Question

The back wall of an aquarium is a mirror that is 30cm away from the front wall. The sides of the tank are negligibly thin. A fish is swimming midway between the front and back walls.

a) An image of the fish appears behind the mirror. How far does this image appear to be from the front wall of the aquarium?

b) Would the refractive index of the liquid have to be larger or smaller in order for the image of the fish to appear in front of the mirror, rather than behind it? Find the limiting value of this refractive index. Initially, water is there in the aquarium having a refractive index $\dfrac{4}{3}$.

Answer 1

The total distance of the image from the front wall is $45cm$. Also, we know the apparent depth, $x = \dfrac{F}{\mu }$, where,$\mu$ is the refractive index, $F$ is the distance of the image from the viewer. And, for the second case, if the image of the fish has to appear in front of the mirror, the apparent depth will decrease.

Formula used:
If $\mu$ is the refractive index, $F$ is the distance of the image from the viewer, then the apparent depth, $x = \dfrac{F}{\mu }$.

Complete step by step solution:

Suppose, a beam of rays from a point object after refraction reaches our eyes in another medium. Now, if the refracted rays are produced backward, they meet at a point. It seems that the refracted rays are diverging from the second point. Clearly, we can see that the second point is the image of the first point. The image of any point object is formed in the same way. Thus, we can have a complete image of an object.
Now, if the object is situated in a denser medium and is viewed from a rarer medium, it appears closer to the surface of separation. For example, if we look at a fish inside water in a pond it appears nearer to the surface than the actual position.
Now, it is given that the distance between the front and the back wall is $30cm$ and a fish is swimming midway between the front and back walls.
a) So, when an image of the fish appears behind the mirror, it takes the same distance, $30cm$, from the midway of the aquarium to the outside of the aquarium. So, it reaches up to $15cm$ from the back wall to the outside. So, we can say that the total distance of the image from the front wall is $45cm$.
If the total distance of the image from the viewer is $45cm$ and the refractive index is $\dfrac{4}{3}$ (given in the question), then the apparent depth of the image from the front wall is $\dfrac{{45}}{{\dfrac{4}{3}}} = \dfrac{{45 \times 3}}{4} = 33.75cm$. [As we know the apparent depth, $x = \dfrac{F}{\mu }$, where ,$\mu$ is the refractive index, $F$ is the distance of the image from the viewer]
So, the image appears at the distance $33.75cm$ from the front wall.

b) Now, if the image of the fish has to appear in front of the mirror, the apparent depth will decrease, and for this, it is obvious from the equation that the refractive index has to increase.
So, the limiting case for the apparent depth is $30cm$ and we know that the total distance of the image from the viewer is $45cm$.
So, the refractive index will be $\mu = \dfrac{{45}}{{30}} = \dfrac{3}{2}$

So, the refractive index will increase and it will be $\dfrac{3}{2}$.

Note: If we place an object in a medium having a refractive index ${\mu _1}$, and see the object from air through successive media of thickness ${d_1},{d_2},{d_3},...{d_n}$ having corresponding refractive indices ${\mu _1},{\mu _2},{\mu _3},...{\mu _n}$, then we have, apparent depth = \dfrac{{{d_1}}}{{{\mu _1}}} + \dfrac{{{d_2}}}{{{\mu _2}}} + \dfrac{{{d_3}}}{{{\mu _3}}} + ... + \dfrac{{{d_n}}}{{{\mu _n}}} = \mathop \sum \limits_{i = 1}^n \dfrac{{{d_i}}}{{{\mu _i}}}