Question

A fish in an aquarium with flat sides looks out at a hungry cat. To the fish, does the distance to the cat appear to be less than the actual distance, the same as the actual distance, or more than the actual distance? Explain.

Short answer

The distance for both you and the cat greater than it really is.

Step-by-step solution

Step1:Snell's law

Snell's Law: It relates the refraction angle ${\theta }_2$to the incident angle${\theta }_1$and the indexes of refraction of the incident medium n1 and the refracted medium n2

$n_1\sin {\theta }_1=n_2\sin {\theta }_2$

Angles of incidence and refraction are measured with respect to the normal line at the point where the ray intersects the interface between the two media. Medium 2 is more optically dense (has a higher refractive index) than medium 1 if the refracted ray is closer to the normal than the incident ray. In the same plane are the incident ray, refracted ray, and normal line.

Given data

A fish in an aquarium looks out at a cat.

The index of refraction of air is $1.00$.

The index of refraction of water is $1.33$

Step3:find Distance

We are asked to determine whether the cat's apparent distance is less than, equal to, or greater than its actual distance to the fish.

Step4:Find cat and fish distance

Consider the paraxial rays that refract from the air into the water and consider the cat to be a point source. Because water has a higher index of refraction than air, the cat's rays are refracted toward the normal at the surface and diverge outward. Extending the outgoing rays backwards results in an image of the cat that is further away from the aquarium than it actually is.

As a result, the distance to the cat appears to be greater than it is.