Question

A small logo is embedded in a thick block of transparent material (n=1.52),3.71 cm beneath the top surface of the block. The block is put under water (n=1.333 ), so there is 2.15 cm of water above the top surface of the block. The logo is viewed from directly above by an observer in air. How far beneath the top surface of the water does the logo appear to be?

Answer 1

The logo appears to be at the same depth beneath the top surface of the water as it is beneath the top surface of the block.

Step-by-step explanation:

To determine how far beneath the top surface of the water the logo appears to be, we can use Snell's law. The equation for Snell's law is:

n1*sin(theta1) = n2*sin(theta2)

where n1 and n2 are the refractive indices of the two materials, and theta1 and theta2 are the angles of incidence and refraction, respectively. Since the observer is in air, the angle of incidence is zero degrees, and the angle of refraction can be calculated using Snell's law:

sin(theta2) = (n1/n2)*sin(theta1)

sin(theta2) = (1.0003/1.333)*sin(0)

sin(theta2) = 0

Therefore, the angle of refraction is zero degrees, which means the logo appears to be at the same depth beneath the top surface of the water as it is beneath the top surface of the block, which is 3.71 cm.