Question

An object of height 2.5 cm is kept at a distance of 12 cm away from concave mirror of focal length 20cm. Calculate the size and the nature of image formed. ​

Answer 1

Given:

The height of the object(O) = 2.5 cm.

The distance of the object(u) = 12 cm.

The focal length of the mirror(f) = 20 cm.

To find:

• The size of the image.
• The nature of the image.

Solution:

Let, the distance of the image from the mirror be, 'v'.

The magnificent of the image be, 'm'.

The size of the image be, 'I'.

f= 20 cm

u = 12 cm

From the formula we get that,

$\dfrac{1}{f} = (\dfrac{1}{u} + \dfrac{1}{v})$

$or, \dfrac{1}{20} = (\dfrac{1}{12}+\dfrac{1}{v})$

$or, (\dfrac{1}{12} + \dfrac{1}{v}) = \dfrac{1}{20}$

$or, \dfrac{1}{v}= ( \dfrac{1}{20} - \dfrac{1}{12})$

$or, \dfrac{1}{v} = (\dfrac{3-5}{60})$

$or, \dfrac{1}{v}= \dfrac{-2}{60}$

$or, \dfrac{1}{v} = \dfrac{-1}{30}$

or, v = -30 cm.

∴ The image has formed at a distance of 30 cm behind the mirror.

From the formula we get that,

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

$or, m = \dfrac{-(-30)}{12}$

$or, m = \dfrac{30}{12}$

or, m = 2.5 [Magnificent is a ratio so, it has no unit.]

Since, the magnificent of the image is positive so, the image is virtual, erect and upright.

From the formula we get that,

$m = \dfrac{I}{O}$

$or, 2.5 = \dfrac{I}{2.5}$

$or, \dfrac{I}{2.5}= 2.5$

or, I = (2.5×2.5)

or, I = 6.25 cm.

∴ The size of the image is 6.25 cm.

Answer:

             The size of the image is 6.25 cm.

The image is virtual, erect and upright.

Answer 2

Answer:

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