Question

Two thin similar convex glass pieces are joined together front to front with its rear portion silvered such that a sharp image is formed $20\text{cm}$ from the mirror. When the air between the glass pieces is replaced by water $({\mu }_w=4/3)$, then the image formed from the mirror is at a distance of
[Assume, glass pieces have same refractive index as water]

• A. $8\text{cm}$
• B. $10\text{cm}$
• C. $6\text{cm}$
• D. $12\text{cm}$

Answer 1

The correct option is D $12\text{cm}$

We know that,

Power of silvered lens will be
$$P=2P_L+P_m$$
where , $P_L$ is power of lens and $P_m$ is power of mirror.
Since the rays arer parallel to the pricipal axis, image will be formed at focus.
Therefore for concave mirror $f=-20\text{cm},R=40\text{m}$

$\therefore$ Power of mirror , $P_m=-\frac{1}{f}$

Substituting the given data we get,

$P_m=-\frac{1}{-0.2}=5D$

Focal length of lens $f_L=\frac{R}{2(\mu -1)}$

Substituting the given data we get,

$f_L=\frac{40}{2\times (\frac{4}{3}-1)}=60\text{cm}$

Power of lens , $P_L=\frac{1}{f_L}$

$\therefore P_L=\frac{100}{60}\text{D}$

$\therefore$ Equivalent power $=2P_L+P_m$

$P=2\times \frac{100}{60}+5=\frac{25}{3}\text{D}$

Net focal length of the combination, $f=\frac{1}{P}$
$\Rightarrow f=\frac{3}{25}\times 100=12\text{cm}$

Hence, option (d) is the correct answer.

Why this question ? Bottom line: The parallel rays after reflection from the mirror will interect at focus, i.e image is formed at focus.