Question

An object is placed in front of convex mirror of radius of curvature 40cm at a distance of 10cm. find the position, nature and magnification of mirror.​

Answer 1

${\huge{\boxed{\mathcal{\red{Answer}}}}}$

Image distance = 20cm.

Nature = virtual, erect and diminished

Magnification = 2

${\huge{\boxed{\mathcal{\red{Explanation}}}}}$

It is given that :

Radius of curvature = 40cm.

Image distance = v = ?

Object distance = u = 10cm.

Focal length = ?

Magnification = ?

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Since :

Radius of curvature = 2 × Focal length, thus

$40 = 2f \\ \\ f = \frac{40}{2} = 20cm$

Thus focal length of mirror is 20cm.

Now according to mirror formula:

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

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

After inputting the given values we get :

$20 = \frac{ - 10v}{10 - v}$

$20(10 - v) = - 10v$

$200 - 20v = - 10v \\ 200 = - 10v + 20v \\ 200 = 10v \\ v = 20cm$

Thus image will form at a distance of 20cm from pole of mirror.

Since convex mirror always forms a virtual, erect and diminished image, thus nature of this image will be same.

Now magnification is called the ratio of image distance and object distance. Therefore magnification in this case will be :

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

$m = \frac{ - 20}{10}$

$m = - 2cm$

Thus we get magnification of the image is -2cm.

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BY SCIVIBHANSHU

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