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.

Solution

Given,

$Object distance, u = - 12 cm Focal length, f = + 15 cm Size of the object, O = 4.5 cm$

Now using the mirror formula,
$\frac{1}{u} + \frac{1}{v} = \frac{1}{f}$
we have,

$\frac{1}{v} = \frac{1}{f} - \frac{1}{u} = \frac{1}{15} + \frac{1}{12} = \frac{4 + 5}{60} = \frac{9}{60} ∴ v = 60 / 9 = 6.7 cm$

∴ Image is formed at 6.7 cm at the back of the mirror ( because v is positive)
Now,
Magnification, $m = \frac{I}{O} = - \frac{v}{u}$

$∴$ $\frac{I}{4.5} = - \frac{6.7}{- 12} I = \frac{6.7 \times 4.5}{12} = 2.5 cm .$

Hence, the image formed is erect, and of course virtual.

As, needle is moved farther away from the mirror then as a result, image moves away from the mirror (upto F) and keeps on decreasing in size.

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Wave Optics

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.

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