Question

An air bubble in a glass slab with refractive index $ 1.5 $ (near normal incidence) is $ 5cm $ deep when viewed from one surface and $ 3cm $ deep when viewed from the opposite face. The thickness $ {\text{(in cm)}} $ of the slab is
(A) 12
(B) 16
(C) 8
(D) 10

Answer 1

Hint: We need to substitute the values of apparent depths from both sides of the glass slab. Thickness of the glass slab will be given by the sum of the actual depths for both the faces of the slab.

Formula Used: The following formulas are used to solve this question.
 $\Rightarrow {X_{app}} = \dfrac{{{X_R}}}{\mu } $ where $ {X_{app}} $ is the apparent depth, $ {X_R} $ is the actual depth and $ \mu $ is the refractive index.

Complete step by step answer
Given in the question, an air bubble in a glass slab has a refractive index $ \mu = 1.5 $.
We know that light refracts when it passes from one medium to another. This change in density of two mediums results in the phenomenon where we are unable to see the real depth of a substance in a medium. Instead the depth that is measured, gives the apparent depth.
We know that, $ {X_{app}} = \dfrac{{{X_R}}}{\mu } $ where $ {X_{app}} $ is the apparent depth, $ {X_R} $ is the actual depth and $ \mu $ is the refractive index.
It is given in the question that the bubble appears to be $ 5cm $ deep when viewed from one surface.
 $ \therefore $ It is understood that the apparent depth from this surface is given by $ {X_{app}}_1 $ where $ {X_{app}}_1 = 5cm $.
Substituting the values, $ \mu = 1.5 $ and $ {X_{app1}} = 5cm $ in the above equation, for surface 1, we get,
 $ {X_{{R_1}}} = \mu {X_{app}}_1 $
 $ \Rightarrow {X_{{R_1}}} = 1.5 \times 5 = 7.5cm $
For the opposite surface, the bubble appears to be $ 3cm $ deep when viewed from one surface.
 $ \therefore $ It is understood that the apparent depth from this surface is given by $ {X_{app}}_2 $ where $ {X_{app}}_2 = 3cm $.
Substituting the values, $ \mu = 1.5 $ and $ {X_{app2}} = 3cm $ in the above equation, for surface 2, we get,
 $ {X_{{R_2}}} = \mu {X_{app}}_2 $
 $ \Rightarrow {X_{{R_2}}} = 1.5 \times 3 = 4.5cm $
Thus the thickness of the slab is given by the sum of the actual depths of the air bubble from the opposite side.
Let the thickness be $ d $.
 $ \therefore d = {X_{{R_1}}} + {X_{{R_2}}} $
 $ \Rightarrow d = 7.5 + 4.5 = 12cm. $
Thus the thickness of the slab is $ 12cm. $

$ \therefore $ The correct answer is Option A.

Note
We can also solve this question in another way as below.
Let the total thickness of the slab be $ t $. Let $ x $ be the distance of the air bubble from one surface. Thus the distance of the air bubble from the opposite side be $ \left( {t - x} \right) $.
Given that, the total apparent thickness is $ t' = 5 + 3 = 8cm. $
Thus, the total length $ \dfrac{x}{\mu } + \dfrac{{\left( {t - x} \right)}}{\mu } = 8 $
Given $ \mu = 1.5 $, thus,
 $ x + t - x = 8 \times 1.5 $
 $ \Rightarrow t = 12cm $
Thus the thickness of the slab is $ 12cm. $
 $ \therefore $ The correct answer is Option A.