Question

A fish swims 10 cm from the front wall of an aquarium that is 35cm wide. The front wall of the aquarium is glass with negligible thickness, but the back wall is a plane mirror. A person looks through the front wall and watches both the fish and its reflection in the mirror.Part A:What is the apparent distance from the front wall of the aquarium to the fish?Part B: What is the apparent distance from the front wall of the aquarium to the image of the fish in the mirror?

Answer 1

El problema es un caso generalmente tipico en optica concerniente a Apparent depth vs real depth

We see the objects closer than their real depth to the surface. We see objects only if the rays coming from them reaches our eyes.

The equation is given by,

$D_a = \frac{D_r}{\eta}$

Where,

$D_a =$Apparenth depth

$D_r =$Real depth

$\eta =$Refractive index of the medium of object

For water $\eta$ is equal to 1.33

I attach an image of the theory that could help clarify the measurements.

We have,

$D_a = \frac{D_r}{\eta}$

$D_a = \frac{10}{1.333}$

$D_a = 7.5cm$

Therefore the apparent distance between the front wall of the aquarium to the fish is 7.5cm

B) The distance between fish and mirror is given by,

$d=35-10= 25$

So we have that real distance from the front wall of to image of fish is

$dr=25+35=60cm$

Applying our equation we have that,

$D_a = \frac{D_r}{\eta}$

$D_a = \frac{60}{1.333}$

$D_a = 45.1cm$

Therefore the apparent distance from the front wall of the aquarium to the image of the fish is 45.1cm