Question

g A point source of light is submerged 2.5 m below the surface of a lake and emits rays in all directions. On the surface of the lake, directly above the source, the area illuminated is a circle. What is the maximum radius that this circle could have

Answer 1

Answer:

$r= 2.5 m tan(48.61) = 2.84m$

Explanation:

For this case the situation is explained in the figure attached.

We can find the maximum radius r with the angle of incidence or critical angle (xi). If we have that the angle is higher than the light we will see a reflecting property. We can assume that the refracted angle for this case is 90 and if we use the following trigonometric property we have:

$tan (x_i) = \frac{r}{d}$

And if we solve for r we got:

$r= d tan(x_i)$

The value of d on this case represent the depth of the watr d = 2.5 m

And we can find the value for the angle $x_i$ using the Snell's Law like this:

$n_{air} sin (90) = n_{water} x_i$

$sin (x_i)= \frac{n_{air} sin (90)}{n_{water}}= \frac{1*1}{1.333}= 0.75$

And we can use the arcsin in order to find the angle like this:

$x_i = arcsin(0.75) =48.61$

And now we can find the maximum radius replacing into

$r= d tan(x_i)$

And we got:

$r= 2.5 m tan(48.61) = 2.84m$