Question

An object is placed at a distance of $16cm$ from a convex mirror of focal length $20cm$, the position of image with its nature is

• A. Virtual, erect behind the mirror at 8.9 cm
• B. Virtual, erect behind the mirror at 89 cm
• C. Real, erect behind the mirror at 8.9 cm
• D. Virtual, inverted behind the mirror at 89 cm

Answer 1

The correct option is A Virtual, erect behind the mirror at 8.9 cm

Given:

Object distance $u=-16cm$

Focal length $f=+20cm$ (+ve for convex mirror)

By mirror formula

$\Rightarrow \frac{1}{v}+\frac{1}{4}=\frac{1}{f}$

$\Rightarrow \frac{1}{v}-\frac{1}{16}=\frac{1}{20}$

$\Rightarrow \frac{1}{v}=\frac{1}{20}+\frac{1}{16}$

$\therefore \frac{1}{v}=\frac{4+5}{80}$

$\therefore v=\frac{80}{9}=+8.9cm$

Magnification $\left(m\right)=\frac{-v}{u}$

$=\frac{-8.9}{-16}$

$=+0.556$

$m=$ +ve, therefore erect and virtual image is formed. And image is formed at $8.9cm$ behind the mirror.

Hence, (A) is correct option.