Question

A spherical mirror forms an erect image three times the linear size of the object. If the distance between the object and the image is 80 cm, the focal length of the mirror is

• A. $15cm$
• B. $-15cm$
• C. $-30cm$
• D. $40cm$

Answer 1

The correct option is C $-30cm$

$\text{Given}=\frac{v\left(\mathrm{Tmag}e\text{distance}\right)}{u\left(\text{object distance}\right)}=3$

$\begin{array}{c}\Rightarrow V=34\\ V+4=80\\ \Rightarrow 34+4=80\\ 4=20cm,V=60cm\\ \text{The image is erect, therefore it must}\\ \text{be virtual}\\ \text{and,}V=60cm,u=-20cm\end{array}$

$\begin{array}{c}\text{Mirror formula}\\ \Rightarrow \frac{1}{v}+\frac{1}{4}=\frac{1}{f}\\ \therefore \frac{1}{60}-\frac{1}{20}=\frac{1}{f}\\ \therefore \frac{20-60}{60\cdot 20}=\frac{1}{f}\\ \Rightarrow f=-30cm\end{array}$

$\begin{array}{c}\text{Thus, the mirror is concave}\\ \text{focal length}=-30cm\\ \text{Hence, (c) is correct option.}\end{array}$