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A small illuminated bulb is at the bottom of a tank, containing a liquid of refractive index $\mu $ up to a height $H$. Find the expression for the diameter of an opaque disc, floating symmetrically on the liquid surface in order to cut- off the light from the bulb.

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Last updated date: 22nd Feb 2024
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Hint: This bulb shall illuminate the area in the form of a cone. The height of the cone shall be equal to the height of the liquid in the tank and the radius of the base is equal to the radius of the tank.
Formula Used: The formulae used in the solution are given here.
When a ray is incident on a surface at angle of incidence ${90^ \circ }$, mathematically,
$\mu = \dfrac{{\sin {{90}^ \circ }}}{{\sin c}}$ where $\mu $ is the refractive index and $c$ is the critical angle.
We know that, for any angle $\theta $, it can be said,
${\sin ^2}\theta + {\cos ^2}\theta = 1$
$ \Rightarrow {\cos ^2}\theta = 1 - {\sin ^2}\theta $

Complete Step by Step Solution: Total internal reflection is defined as the phenomenon which occurs when the light rays travel from a more optically denser medium to a less optically denser medium.
A ray of light passes from a medium of water to that of air. Light rays will be refracted at the junction separating the two media. Since it passes from a medium of a higher refractive index to that having a lower refractive index, the refracted light ray bends away from the normal. At a specific angle of incidence, the incident ray of light is refracted in such a way that it passes along the surface of the water. This particular angle of incidence is called the critical angle. Here the angle of refraction is 90 degrees. When the angle of incidence is greater than the critical angle, the incident ray is reflected back to the medium. We call this phenomenon total internal reflection.
When a ray is incident on a surface at angle of incidence ${90^ \circ }$, mathematically,
$\mu = \dfrac{{\sin {{90}^ \circ }}}{{\sin c}}$ where $\mu $ is the refractive index and $c$ is the critical angle.
$ \Rightarrow \sin c = \dfrac{1}{\mu }$.
We know that, for any angle $\theta $, it can be said,
${\sin ^2}\theta + {\cos ^2}\theta = 1$
$ \Rightarrow {\cos ^2}\theta = 1 - {\sin ^2}\theta $
So, when $\theta = c$, we have $\cos c = \sqrt {1 - {{\sin }^2}c} $.
Substituting, $\sin c = \dfrac{1}{\mu }$, we get,
$\cos c = \sqrt {1 - {{\left( {\dfrac{1}{\mu }} \right)}^2}} = \dfrac{{\sqrt {{\mu ^2} - 1} }}{\mu }$.
It has been given that a small illuminated bulb is at the bottom of a tank, containing a liquid of refractive index $\mu $ up to a height $H$.
This bulb shall illuminate the area in the form of a cone. The height of the cone shall be equal to the height of the liquid in the tank and the radius of the base is equal to the radius of the tank.
Let this radius be $R$.
Now for $\theta = c$,
$\tan c = \dfrac{R}{H}$.
Thus, $R = H\tan c$.
We have seen that, $\sin c = \dfrac{1}{\mu }$ and $\cos c = \dfrac{{\sqrt {{\mu ^2} - 1} }}{\mu }$,
$\tan c = \dfrac{{\sin c}}{{\cos c}} = \dfrac{{\dfrac{1}{\mu }}}{{\dfrac{{\sqrt {{\mu ^2} - 1} }}{\mu }}}$
Simplifying the equation, we get, $\tan c = \dfrac{1}{{\sqrt {{\mu ^2} - 1} }}$. Thus, the radius $R = \dfrac{H}{{\sqrt {{\mu ^2} - 1} }}$.
Now the diameter of the tank is twice the radius, $D = 2R = \dfrac{{2H}}{{\sqrt {{\mu ^2} - 1} }}$.

Note: We can also calculate the area visible from the information provided. Area visible is the area of the circle with radius $R$.
$Area = \pi {R^2} = \dfrac{{\pi H}}{{\sqrt {{\mu ^2} - 1} }}$.