Question

An object of height 4 cm is kept at a distance of 30 cm from a concave lens. Use lens formula to determine the image distance, nature and size of the image formed if focal length of the lens is 15 cm. (3)

Answer 1

We haveHeight of object, h1= 4 cmFocal length of lens, f = -15 cmObject distance, u = -30 cmThus the image will be formed in front of the lens at a distance of 10 cm from the lens, virtual and erect of size 1.33 cm.

Answer 2

Given :

Height of object ( h ) = + 4 cm

Object distance ( u ) = - 30 cm

( negative sign indicates that the image is kept on the left side in front of the lens ).

Focal length of lens ( f ) = - 15 cm

To find :

1. Image distance ( v )
2. Nature of image produced .
3. Height of image ( h' )

Explanation :

We know , the lens formula , $\frac{1}{v}$ - $\frac{1}{u}$ = $\frac{1}{f}$ ; or

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

                  therefore ,         v = $\frac{u f}{u + f}$

substituting the values we get ,

                                           v = $\frac{( - 15)(- 30)}{( -15) + ( -30)}$

therefore , v = - 10 cm

Now ,

magnification ,                m = $\frac{v}{u} = \frac{h'}{h}$

substituting the values  m = $\frac{- 10}{ - 30} = \frac{h'}{4}$

        therefore , h' = $\frac{4}{3}$ = +1.33 cm

Final Answer :

1. The image is formed at 10 cm in front of the lens between optic center and focus .
2. The image is virtual and erect , also , diminished with respect to the object .
3. The size or height of the image formed is 1.33 cm .