Question

An opaque cylindrical tank with an open top has a diameter of 3.00m and is completely filled with water. When the setting sun reaches an angle of 37 degrees above the horizon, sunlight ceases to illuminate any part of the bottom of the tank. How deep is the tank?

Answer 1

Hint: We will first find out the angle of incidence and the value of $\sin r$ with the help of Snell’s Law. Then we will find out the values of the trigonometric functions sin, cos and tan in order to solve this solution further. Refer to the figure as well.

Formula used: $\mu = \dfrac{{\sin i}}{{\sin r}}$, $\cos \theta = \sqrt {1 - {{\sin }^2}\theta } $, $\tan \theta = \dfrac{{\sin \theta }}{{\cos \theta }}$

Complete Step-by-Step solution:



Since the sunlight is unable to illuminate any part of the bottom of the tank, this means that the sunlight is reaching the end corner of the tank. (refer to the figure)
The angle made by the sunlight above the horizon is 37 degrees.
The ray is coming from the rare medium to the denser medium, hence it will be bent towards the normal.
The angle of incidence in such a case will be i.
The diameter of the tank is given as 3m.
Thus, from the figure it is clear that the angle between the normal and the horizon is 90 degrees. So, the angle of incidence will be-
$
   \Rightarrow i + 37^\circ = 90^\circ \\
    \\
   \Rightarrow i = 53^\circ \\
$
Let the depth of the tank be h.
Let the angle of refraction be r.
As we know that according to the Snell’s Law $\mu = \dfrac{{\sin i}}{{\sin r}}$.
Since the tank was filled with water, we know that the refractive index of water is $\mu = \dfrac{4}{3}$.
Placing the values, we will get-
$
   \Rightarrow \dfrac{4}{3} = \dfrac{{\sin i}}{{\sin r}} \\
    \\
   \Rightarrow \dfrac{4}{3} = \dfrac{{\sin 53^\circ }}{{\sin r}} \\
$
Since we know that the value of $\sin 53^\circ = \dfrac{4}{5}$ , we will put this value in the above equation-
$
   \Rightarrow \dfrac{4}{3} = \dfrac{{\sin 53^\circ }}{{\sin r}} \\
    \\
   \Rightarrow \dfrac{4}{3} = \dfrac{{\dfrac{4}{5}}}{{\sin r}} \\
    \\
   \Rightarrow \dfrac{4}{3} = \dfrac{4}{{5\sin r}} \\
    \\
   \Rightarrow \sin r = \dfrac{3}{5} \\
$
As we know that $\cos \theta = \sqrt {1 - {{\sin }^2}\theta } $. So
$
   \Rightarrow \cos r = \sqrt {1 - {{\sin }^2}r} \\
    \\
   \Rightarrow \cos r = \sqrt {1 - \dfrac{9}{{25}}} \\
    \\
   \Rightarrow \cos r = \sqrt {\dfrac{{16}}{{25}}} \\
    \\
   \Rightarrow \cos r = \dfrac{4}{5} \\
$
As we know that $\tan \theta = \dfrac{{\sin \theta }}{{\cos \theta }}$
$
   \Rightarrow \tan r = \dfrac{{\sin r}}{{\cos r}} \\
    \\
   \Rightarrow \tan r = \dfrac{{\dfrac{3}{5}}}{{\dfrac{4}{5}}} \\
    \\
   \Rightarrow \tan r = \dfrac{3}{4} \\
$
In triangle ABC, the value of $\tan r$ will be-
$ \Rightarrow \tan r = \dfrac{3}{h}$ (as $\tan \theta = \dfrac{{perpendicular}}{{base}}$)
$
   \Rightarrow \dfrac{3}{4} = \dfrac{3}{h} \\
    \\
   \Rightarrow 3h = 12 \\
    \\
   \Rightarrow h = 4m \\
$
Hence, the depth of the tank is 4m.

Note: Snell 's law informs us the degree of refraction and relationship between the angle of incidence, refractive angle and refractive indices of a given media pair. We know that as light transits from one medium to another, it experiences refraction or bending. The bend degree is estimated by Snell's law. It is also known as refraction law.