Question

7. A fisherman sees a lobster cage on the sea floor and tries to retrieve it by lowering a winch directly

over where the cage appears to be. The winch lands behind the cage.
a) Explain why the cage isn't exactly where he thought it was.
b) The fisherman uses an underwater camera to get a better view. Light travels from the
lobster cage and hits the camera lens at an angle of 37.2° to the normal. If water has a
refractive index of 1.40 and the lens has a refractive index of 1.49, at what angle to the
normal will the refracted ray in the camera lens be?

Answer 1

Step-by-step explanation:

a) The fisherman sees the cage at a different location than where it actually is because of the refraction of light in water. When light travels from air into water, its speed decreases, causing the light to bend or refract. This means that the light entering the fisherman's eyes is coming from a slightly different direction than the actual position of the cage, making it appear as if the cage is in a different location.

b) To calculate the angle of the refracted ray in the camera lens, we can use Snell's law, which states that the ratio of the sine of the angle of incidence to the sine of the angle of refraction is equal to the ratio of the refractive indices of the two materials.

Using this formula, we can calculate the angle of refraction in the lens:

sin(θ2) / sin(θ1) = n2 / n1

where n1 is the refractive index of water (1.40), n2 is the refractive index of the lens (1.49), and θ1 and θ2 are the angles of incidence and refraction, respectively.

Solving for θ2, we find:

θ2 = sin^-1(sin(θ1) * n1 / n2) = sin^-1(sin(37.2°) * 1.40 / 1.49) = 35.0°.

So, the refracted ray in the camera lens will be at an angle of 35.0° to the normal.