Question

A silvered, hollow glass sphere (mirror),has a round hole as shown in the figure. Radius of round hole makes $\alpha$ (acute) angle at the centre. Into this round hole a parallel beam of ray falls, perpendicular to the plane of the hole. Part of rays having undergone one reflection, (Portion crossing the smaller circle of radius $r_2$) exits the sphere back through the hole. (Assume 100% reflection). Choose the correct option.

• A.
  $r_2=R\sin \left(\frac{\alpha }{2}\right)$
• B.
  $r_1=3r_2-\frac{4r_2^3}{R^2}$
• C.
  $r_1=\frac{R^2}{r_2}\sin \left(\alpha \right)$
• D. Fraction of the power of the incoming beam, which exit through the hole after one reflection is ${\left[\frac{\sin \left(\alpha /3\right)}{\sin \alpha }\right]}^2.$

Answer 1

Answer: (B), (D)

Fraction of the power of the incoming beam, which exit through the hole after one reflection is ${\left[\frac{\sin \left(\alpha /3\right)}{\sin \alpha }\right]}^2.$


From the given figure, $\sin \theta =\frac{r_2}{R},\sin \alpha =\frac{r_1}{R}$

Also, $90-\alpha +\theta +\theta +\theta =90$

$\Rightarrow \theta =\frac{\alpha }{3}$

$\therefore \sin 3\theta =\frac{r_1}{R}$

Since, $\sin 3\theta =3\sin \theta -4{\sin }^3\theta$

$\frac{r_1}{R}=\frac{3r_2}{R}-4{\left(\frac{r_2}{R}\right)}^3{r}_1=3r_2-\frac{4r_2^3}{R^2}\text{Fraction of power (P)}=\frac{\pi r_2^2}{\pi r_1^2}={\left(\frac{\sin \theta }{\sin \alpha }\right)}^2$

$\Rightarrow \text{P}={\left(\frac{\sin \frac{\alpha }{3}}{\sin \alpha }\right)}^2$