A tank is filled with a liquid of refractive index \( \sqrt{2} \), up to a height of 30 cm. A tiny bulb is glowing at the bottom of the tank. Calculate the diameter of an opaque disc floating symmetrically on the liquid surface that can cut off completely the light from the bulb that comes out of the liquid surface.
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The critical angle for total internal reflection depends on the refractive index of the medium. When the angle of incidence exceeds this critical angle, the light is totally reflected within the medium. In this case, the disc must be wide enough to block all light coming from the bulb.
We are given the following data:
\begin{itemize}
\item Refractive index of the liquid, \( n_1 = \sqrt{2} \)
\item Refractive index of air, \( n_2 = 1 \)
\item Height of the liquid, \( h = 30 \, \text{cm} \)
\end{itemize}
The light emitted by the bulb will refract at the liquid surface. For the light to be blocked by the disc, it must undergo total internal reflection at the liquid surface. This happens when the angle of incidence exceeds the critical angle.
The critical angle \( \theta_c \) is given by:
\[
\sin \theta_c = \frac{n_2}{n_1}
\]
Substituting the values of \( n_2 \) and \( n_1 \):
\[
\sin \theta_c = \frac{1}{\sqrt{2}}
\]
\[
\theta_c = \sin^{-1} \left( \frac{1}{\sqrt{2}} \right) = 45^\circ
\]
Now, to block the light, the disc must cover all the rays exiting from the bulb. The rays will refract at an angle of \( \theta_c \) at the liquid surface. The distance from the bulb to the liquid surface is 30 cm, so the light will spread out in a circular region with radius \( r \).
Using the geometry of the situation:
\[
\tan \theta_c = \frac{r}{h}
\]
where \( r \) is the radius of the circle, and \( h = 30 \, \text{cm} \) is the distance from the bulb to the surface.
Since \( \tan 45^\circ = 1 \), we have:
\[
1 = \frac{r}{30}
\]
\[
r = 30 \, \text{cm}
\]
Thus, the radius of the disc is \( 30 \, \text{cm} \), and therefore the diameter of the disc is:
\[
d = 2r = 2 \times 30 = 60 \, \text{cm}
\]