Question

An object is placed at 5 cm in front of a concave mirror of radius of curvature 15 cm. Find the position, nature, and magnification of the image in each case.

Answer 1

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We have been given that :-

• Object distance = 5cm
• Focal length = 15/2 = 7.5cm
• Image distance = ?
• Magnification = ?

Now according to mirror formula :

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

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

After inputting the known values we get :-

$7.5 = \frac{ - 5v}{5 - v}$

$37.5 - 7.5v = - 5v$

$37.5 = - 5v + 7.5v$

$37.5 = 2.5v$

$v = \frac{37.5}{2.5}$

$v = 15cm$

Therefore the image will form at 15cm away from pole of mirror.

Now we know magnification is ratio of image distance to object distance.

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

After inputting the values we get :-

$m = \frac{ - 15}{5}$

$m = - 3cm$

The image formed will be virtual, erect and magnified.

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BY SCIVIBHANSHU

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