The nature of the image, formed when an object of 3 cm height is placed at a distance of 10 cm in front of a converging mirror of focal length of 20 cm, will be [[1]] and [[2]].

Solution:

The nature of the image, formed when an object of 3cm height is placed at a distance of 10 cm in front of a converging mirror of focal length of 20 cm, will be virtual and erect.
Given,
Distance of the object from the mirror,

u = - 10 cm

(for a concave mirror)
The focal length (

f

) of the mirror,

f = - 20 cm

Height of the object

{, h}_{0} = 3 cm

By using the mirror formula we know that,

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

\frac{1}{v} + \frac{1}{(- 10)} = \frac{1}{(- 20)}

\frac{1}{v} = \frac{1}{(- 20)} - \frac{1}{(- 10)}

\frac{1}{v} = \frac{1}{20}

v = + 20 cm

Now, by using the magnification formula, we know that

m = - \frac{v}{u}

m = - \frac{20}{(- 10)}

m = \frac{20}{10}

m = + 2

So, the

magnification

of the image is

+ 2

, and the positive sign implies that the image is virtual and erect.

The nature of the image, formed when an object of $3 cm$ height is placed at a distance of $10 cm$ in front of a converging mirror of focal length of $20 cm$ , will be [[1]] and [[2]].

Solution:

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