Question

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.​

Answer 1

${\bf{\green{〰Solution〰}}}$

Size of the needle, h = 4.5 cm

Object distance, u = – 12 cm

Focal length of the convex mirror, f = 15 cm

Mirror formula =

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

Where,

Where,

f is the focal length

u is object distance

v is the image distance

$\begin{gathered}\sf \frac{1}{15} = \frac{1}{ - 12} + \frac{1}{v} \\ \sf \frac{1}{v} = \frac{1}{15} - \frac{1}{ - 12} \\ \frac{1}{v} = \frac{9}{60} \\\sf v \: = 6.67 \: cm\end{gathered}$

The magnification is given by

m= h'/h = -v/u

h' = -v/v×h

h' = -6.67/-17×4.5

h' = 2.5 cm

M = h'/h = 2.5/4.5 cm

⇒ m = 0.56

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$\bf A N S W E R$

Magnification is 0.56. And, the height of the image is 2.5 cm. The image is virtual.