# Q 9.2   A 4.5 cm needle is placed 12 cm away from a convex mirror of focal length 15 cm. Give the location of the image and the magnification. Describe what happens as the needle is moved farther from the mirror.

Given, the height of needle, h = 4.5 cm

distance of object = 12 cm

focal length of convex mirror =  15 cm.

Let the distance of the image be v

Now as we know

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

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

$\frac{1}{v}=\frac{1}{15} - \frac{1}{-12}$

$\frac{1}{v}=\frac{1}{15} + \frac{1}{12}$

v = 6.7 cm

Hence the distance of the image is 6.7 cm from the mirror and it is on the other side of the mirror.

Now, let the size of the image be h'

so.

$m=-\frac{v}{u} = \frac{h'}{h}$

$h'= -\frac{v}{u}*h$

$h'= -\frac{6.7}{-12}*4.5$

$h'= 2.5 cm$

Hence the size of the image is 2.5 cm. positive sign implies the image is erect, virtual and diminished.

magnification of the image =    $\frac{h'}{h}$     =     $\frac{2.5}{4.5}$     =   0.56

m = 0.56

The image will also move away from the mirror if we move the needle away from the mirror, and the size of the image will decrease gradually.

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