Question

An object of height $1cm$ is kept perpendicular to the principal axis of a convex mirror of radius of curvature $20cm$. If the distance of the object from the mirror is $20cm$ then the distance (in cm) between heads of the image and the object will be:

• A. $\sqrt{\frac{6404}{9}}$
• B. $\sqrt{\frac{6414}{9}}$
• C. $\frac{40}{3}$
• D. none of these

Answer 1

The correct option is A $\sqrt{\frac{6404}{9}}$

Use co-ordinate system position of object head with respect to pole of convex mirror is $(-20,1)$

$f=+\frac{20}{2}=+10cm\frac{1}{v}+\frac{1}{u}=\frac{1}{f}\frac{1}{v}=\frac{1}{f}-\frac{1}{u}=\frac{1}{10}+\frac{1}{20}\therefore v=\frac{200}{30}=\frac{20}{3}cm$

Now, $\frac{-v}{u}=\frac{H_i}{H_o}or{H}_i=\frac{-20}{3\times -20}\times 1=+\frac{1}{3}$

$\therefore$ co-ordinates of head of image is $(\frac{20}{3},+\frac{1}{3})$

$\therefore$ distance between two heads

$=\sqrt{{(-20-\frac{20}{3})}^2+{\{1-(+\frac{1}{3})\}}^2}=\sqrt{\frac{6400}{9}+\frac{4}{9}}=\sqrt{\frac{6404}{9}}$