Day: February 7, 2024

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202402071059 Solution to 1980-CE-PHY-II-16

A light ray passes through a spherical air bubble in water. Which of the following represents the path of the emergent ray?

A. P
B. Q
C. R
D. S

Roughwork.

Refractive (/refraction) index of a pure substance, with some exceptions, is proportional to its density; and of a mixture of substances, as a rule, to its mass concentration.

Thus water, higher in density than air, should have a larger refractive index than air:

n_{\textrm{water}}>n_{\textrm{air}};

alternatively, as light travels in air faster than in water, from

\displaystyle{v=\frac{c}{n}},

it can be seen also.

Let \theta_{1} and \theta_{2} be the angle of incidence and of refraction when light is refracted through the first interface; and \theta_{2} and \theta_{3}, when through the second interface.

Assuming WLOG n_1\geqslant n_2 and noting \sin \theta \geqslant 0 (non-negative) for \theta\in \big[ 0,\frac{\pi}{2}\big]. By Snell’s Law, write

\begin{aligned} n_1\sin\theta_1 & = n_2\sin\theta_2 \\ \frac{\sin\theta_1}{\sin\theta_2} & = \frac{n_2}{n_1}\leqslant 1 \\ \sin\theta_1 & \leqslant \sin\theta_2 \\ \theta_1 & \leqslant \theta_2 \\ \end{aligned}

\therefore \quad n_1\geqslant n_2\Longrightarrow \theta_2\geqslant \theta_1

\therefore\quad n_{\textrm{water}}>n_{\textrm{air}}\Longrightarrow \theta_{\textrm{air}}>\theta_{\textrm{water}}

Answer. Path P.

Reverse the direction of incidence and emergence:

Light ray along path: P bends away from and then towards the normal; Q bends towards the normal once and again; R bends away from the normal once and again; S bends towards and then away from the normal. Wits have it that Q and R are unlikely. So we turn our attention to light rays P and S.

Fussy enough, path P is symmetric with respect to straight line L_P; but there across path S is no line of symmetry were it to be L_S.

This problem is not to be attempted.