Question

A $4.5cm$ needle is placed at $12cm$ away from a convex mirror of a focal length of $15cm$ Give the location of the image and the magnification. Describe what happened as the needle is moved further from the mirror.

Answer 1

Step 1. Given data:

1. Height of the needle $\left(h_1\right)$ $=4.5cm$
2. Object distance $\left(u\right)=-12cm$
3. Focal length of the convex mirror $\left(f\right)=15cm$
4. Let image distance = $v$

Step 2. Formula used:

The value of $v$ can be obtained using the mirror formula

$$\begin{gathered} \raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$u$}\right.+\raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$v$}\right.=\raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$f$}\right. \\ \raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$v$}\right.=\raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$f$}\right.-\raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$u$}\right.........\left(1\right) \end{gathered}$$

Where $v=$image distance, $u=$object distance, $f=$focal length.

Also, magnification $m=\raisebox{1ex}{$h_2$}\!\left/ \!\raisebox{-1ex}{$h_1$}\right.=\raisebox{1ex}{$-v$}\!\left/ \!\raisebox{-1ex}{$u$}\right..........\left(2\right)$

Where $h_2=$size of image, $h_1=$height of object,

Step 3. Calculations:

Image distance:

Now putting the values in equation $\left(1\right)$

$$\begin{gathered} \raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$v$}\right.=\raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$15$}\right.-\raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$\left(-12\right)$}\right. \\ =\raisebox{1ex}{$(4+5)$}\!\left/ \!\raisebox{-1ex}{$60$}\right.=\raisebox{1ex}{$9$}\!\left/ \!\raisebox{-1ex}{$60$}\right. \\ v=\raisebox{1ex}{$60$}\!\left/ \!\raisebox{-1ex}{$9$}\right.=6.7cm \end{gathered}$$

Hence, the image of the needle is $6.7cm$ away from the mirror. Also, it is on the other side of the mirror.

Image size:

The image size is given by the magnification formula:

image is formed $6.7cm$ behind the convex mirror. it must be virtual and erect.

if $h_2$ is size of image , then

$m=\raisebox{1ex}{$h_2$}\!\left/ \!\raisebox{-1ex}{$h_1$}\right.=\raisebox{1ex}{$-v$}\!\left/ \!\raisebox{-1ex}{$u$}\right.$

$m=\raisebox{1ex}{$-6.7$}\!\left/ \!\raisebox{-1ex}{$\left(-12\right)$}\right.=0.558$

Now $$\begin{gathered} \raisebox{1ex}{$h_2$}\!\left/ \!\raisebox{-1ex}{$h_1$}\right.=0.558 \\ {h}_2=0.558\times 4.5 \\ {h}_2=2.5cm \end{gathered}$$

Hence magnification $m=\raisebox{1ex}{$h_2$}\!\left/ \!\raisebox{-1ex}{$h_1$}\right.=\raisebox{1ex}{$2.5$}\!\left/ \!\raisebox{-1ex}{$4.5$}\right.=0.56$

If the needle is moved farther from the mirror, the image will also move away from the mirror, and the size of the image will reduce gradually.

Thus, (i) Image distance $\left(v\right)=6.7cm$ away from the mirror.

(ii) Image size $\left(h_2\right)=2.5cm$

(iii) If the needle is moved farther from the mirror, the image will also move away from the mirror, and the size of the image will reduce gradually.