Question

An air bubble in a glass slab (${\mu _g}\, = \,1.5$) is $6cm$ deep as viewed from one face and $4cm$ as viewed from the other face. The thickness of glass slab is:
A. $10cm$
B. $15cm$
C. $6.67cm$
D. $11.67cm$

Answer 1

Hint: Refractive index is the measure of the bending of a ray of light when passing from one medium into another medium.We are unable to see the actual depth of the substance in a medium due to the refraction of light when it travels from one medium to another medium. The thickness of the glass slab can be calculated by taking the sum of the real depths for both the faces of the glass slab.

Complete step by step solution:
According to the question, the air bubble is inside a glass slab of refractive index,${\mu _g}\, = \,1.5$.
Now, as we know due to refraction, we are unable to measure the real depth of the substance in a medium. So, the depth measured here is the apparent depth.
Refractive index can be expressed in terms of actual and apparent depth as
${\mu _g}\, = \,\dfrac{{{x_R}}}{{{x_{Ap}}}}$……(i)
Here,
Refractive index,${\mu _g}\, = \,1.5$is a unitless quantity.
Real depth is given by${x_R}$ which is measured in$cm$.
Apparent depth is given by ${x_{Ap}}$ which is measured in$cm$.
Now, according to the question:
The air bubble appears to be $6cm$deep when viewed from one face and $4cm$ deep when viewed from the other face.
Hence, $6cm$and $4cm$ are the apparent depths of the air bubble as viewed from one face (Face$1$) and another face (Face$2$) respectively.
Now, for Face $1$
It is understood that the apparent depth from this face is given by${x_{Ap1}}$where ${x_{Ap1\,}}\, = \,6\,cm$.
Substituting the values of refractive index${\mu _g}\, = \,1.5$and apparent depth${x_{Ap1\,}}\, = \,\,6\,cm$in equation (i), for Face$1$
For finding the real depth for Face$1$
${\mu _g}\, = \,\dfrac{{{x_{R1}}}}{{{x_{Ap1}}}}$
$ \Rightarrow \,1.5\, = \,\dfrac{{{x_{R1}}}}{6}$
$ \Rightarrow \,{x_{R1}}\, = \,\,6\, \times \,1.5\,cm$
$ \Rightarrow \,{x_{R1}}\, = \,9\,cm$
Therefore, the real depth of air bubbles from Face $1$ is given by ${x_{R1}}\, = \,9\,cm$.
Now, for Face $2$
It is understood that the apparent depth from this face is given by ${x_{Ap2}}$ where ${x_{Ap2\,}}\, = \,\,4\,cm$.
Substituting the values of refractive index ${\mu _g}\, = \,1.5$ and apparent depth ${x_{Ap2\,}}\, = \,\,4\,cm$ in equation (i), for Face$2$
For finding the real depth for Face$2$
${\mu _g}\, = \,\dfrac{{{x_{R2}}}}{{{x_{Ap2}}}}$
$ \Rightarrow \,1.5\, = \,\dfrac{{{x_{R2}}}}{4}$
$ \Rightarrow \,{x_{R2}}\, = \,\,4\, \times \,1.5\,cm$
$ \Rightarrow \,{x_{R2}}\, = \,6\,cm$
Therefore, the real depth of air bubbles from Face $2$ is given by ${x_{R2}}\, = \,6\,cm$.
Thus the thickness of the glass slab is given by the sum of the real depths of the air bubble from both faces (Face$1$ and Face$2$).
Let the thickness of glass slab be $t$ in$cm$,
$t\, = \,{x_{R1}}\, + \,{x_{R2}}$
$ \Rightarrow \,t\, = \,9 + \,6\,cm$
$ \Rightarrow \,t\, = \,15\,cm$
Hence, the thickness of the glass slab is given by $t\, = \,15\,cm$.
Therefore, option (B) $15\,cm$ is the correct option.

Note:
An object placed in a denser medium when viewed from a rarer medium appears to be at a depth less than the real depth which is known as apparent depth but when the object in the denser medium is viewed along the normal from a rarer medium, the apparent shift in depth is zero.