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Question: A jar of height h is filled with a transparent liquid of refractive index μ. At the center of the ja...

A jar of height h is filled with a transparent liquid of refractive index μ. At the center of the jar on the bottom surface is a dot. Find the minimum diameter of a disc, such that when it is placed on the top surface symmetrically about the center, dot is invisible.

Explanation

Solution

To solve this question, we have to apply the rule of total internal reflection. Using the condition for this effect, and applying mathematics on the geometry of the figure, we can get the answer.

Formula used: The formula used in this solution is
iC=sin1(1μ)\Rightarrow {i_C} = {\sin ^{ - 1}}\left( {\dfrac{1}{\mu }} \right), here iC{i_C} is the critical angle of incidence for a light ray travelling from a medium of refractive indexμ\mu to the air.

Complete step by step solution:
Let the minimum diameter of the disc be dd. The situation is represented in the figure below.

As we can see that the rays are emerging out of the dot at the centre of the jar. The topmost layer of water is an interface, which is separating two mediums, air and water. So the rays emerging out undergo refraction.
For the dot to be invisible outside the jar, none of the rays should come out of the jar. This is only possible when these rays bounce back into the jar, or they slip along the topmost layer of water.
Or in other words, these rays should undergo total internal reflection.
We know that for this effect to take place, the angle of incidence should be greater or equal to the critical angle of incidence iC{i_C}. For the minimum diameter of the disc, we take it equal to iC{i_C}.
From the figure we can see that
θ=iC\Rightarrow \theta = {i_C} (Alternate interior angles) …………….(i)
As the dot is present at the centre of the jar, so in the above figure
AN=BN=d/2\Rightarrow AN = BN = d/2 ………………..(ii)
In triangle BON
sinθ=BNOB\Rightarrow \sin \theta = \dfrac{{BN}}{{OB}} ………………..(iii)
By Pythagoras theorem, we have
OB2=BN2+ON2\Rightarrow O{B^2} = B{N^2} + O{N^2}
OB2=(d2)2+h2\Rightarrow O{B^2} = {\left( {\dfrac{d}{2}} \right)^2} + {h^2}
Taking under root, we get
OB=d42+h2\Rightarrow OB = \sqrt {{{\dfrac{d}{4}}^2} + {h^2}}
OB=12d2+4h2\Rightarrow OB = \dfrac{1}{2}\sqrt {{d^2} + 4{h^2}} …………………….(iv)
Putting (ii) and (iv) in (iii)
sinθ=d/212d2+4h2\Rightarrow \sin \theta = \dfrac{{d/2}}{{\dfrac{1}{2}\sqrt {{d^2} + 4{h^2}} }}
sinθ=dd2+4h2\Rightarrow \sin \theta = \dfrac{d}{{\sqrt {{d^2} + 4{h^2}} }}
From (i)
siniC=dd2+4h2\Rightarrow \sin {i_C} = \dfrac{d}{{\sqrt {{d^2} + 4{h^2}} }} …………………..(v)
We know that the critical angle of incidence is given by
iC=sin1(1μ)\Rightarrow {i_C} = {\sin ^{ - 1}}\left( {\dfrac{1}{\mu }} \right)
Taking sine both the sides
siniC=1μ\Rightarrow \sin {i_C} = \dfrac{1}{\mu } ………………….(vi)
Putting (vi) in (v)
1μ=dd2+4h2\Rightarrow \dfrac{1}{\mu } = \dfrac{d}{{\sqrt {{d^2} + 4{h^2}} }}
Taking square on both the sides
1μ2=d2d2+4h2\Rightarrow \dfrac{1}{{{\mu ^2}}} = \dfrac{{{d^2}}}{{{d^2} + 4{h^2}}}
On cross multiplying, we get
d2+4h2=μ2d2\Rightarrow {d^2} + 4{h^2} = {\mu ^2}{d^2}
Subtractingd2{d^2} from both sides
4h2=μ2d2d2\Rightarrow 4{h^2} = {\mu ^2}{d^2} - {d^2}
4h2=d2(μ21)\Rightarrow 4{h^2} = {d^2}\left( {{\mu ^2} - 1} \right)
Dividing by (μ21)\left( {{\mu ^2} - 1} \right)
4h2(1μ2)=d2\Rightarrow \dfrac{{4{h^2}}}{{\left( {1 - {\mu ^2}} \right)}} = {d^2}
Finally, taking square root on both the sides, we get
d=2h1μ2\Rightarrow d = \dfrac{{2h}}{{\sqrt {1 - {\mu ^2}} }}
Hence, the minimum required diameter of the disc is equal to2h1μ2\dfrac{{2h}}{{\sqrt {1 - {\mu ^2}} }}.

Note:
Before using the effect of total internal reflection, always check its condition. The condition for total internal reflection to take place is that the light ray should be travelling from an optically denser medium to an optically rarer medium. In this question, the light ray from the dot was travelling from water, an optically denser medium to air, an optically rarer medium. So we could make use of this phenomenon.