Question

A cubical block of glass, refractive index $1.5$, has a spherical cavity of radius $r=9cm$ inside it as shown in Fig. A luminous point object $O$ is at a distance of $18cm$ from the cube (see figure). What is the apparent position of $O$ as seen from $A$?

• A. $17cm$, left of $S_4$
• B. $25cm$, right of $S_4$
• C. $13cm$, left of $S_4$
• D. $10cm$, left of $S_4$

Answer 1

The correct option is A $17cm$, left of $S_4$

We have to consider four refractions at $S_1,S_2,S_3$, and $S_4$, respectively. At each refraction, we will apply single surface refraction equation.
For refractive at first surface $S_1$:
$\frac{3/2}{v_1}-\frac{1}{(-18)}=0$
$v_1=-27cm$
First image lies to the left of $S_1$
For refractive at second surface $S_2$:
$\frac{1}{v_2}-\frac{3/2}{-(27+9)}=\frac{(1-3/2)}{+9}$
$v_2=-\frac{72}{7}cm$
Note that origin of Cartesian coordinate system lies at vertex of surface $S_2$. The object distance is $(27+9)cm$. The second image lies to left of $S_2$.
For refractive at third surface $S_3$:
$u_3=-(\frac{72}{7}+8)=-\frac{198}{7}$
$\frac{1.5}{v_3}-\frac{1}{(-198/7)}=\frac{(1.5-1)}{(-9)}$
$v_3=-16.5cm$
For refractive at fourth surface $S_4$:
$u_4=-(16.5+9)=-25.5cm$
$\frac{1}{v_4}-\frac{3/2}{(-25.5)}=\frac{(1-3/2)}{\infty }=0$
$v_4=-17cm$
The final image lies at $17cm$ to the left of surface $S_4$.