Question

Statement $1$: A ray is incident from outside on a glass sphere surrounded by air as shown in figure. $AB$ is the diameter of sphere. The given incident ray may suffer total internal reflection at the second interface.


Statement $II$: For a ray going from a denser to rarer medium, the ray may suffer total internal reflection.
For the given statements, choose the correct option:

• A. Statement $I$ and $II$ both are true and $II$ is a correct explanation of $1$.
• B. Statement $I$ and $II$ both are true, but $II$ is not a correct explanation of $I$
• C. Statement $I$ is false, but statement $II$ is true
• D. Statement $I$ and statement $II$ both are false.

Answer 1

The correct option is C Statement $I$ is false, but statement $II$ is true

From the symmetry of figure about centre of sphere, angle of refraction at first interface is equal to angle of incidence at second interface.


In $\Delta ACD,$
$AC=CD$
$\Rightarrow \angle CAD=\angle ADC$
$\therefore \angle CAD=\angle ADC=r$
Again, from principle of reversibility of light, the angle of emergence at second surface equals the angle of incidence at first.
$({\mu }_{air}\sin i={\mu }_{glass}\sin r)$
T.I.R at second surface will occur if, $r>{\theta }_c$

Maximum possible angle of incidence at $A$ is,
$i={90}^{\circ }$
Hence, for $i={90}^{\circ },r={\theta }_c$
$\Rightarrow$ Since $i>{90}^{\circ }$ is not possible at incident point $A$ shown in figure, thus $r>{\theta }_c$ will be never achieved
$\therefore$ statement $I$ is false.
Statement $II$ is true, because T.I.R will happen when $i>{\theta }_c$ at interface is met.

Why this question? Caution: From the geometry of given figure, point $A$ is at the diametric end of sphere, and $i>{90}^{\circ }$ is not possible at point $A$.