Question

An object of height 9 cm is kept at a distance of 24 cm in front of a concave mirror. The focal length of the mirror is 16 cm. At what distance from the mirror should a screen be kept so as to get a clear image? What will be the size and nature of the image?
Just tell me the formula

Answer 1

Answer:

Image distance = -48 cm

Image size = -18 cm

Nature = Real and inverted

Explanation:

Given:

• Height of the object = 9 cm
• Object distance =  24 cm
• Focal length of the mirror = 16 cm

To Find:

• Image distance
• Size and nature of the image

Solution:

Given that it is concave mirror. By sign convention,

Object distance = -u = -24 cm

Focal length = -f = -16 cm

By mirror formula we know that,

$\tt \dfrac{1}{f} =\dfrac{1}{v} + \dfrac{1}{u}$

where f is the focal length, u is the object distance, v is the image distance

Substituting the data we get,

$\tt \dfrac{1}{-16} =\dfrac{1}{-24} + \dfrac{1}{u}$

$\tt \dfrac{1}{v} =\dfrac{1}{-16} +\dfrac{1}{24}$

$\tt \dfrac{1}{v}=\dfrac{-1}{48}$

v = -48 cm

Hence the screen must be kept at a distance of 48 cm from the concave mirror.

Now we know that,

$\tt m=\dfrac{-v}{u} =\dfrac{Image\:height}{Object\:height}$

where m is the magnification

$\tt m=\dfrac{48}{-24} =\dfrac{h}{9}$

h = -18 cm

The height/size of the image is -18 cm and the negative sign indicates that image is real and inverted.