Question

A fish is located at a distance of 10 cm from the wall of the fish pond. The thickness of the glass wall is 2 cm. Find the apparent position of the fish. `("Given"  ""_aµ_g = 3/2  "and"  ""_aµ_w = 4/3)`

Answer 1

Given:

Distance of the fish from the inner wall of the pond, u_w = 10 cm

Thickness of the glass wall, t_g = 2 cm

Refractive index of glass relative to air, µ_g = `3/2` = 1.5

Refractive index of water relative to air, µ_w = `4/3` ≈ 1.33 (approx.)

To Find:

The apparent position of the fish as seen by the observer’s eye outside the pond.

Formula:

When light passes from water through glass and then to air, the apparent distance of the fish (as seen from the outside) can be found by considering the refraction at each interface.

1. Apparent distance of the fish from the inner surface of the glass (water-glass interface) is:

`u_w^' = u_w/mu_w`

2. The actual thickness of the glass is (t_g), but the apparent thickness of the glass from outside (glass-air interface) is:

`t_g^' = t_g/mu_g`

3. Therefore, the total apparent distance of the fish from the outer surface of the glass (wall) is:

`"d"_"apparent" = u_w^' + t_g^' = u_w/mu_w + t_g/mu_g`

Calculation:

1. Apparent distance of fish from water-glass interface:

`u_w^' = 10/(4/3)`

= `10 xx 3/4`

= 7.5 cm

2. Apparent thickness of glass wall as seen from outside:

`t_g^' = 2/(3/2)`

= `2xx 2/3`

≈ `4/3`

≈ 1.33 cm (approx.)

3. Total apparent distance of fish from the outside surface of the glass:

`"d"_"apparent"` = 7.5 + 1.33 = 8.83 cm

∴ The apparent position of the fish from the outside of the glass wall is approximately 8.83 cm.