If an object of 10 cm height is placed at a distance of 36 cm from a concave mirror of focal length 12 cm, then the position of the image, nature and height of the image will be?

A
Front of the mirror; real and inverted; small in size
B
Behind of the mirror; virtual and erect; large in size
C
Behind of the mirror; virtual and erect; small in size
D
Front of the mirror; real and inverted; large in size

Solution:

If an object of

10 cm

height is placed at a distance of

36 cm

from a concave mirror of focal length of

12 cm

, then the position of the image, nature and height of the image will be at the front of the mirror, real and inverted and small in size.
The Focal length of the concave mirror,

f = - 12 cm

Distance of the object from the mirror,

u = - 36 cm

Using the mirror formula,

1 / v + 1 / u = 1 / f

1 / v = 1 / f - 1 / u

\frac{1}{v} = \frac{1}{36} - \frac{1}{12} = - \frac{2}{36} = - \frac{1}{18}

\frac{1}{v} = - \frac{2}{36}

\frac{1}{v} = - \frac{1}{18} cm

v = - 18 cm

The image position is

- 18 cm

. The image is in front of the mirror.
Now using the magnification formula;

m = \frac{h_{2}}{h_{1}}

m = - \frac{v}{u}

\frac{h_{2}}{10} = \frac{- 18}{- 36}

h_{2} = - \frac{10}{2}

h_{2} = - 5 cm

Here the height

h_{2}

of the image is less than the height of h1, which is

- 5 cm

. Hence the image formed will be real and inverted.
To find the nature, we use the magnification formula:

m = - \frac{v}{u}

m = - \frac{- 18}{36}

m = - \frac{1}{2}

m = - 0.5

Since the magnification is -0.5, the image formed is small in size.

If an object of $10 cm$ height is placed at a distance of $36 cm$ from a concave mirror of focal length $12 cm$ , then the position of the image, nature and height of the image will be?

Solution:

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