Question

hi good afternoon pls I need solution to a problem the question goes like this : The horizontal floor of a water reservoir is said to be 1.0m deep when viewed vertically from above .if the refractive index of water is 1.35 calculate the real depth of the reservoir a)2.35mb)1.35mc)1.00md)0.35m

Answer 1

$\begin{gathered} n1sin\theta1=n2sin\theta2 \\ 1sin(90)=1.35sin\theta2 \\ \theta2=47.79\degree \end{gathered}$

First, you have to use Snell's law to calculate the refractive angle. Remember the light enters at a 90 angle and the refractive index of air is 1.

$\begin{gathered} h\cdot sin\theta2=1m \\ h=\frac{1}{sin(47.79)}=1.35m \end{gathered}$