To a fish in an aquarium, the $4.00-\mathrm{mm}$-thick walls appear to be only $3.50\mathrm{mm}$thick. What is the index of refraction of the walls?

Scattering from either a datum plane can create a picture. For refraction from such a surface, the object and image distances are linked by

$s^{\mathrm{\prime }}=-\frac{n_2}{n_1}s$

The negative sign indicates that we are coping with a virtual image. $n_1$is the index of refraction of the object's medium, $n_2$is the reflectance of the observer's medium, and distance The focal length and the image distance, being measured from the edge.

• $t=4.00\mathrm{mm}$is the actual thickness of the walls.
• $t^{\mathrm{\prime }}=3.50\mathrm{mm}$is the apparent thickness of the walls.
• The refractive index of water is $n_{\text{water}}=1.33$
  

We're smart enough to figure out just what the walls' index of refraction is.

Calculation includes the picture depth of the interior of both the aquarium walls to the real distance here to fish's eyes.

$s_{\text{int}}^{\mathrm{\prime }}=\left(\frac{n_{\text{water}}}{n_{\text{walls}}}\right)s_{\text{int}}$

Equation was applied the real image of the exterior of both the aquarium walls to the real distance here to fish's eyes.

$s_{\text{ext}}^{\mathrm{\prime }}=\left(\frac{n_{\text{water}}}{n_{\text{walls}}}\right)s_{\text{ext}}$

The difference between the focal length of its overall design dictates the seeming thickness of a walls. So

$t^{\mathrm{\prime }}=s_{\mathrm{ext}}^{\mathrm{\prime }}-s_{\mathrm{int}}^{\mathrm{\prime }}$

$=\left(\frac{n_{\text{water}}}{n_{\text{walls}}}\right)s_{\text{ext}}-\left(\frac{n_{\text{water}}}{n_{\text{walls}}}\right)s_{\text{int}}$

$=\left(s_{\text{ext}}-s_{\text{int}}\right)\frac{n_{\text{water}}}{n_{\text{walls}}}$

the exact length of the panels,

$t^{\mathrm{\prime }}=t\left(\frac{n_{\text{water}}}{n_{\text{walls}}}\right)$

$n_{\text{walls}}=\left(\frac{t}{t^{\mathrm{\prime }}}\right)n_{\text{water}}$

$n_{\text{walls}}=\left(\frac{4.00\mathrm{mm}}{3.50\mathrm{mm}}\right)$

$=1.52$