
A fish in water (refractive index n) looks at a bird vertically above in the air. If y is the height of the bird and x is the depth of the fish from the surface, then distance of the bird as estimated by the fish is
A. \[x + y\left( {1 + \dfrac{1}{{\rm{n}}}} \right)\]
B. \[y + x\left( {1 - \dfrac{1}{{\rm{n}}}} \right)\]
C. \[x + y\left( {1 - \dfrac{1}{{\rm{n}}}} \right)\]
D. \[x + {\rm{n}}y\]
Answer
559.8k+ views
Hint: We will be using the relationship between normal shift and actual height when there are two refracting mediums with different refractive indexes. Also, the basic analogy of the distance of a body from another body will be used.
Complete step by step solution:
It is given that the bird is in air, and fish is in water. So it can be easily understood that the bird is in a rarer medium, and fish is in a denser medium because the refractive index of water is higher than the refracting index of air.
We can write the expression for the normal shift of a body when it is in the denser medium as below:
\[D = \left( {{\rm{n}} - 1} \right)h\]
Here n is the refractive index of water, and h is the height of the bird from the surface, which is equal to the bird's height from the surface.
\[h = y\]
Substitute y for h in the above equation.
\[D = \left( {{\rm{n}} - 1} \right)y\]
The bird's height from the surface of the water is equal to the summation of the normal shift and height of the bird from the surface.
$
H = y + \left( {{\rm{n}} - 1} \right)y\\
\implies H = {\rm{n}}y
$
Here H is the height of the bird from the water surface.
The actual distance between bird and fish will become the summation of H and fish depth from the surface.
\[{\rm{Distance = }}x + {\rm{n}}y\]
Therefore, the total distance between bird and fish is given by the relation \[\left( {x + {\rm{n}}y} \right)\]
So, the correct answer is “Option D”.
Note:
Take extra care while calculating the value of the normal shift and the bird's height from the surface of the water.
The bird's height from the surface of the water is equal to the summation of the normal shift and height of the bird from the surface.
Do not forget to add the bird's height from the water's surface in the final expression of the total distance between the bird and the fish.
Complete step by step solution:
It is given that the bird is in air, and fish is in water. So it can be easily understood that the bird is in a rarer medium, and fish is in a denser medium because the refractive index of water is higher than the refracting index of air.
We can write the expression for the normal shift of a body when it is in the denser medium as below:
\[D = \left( {{\rm{n}} - 1} \right)h\]
Here n is the refractive index of water, and h is the height of the bird from the surface, which is equal to the bird's height from the surface.
\[h = y\]
Substitute y for h in the above equation.
\[D = \left( {{\rm{n}} - 1} \right)y\]
The bird's height from the surface of the water is equal to the summation of the normal shift and height of the bird from the surface.
$
H = y + \left( {{\rm{n}} - 1} \right)y\\
\implies H = {\rm{n}}y
$
Here H is the height of the bird from the water surface.
The actual distance between bird and fish will become the summation of H and fish depth from the surface.
\[{\rm{Distance = }}x + {\rm{n}}y\]
Therefore, the total distance between bird and fish is given by the relation \[\left( {x + {\rm{n}}y} \right)\]
So, the correct answer is “Option D”.
Note:
Take extra care while calculating the value of the normal shift and the bird's height from the surface of the water.
The bird's height from the surface of the water is equal to the summation of the normal shift and height of the bird from the surface.
Do not forget to add the bird's height from the water's surface in the final expression of the total distance between the bird and the fish.
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