Question

A transparent cube of \[15{\rm{ cm}}\] edge contains a small air bubble. Its apparent depth when viewed through one face is \[6{\rm{ cm}}\] and when viewed through the opposite face is \[4{\rm{ cm}}\]. Then the refractive index of the material of the cube is:
(1) 2.0
(2) 1.5
(3) 1.6
(4) 2.5

Answer 1

Hint: Refractive index of the material of the cube is equal to the ratio of real depth and apparent depth of the bubble. We will write the individual expressions for the refractive index of cube material when viewed from both the faces.

Complete step by step answer:
Given:
The edge of the given transparent cube is \[a = 15{\rm{ cm}}\].
The apparent depth of the bubble when viewed through one face is \[{d_1} = 6{\rm{ cm}}\].
The apparent depth of the bubble when viewed through the opposite side \[{d_2} = 4{\rm{ cm}}\].
We are required to find the refractive index of the material of the cube.
Let us assume the real depth of the bubble when viewed from one face is x.
Let us write the expression for the refractive index of the material of the cube in terms of real depth and apparent depth when viewed through one face.
\[\mu = \dfrac{x}{{{d_1}}}\]
On substituting \[6{\rm{ cm}}\] for \[{d_1}\] in the above expression, we get:
\[\mu = \dfrac{x}{{6{\rm{ cm}}}}\]……(1)
We know that the edge of the cube is a and real depth of the cube, when viewed from one face, is x so that we can write the real depth of the bubble as the difference of edge a and x.
We can write the expression for the refractive index of the material of cube in terms of real depth and apparent depth when viewed through the opposite face
\[\mu = \dfrac{{a - x}}{{{d_2}}}\]
On substituting \[15{\rm{ cm}}\] for a and \[4{\rm{ cm}}\] for \[{d_2}\] in the above expression, we get:
\[\mu = \dfrac{{15{\rm{ cm}} - x}}{{4{\rm{ cm}}}}\]……(2)
We know that the refractive index of a cube obtained through one face must be equal to the refractive index obtained when viewed through another face. Therefore, we can equate equation (1) and equation (2).
\[\begin{array}{l}
\dfrac{x}{{6{\rm{ cm}}}} = \dfrac{{15{\rm{ cm}} - x}}{{4{\rm{ cm}}}}\\
x = 9{\rm{ cm}}
\end{array}\]
On substituting \[9{\rm{ cm}}\] for x in equation (1), we get:
\[\begin{array}{l}
\mu = \dfrac{{9{\rm{ cm}}}}{{6{\rm{ cm}}}}\\
 = 1.5 \\
\end{array}\]
Therefore, the refractive index of the material of the cube is 1.5, and option (2) is correct.

Note:When we see an object situated in denser medium its real depth decreases due to the phenomena of refraction as a result of which apparent depth will become lesser than the real depth of the object.