Question

An object 4cm in size is placed at a distance of 25cm from a concave mirror of focal length 15cm. Find the position, nature, and height of the image

Answer 1

Given:

Focal length =15cm

Object distance =25cm

Object height  =4cm

To Find:

The position, nature, and height of the image.

Solution:

We can simply solve this numerical problem by using the following process.

Now,

Focal length (f) =15cm

Object distance (u) =25cm

Object height (h₁)  =4cm

Image distance (v) = ?

Magnification (m) =?

[We have to find out at what distance the screen should be placed so that a sharp image of the object can be obtained.

We know that the object distance, u is always negative and here, the mirror is a concave mirror. So, the focal length is also considered as negative.]

From mirror formula,

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

$\frac{1}{-25} =\frac{1}{-15} +\frac{1}{v}$

$\frac{1}{v}= \frac{1}{-15}+\frac{1}{25}$

$v= -37.5 cm$

Magnification (m) = $- \frac{v}{u} = \frac{h_{1} }{h_{2} } }$

m = $-\frac{(-37.5)}{25} =\frac{h_{1} }{4}$

h₁ =$- \frac{37.5 X 4}{25}$

h₁ = -6 cm

Hence,

To get a sharp image, v = 37.5 cm

The height of the image is 4cm

The image is real, inverted and magnified.

Therefore, the nature of the image is real and inverted and the height of the image is 4 cm.