Question

The distance between a real object and its image in a convex mirror of focal length $12\text{cm}$ is $32\text{cm}$. Find the size of image if the object size is $1\text{cm}$.

• A. $\frac{2}{3}\text{cm}$
• B. $\frac{1}{3}\text{cm}$
• C. $1\text{cm}$
• D. $\frac{1}{4}\text{cm}$

Answer 1

Answer: (B)

Step 1: Given that:

Distance between object and image = $32cm$

Focal length($f$) of the convex mirror = $+12cm$

If u be the object distance and v be the image distance from the convex mirror then,

$u+v=32cm$

$v=32-u$

Size of the object

($h_o$) = $1cm$

Note: The focal length of a convex mirror is taken as positive as it lies right to the pole of the mirror.)

Step 2: Formula used:

The mirror formula is;

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

Where $u$ = object distance, $v$ = image distance and $f$ = focal length of the mirror

Magnification is given by;

$m=\frac{h_i}{h_o}=-\frac{v}{u}$

Step 3: Calculation of the object distance:

Using the mirror formula;

$\frac{1}{32-u}+\frac{1}{-u}=\frac{1}{+12cm}$

$\frac{1}{32-u}-\frac{1}{u}=\frac{1}{12}\frac{u-(32-u)}{u(32-u)}=\frac{1}{12}$

$\frac{2u-32}{32u-u^2}=\frac{1}{12}$

$2u\times 12-32\times 12=32u-u^2$

$\Rightarrow u^2-32u+24u-32\times 12=0$

$\Rightarrow u^2-8u-384=0$

$\Rightarrow (u^2-24u)+(16u-384)=0$

$\Rightarrow u(u-24)+16(u-24)=0$

$\Rightarrow (u-24)(u+16)=0$

$When,(u-24)\ne 0$

$(u+16)=0$

$u=-16cm$

Thus,

$v=32-(-16cm)$

$v=48cm$

$Andif,(u+16)\ne 0$

$u=24cm$

Thus,

$v=32-24cm$

$v=8cm$

Step 4: Calculation of the size of the image:

For u= 24cm

$\frac{h_i}{1cm}=-\frac{48cm}{-16cm}$

$\frac{h_i}{1cm}=\frac{3}{1}$

$h_i=3cm$

And

for, u= 24cm

$\frac{h_i}{1cm}=-\frac{8cm}{-24cm}$

$h_i=\frac{1}{3}cm$

Thus,

the size of the image is $\frac{1}{3}cm$ .