Image by Refraction through Flat Surface

[Tom MS]

Homework Statement

A fish watcher at point P watches a fish through a glass wall of a fish tank. The watcher is level with the fish; the index of refraction of the glass is 8/5, and that of the water is 4/3. The distances are d₁=8.0 cm, d₂=3.0 cm, and d₃=6.8 cm (a) To the fish, how far away does the watcher appear to be?

Homework Equations

n₁*sin(θ₁) = n₂*sin(θ₂)

The Attempt at a Solution

I got the correct answer of 20.0 cm, but only when I used very small angles to form the image. I basically chose an arbitrary angle of light coming from the watcher and, using trig, calculated the final angle of refraction from the glass-water interface. I then traced it backwards to the central axis.

As the "arbitrary initial angle approached 0°, the image distance from the fish got closer to 20.0 cm. This would seem to indicate snell's law becomes more accurate with small angles or that the image is smeared out. I couldn't find anything about this when I looked it up. Could someone explain this to me?

 

[TSny]

I basically chose an arbitrary angle of light coming from the watcher and, using trig, calculated the final angle of refraction from the glass-water interface. I then traced it backwards to the central axis.

As the "arbitrary initial angle approached 0°, the image distance from the fish got closer to 20.0 cm. .

Sounds good.

This would seem to indicate snell's law becomes more accurate with small angles or that the image is smeared out.

Snells law is accurate for all angles. But, you are right that the image would become blurred if the fish's eye could pick up rays for larger angles. For the case where the size of the watcher is small and the fish is very close to the normal line from the watcher, you can see that the only rays that reach the eye of the fish are rays with small angles of incidence.