Question

In Fig. a fish watcher watches a fish through a $3.0cm$ thick glass wall of a fish tank. The watcher is in level with the fish; the index of refraction of the glass is $8/5$ and that of the water is $4/3$.
a. To the fish, how far away does the watcher appear to be?
b. To the watcher, how far away does the fish appear to be?

Answer 1

. $d_1=9.4cm$ (watcher to glass)

$d_2=4.6cm$ (glass width)

$d_3=77cm$ (glass to fish)

light passes straight through the water and the tank at no angle.

optical thickness and the thickness are not same.

so, $\left(a\right)9.4cm+\frac{8\times 4.6cm}{5}+4\times \frac{7.7}{3}$

$=9.4+7.36cm+10.27cm$

$=27.0cm$

same will be true the opposite way meaning .

the answer to $\underset{–}{b}$ is the same as answer to $\underset{–}{a}$