Question

Amber (n = 1.546) is a transparent brown-yellow fossil resin. An insect, trapped and preserved within the amber, appears to be 1.8 cm beneath the surface when viewed directly from above. How far below the surface is the insect actually located?

Answer 1

You'll want to draw a diagram to follow along. Light projects upward from the insect at an angle q1 from normal, crosses the air/amber interface a distance x from the normal, and continues at an angle q2 to the eye. (Draw in rays symmetric about the normal to these for the light going to the other eye.) From Snell's law
(sin q1) / n1 = (sin q2) / n2
where n1=1.546 is the index of refraction of amber and n2 is the index of refraction in air. Let y be the depth of the insect below the surface, and y0=1.8 cm the apparent depth of the insect. Extend the rays from the eye to the surface of the amber until they meet at the apparent depth of the insect. Notice that
tan q1 = x / y
tan q2 = x / y0
y / y0 = (tan q2) / (tan q1)
= (sin q2 / sin q1) / (cos q1 / cos q2)
Substituting from Snell's law,
y = y0 (n2 / n1) (cos q1 / cos q2)
If the observer is sufficiently far from the amber, then cos q1 ~ cos q2 ~ 1, and
y ~ y0 (n2 / n1) = 1.8 cm * 1.546 m