
A fish is 30cm below the water surface in a still pond. What is its apparent depth from the water surface if viewed from the top? Take the refractive index of water to be $\dfrac{4}{3}$.
Answer
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Hint: Here, we need to calculate the apparent depth resulting due to the refraction of light. We know that the refractive index of a medium can be calculated as the ratio of real depth to apparent depth.
That is,
$\mu = \dfrac{{{\text{real}}\,{\text{depth}}}}{{{\text{apparent}}\,{\text{depth}}}}$
Using this formula we can calculate the apparent depth at which the object will appear.
Complete step by step answer:
It is given that a fish is $30\,cm$ below the water surface in a still pond.
Which means, the real depth of the fish is $d = 30\,cm$
The refractive index of water is given as $\dfrac{4}{3}$
When a ray of light goes from one medium to another the ray will bend. The refractive index is a measure of this bending of light.
We need to calculate the apparent depth resulting due to refraction of light.
We know that the refractive index of a medium can be calculated as the ratio of real depth to apparent depth.
That is,
$\mu = \dfrac{{{\text{real}}\,{\text{depth}}}}{{{\text{apparent}}\,{\text{depth}}}}$
We already have $\mu = \dfrac{4}{3}$. Let us substitute this in the above equation, then we get,
$\Rightarrow \dfrac{4}{3} = \dfrac{{{\text{real}}\,{\text{depth}}}}{{{\text{apparent}}\,{\text{depth}}}}$
Now let us take apparent depth to the left hand side. Then we get,
$\Rightarrow {\text{apparent}}\,{\text{depth}} = \dfrac{{{\text{real}}\,{\text{depth}}}}{{\dfrac{4}{3}}}$
On substituting the value of real depth we get,
$\Rightarrow {\text{apparent}}\,{\text{depth}} = \dfrac{{30\,cm}}{{\dfrac{4}{3}}}$
$\therefore {\text{apparent}}\,{\text{depth}} = 22.5\,cm$
Therefore, the fish will appear at a depth $22.5\,cm$ from the top.
Note:
Don’t confuse between real depth and apparent depth. Real depth is the original distance of the object from the water surface. Whereas the apparent depth is the distance from the surface of the water to the position at which the object will appear to be due to refraction of light. Apparent depth in the denser medium will be always less than the real depth.
That is,
$\mu = \dfrac{{{\text{real}}\,{\text{depth}}}}{{{\text{apparent}}\,{\text{depth}}}}$
Using this formula we can calculate the apparent depth at which the object will appear.
Complete step by step answer:
It is given that a fish is $30\,cm$ below the water surface in a still pond.
Which means, the real depth of the fish is $d = 30\,cm$
The refractive index of water is given as $\dfrac{4}{3}$
When a ray of light goes from one medium to another the ray will bend. The refractive index is a measure of this bending of light.
We need to calculate the apparent depth resulting due to refraction of light.
We know that the refractive index of a medium can be calculated as the ratio of real depth to apparent depth.
That is,
$\mu = \dfrac{{{\text{real}}\,{\text{depth}}}}{{{\text{apparent}}\,{\text{depth}}}}$
We already have $\mu = \dfrac{4}{3}$. Let us substitute this in the above equation, then we get,
$\Rightarrow \dfrac{4}{3} = \dfrac{{{\text{real}}\,{\text{depth}}}}{{{\text{apparent}}\,{\text{depth}}}}$
Now let us take apparent depth to the left hand side. Then we get,
$\Rightarrow {\text{apparent}}\,{\text{depth}} = \dfrac{{{\text{real}}\,{\text{depth}}}}{{\dfrac{4}{3}}}$
On substituting the value of real depth we get,
$\Rightarrow {\text{apparent}}\,{\text{depth}} = \dfrac{{30\,cm}}{{\dfrac{4}{3}}}$
$\therefore {\text{apparent}}\,{\text{depth}} = 22.5\,cm$
Therefore, the fish will appear at a depth $22.5\,cm$ from the top.
Note:
Don’t confuse between real depth and apparent depth. Real depth is the original distance of the object from the water surface. Whereas the apparent depth is the distance from the surface of the water to the position at which the object will appear to be due to refraction of light. Apparent depth in the denser medium will be always less than the real depth.
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