Question

A flashlight is held at the edge of a swimming pool at a height h = 2.5 m such that its beam makes an angle of θ = 38 degrees with respect to the water's surface. The pool is d = 3.75 m deep and the index of refraction for air and water are n1 = 1 and n2 = 1.33, respectively.

What is the horizontal distance, D, from the edge of the pool to the point on the bottom of the pool where the light strikes? Write your answer in m.

Answer 1

Firs, let's find the distance $CE = BO$. From the triangle $\triangle AOB$ expressing the side $BO$, obtain:

$$BO = \dfrac{AB}{\tan\theta} =h\cot\theta = CE$$

Now let's find $EF$. According to the Snell's law:

$$n_1\sin\theta_1 = n_2\sin\theta_2\\ \sin\theta_2 = \dfrac{n_1\sin\theta_1}{n_2}\\ \theta_2 = \arcsin\left(\dfrac{n_1\sin\theta_1}{n_2}\right)$$

From the triangle $\triangle ODF$:

$$EF = OE\tan\theta_2 = d\tan\theta_2 = d\tan\left(\arcsin\left(\dfrac{n_1\sin\theta_1}{n_2}\right) \right)$$

Finally:

$$D = CE + EF = h\cot\theta + d\tan\left(\arcsin\left(\dfrac{n_1\sin\theta_1}{n_2}\right) \right)$$

Obtain the following number:

$$D = 2.5\cot38\degree + 3.75\tan\left(\arcsin\left(\dfrac{1\cdot\sin52\degree}{1.33}\right) \right) \approx 5.83m$$

Answer. 5.83 m.