Question

An object of size 50 cm, placed in front of a spherical mirror at 20 cm, forms an image of size 25 cm above the principal axis. Find out the focal length with sign convention and the type of spherical mirror used?

• A. – 25 $cm$, concave mirror
• B. – 20 $cm$, concave mirror
• C. + 25$cm$, convex mirror
• D. + 20 $cm$, convex mirror

Answer 1

The correct option is D

+ 20 $cm$, convex mirror

Given:
Height of the image $h^{{}^{\prime }}=+25cm$ (positive since image is formed above principal axis)
Object distance $\left(u\right)=-20cm$
Height of the object $\left(h\right)=50cm$
Let $M$ be the magnification, $v$ be the image distance and $f$ be the focal length.

We know that,
$M=\frac{-v}{u}=\frac{h^{\prime }}{h}$

$⟹\frac{25}{50}=\frac{1}{2}=\frac{-v}{u}$

$⟹\frac{-v}{-20}=\frac{1}{2}$

$\therefore$ Image distance $\left(v\right)=+10cm$
(After applying sign convention $v$ is positive so it shows that the image formed is virtual)

Now applying the given values in the mirror formula,

$\frac{1}{v}+\frac{1}{u}=\frac{1}{f}$

$\frac{1}{10}-\frac{1}{20}=\frac{1}{f}$

$\frac{1}{f}=\frac{1}{20}$

$\therefore$ focal length,$f=+20cm$
Since focal lengths are positive and the image is virtual, the mirror is convex.