Question

Consider the shown glass slab having a thin biconvex cavity of air centred at $O$ . A black spot at $P$ will have image ( $\mu _{\text{glass}} = \frac{3}{2}$ , the radius of curvature of each surface of the cavity = $10 \, cm$ )

• A. $7.5cm$ on the right side of $O$
• B. $7.5cm$ on the left side of $O$
• C. $3.75cm$ on the right side of $O$
• D. $3.75cm$ on the left side of $O$

Answer 1

Answer: (D)

$\frac{1}{f}=\left(\mu - 1\right)\left[\frac{1}{1 0} + \frac{1}{1 0}\right]$
$= \left(\frac{2}{3} - 1\right) \frac{2}{1 0}$
$=-\frac{1}{1 5}\Rightarrow f=-15$
This lens will behave like a biconcave lens.
So, u = -5 cm
$\frac{1}{v}-\frac{1}{u}=\frac{1}{f}$
$\Rightarrow \frac{1}{\text{v}} - \frac{1}{- 5} = \frac{1}{- 1 5}$
$-\frac{1}{v}=\frac{1}{1 5}+\frac{1}{5}=\frac{4}{1 5}$
$\Rightarrow v=-3\text{.}75\text{cm}$