Question

An object of size $50\text{cm}$ is placed $20\text{cm}$ in front of a spherical mirror and its image of size $25\text{cm}$ is formed above the principal axis. Find the focal length as per sign convention and the type of spherical mirror used here?

• A. $-25\text{cm}$, concave mirror
• B. $-20\text{cm}$, concave mirror
• C. $+25\text{cm}$, convex mirror
• D. $+20\text{cm}$, convex mirror

Answer 1

The correct option is D $+20\text{cm}$, convex mirror

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

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

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

$⟹\frac{-v}{-20}=\frac{1}{2}$

$\therefore$ Image distance $\left(v\right)=+10\text{cm}$
Positive sign indicates that image is formed right of the mirror.

Now applying 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,$f=+20\text{cm}$
Since focal length is positive and the image is virtual and erect, the mirror must be convex.