Question

In the figure shown radius of curvature of either surface of equi convex half lens is $40\text{cm}$ and refractive index $1.5$. Its one side is silvered. A plane mirror is also placed. A small object $O$ is placed such that there is no parallax between final image formed by lens and mirror. If transverse length of final image formed by silvered lens system is twice that formed by mirror. Choose the correct options:

• A. $a=2.5\text{cm}$
• B. $b=5\text{cm}$
• C. Distance of image from object is $15\text{cm}$
• D. Image formed by silvered lens system is virtual

Answer 1

Answer: (C), (D)

Image formed by silvered lens system is virtual


Distance of object from the plane mirror $\Rightarrow AB=a+b$
$\Rightarrow$ Image will be formed by the plane mirror at a distance of $(a+b)$ behind the mirror.
From the above statement we can say that the polished lens will also form image at the same place.
Distance of image from combination is therefore gievn by :-
$\Rightarrow OC=b+(a+b)$
$OC=a+2b$
Now, according to the question $m=2$ for the polished lens.
$\Rightarrow \frac{OC}{AO}=2$
$\Rightarrow \frac{a+2b}{a}=2$
or, $a=2b$ ..(1)
Now, we can say that the focal length of the polished lens can be calculated by incidenting a parallel beam of light on the combination of the lens and concave mirror. Since, in this process refraction from the lens takes place $2$ times. So, from the formula of power of combination of lens and mirror we can say that, focal length of combination.
$\Rightarrow \frac{1}{F}=\frac{1}{f_m}-\frac{1}{f_l}-\frac{1}{f_l}$
or, $\Rightarrow \frac{1}{F}=\frac{1}{f_m}-\frac{2}{f_l}$
Here, $f_l$ is given by :-
$\frac{1}{f_l}=(1.5-1)(\frac{1}{40}-\frac{1}{-40})$
$\Rightarrow f_l=40\text{cm}$
and $f_m=-20\text{cm}$
$\Rightarrow \frac{1}{F}=\frac{1}{-20}-\frac{2}{40}$
$\frac{1}{F}=\frac{-4}{40}$
$F=-10\text{cm}$
Now, applying mirrors formula for the combination since, the combination is behaving as a concave mirror of focal length $10\text{cm}$
$\frac{1}{F}=\frac{1}{v}+\frac{1}{u}$
$\frac{1}{F}=\frac{1}{OC}+\frac{1}{-OA}$
$\frac{1}{-10}=\frac{1}{(2b+a)}+\frac{1}{-a}$
$\frac{-1}{10}=\frac{1}{2a}-\frac{1}{a}$
$[\because a=2b]$
$\Rightarrow a=5\text{cm}$
and $b=2.5\text{cm}$
Distance of image from the object
$=AC$
$=2(a+b)=2\times (5+2.5)$
$=15\text{cm}$
Imaged formed by silvered lens system is virtual.