_{i}}\)

Here, the refractive index of first medium is \(n_{1}\), the refractive index of second medium is \(n_{2}\), the angle of refraction is \(\theta_{r}\),

and the angle of incidence is \(\theta_{i}\).

The expression for Snell's law is as follows:

$$ \frac{n_{1}}{n_{2}}=\frac{\sin \theta_{r}}{\sin \theta_{i}} $$

Rearrange the above expression for \(\sin \theta_{i}\).

$$ \begin{aligned} n_{1}\left(\sin \theta_{i}\right) &=n_{2}\left(\sin \theta_{r}\right) \\ \sin \theta_{i} &=\frac{n_{2}\left(\sin \theta_{r}\right)}{n_{1}} \end{aligned} $$

The light ray passes from air to water. The medium with higher refractive index is called the denser medium. In an optically denser medium, the light ray bends towards the normal. When the light passes from air to water, the refracted ray in the water will bend towards the normal.

The expression for the incident angle is as follows:

$$ \sin \theta_{i}=\frac{n_{2}\left(\sin \theta_{r}\right)}{n_{1}} $$

Substitute \(1.33\) for \(n_{2}, 1.00\) for \(n_{1}\), and \(31.0^{\circ}\) for \(\theta_{r}\).

$$ \begin{aligned} \sin \theta_{i} &=\frac{(1.33)\left(\sin \left(31.0^{\circ}\right)\right)}{1.00} \\ &=0.685 \\ \theta_{i} &=\sin ^{-1}(0.685) \\ &=43.2^{\circ} \end{aligned} $$

The angle of the sun above the horizon is as follows:

\(\theta^{\prime}=90^{\circ}-\theta_{i}\)

Here, \(\boldsymbol{\theta}^{\prime}\) is the angle of the sun above the horizon level. Substitute \(43.2^{\circ}\) for \(\theta_{i}\).

\(\begin{aligned} \theta^{\prime} &=90^{\circ}-43.2^{\circ} \\ &=46.8^{\circ} \end{aligned}\)

Thus, the angle of sun above the horizon is \(46.8^{\circ}\).

The refractive index of air is 1 and water is \(1.33\). The refractive index of the water is greater than the refractive index of air. Therefore, the incident angle will be greater than refracted angle. The incident angle is the angle between the incident ray and the line that is perpendicular to the surface. Therefore to find

the angle of sun above the horizon, subtract the incident angle from \(90^{\circ}\).