Answer
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Hint To answer this question, we need to trace the path of the light rays as they travel from the fish swimming inside the water to the air. The image of the fish has to be located as seen by the shooter.
Formula Used: The formula used in solving this question is given by
$h' = \dfrac{h}{\mu }$, here $h'$ is the apparent depth of an object residing inside a medium of refractive index $\mu $, and the actual depth of the object is $\mu $.
Complete step by step answer
Let us consider a fish swimming inside the water as shown in the figure below. Let the point F represent the fish.
Suppose that the eye of the shooter is viewing the fish near normally, as shown.
As we can see that the light rays are originating from the fish F to pass through the water and reaching the eye of the observer via air. But they do not travel in a straight line. As can be seen above, the ray FC, on reaching the point C bends away from the normal. This is due to the difference in the optical densities of the air and water, with air being a rarer medium than water. The image of the fish is formed by these bent rays, which appear to be diverging from a point little above the actual position of the fish, that is, the point F. So the image of the fish appears a little shallower than the actual depth of the fish. The apparent depth $h'$ of the fish is related to its actual depth $h$ by
$h' = \dfrac{h}{{{\mu _w}}}$ , where ${\mu _w}$ is the refractive index of the liquid.
As we know that ${\mu _w} = 1.33$ which is greater than one. So, from the above relation the apparent depth is less than the actual depth.
Now, the shooter will aim at the position of the image of the fish to shoot it. As the actual fish is a little deeper, so the target is missed. Hence, it is difficult for a shooter to shoot a fish, which is swimming in water.
Note
We have used the formula for the apparent depth of the fish just for the mathematical explanation of the situation. It is not necessary to use the formula to explain this answer. Only on the basis of the path of the light rays we can explain these types of questions.
Formula Used: The formula used in solving this question is given by
$h' = \dfrac{h}{\mu }$, here $h'$ is the apparent depth of an object residing inside a medium of refractive index $\mu $, and the actual depth of the object is $\mu $.
Complete step by step answer
Let us consider a fish swimming inside the water as shown in the figure below. Let the point F represent the fish.
Suppose that the eye of the shooter is viewing the fish near normally, as shown.
As we can see that the light rays are originating from the fish F to pass through the water and reaching the eye of the observer via air. But they do not travel in a straight line. As can be seen above, the ray FC, on reaching the point C bends away from the normal. This is due to the difference in the optical densities of the air and water, with air being a rarer medium than water. The image of the fish is formed by these bent rays, which appear to be diverging from a point little above the actual position of the fish, that is, the point F. So the image of the fish appears a little shallower than the actual depth of the fish. The apparent depth $h'$ of the fish is related to its actual depth $h$ by
$h' = \dfrac{h}{{{\mu _w}}}$ , where ${\mu _w}$ is the refractive index of the liquid.
As we know that ${\mu _w} = 1.33$ which is greater than one. So, from the above relation the apparent depth is less than the actual depth.
Now, the shooter will aim at the position of the image of the fish to shoot it. As the actual fish is a little deeper, so the target is missed. Hence, it is difficult for a shooter to shoot a fish, which is swimming in water.
Note
We have used the formula for the apparent depth of the fish just for the mathematical explanation of the situation. It is not necessary to use the formula to explain this answer. Only on the basis of the path of the light rays we can explain these types of questions.
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