Question

A 4cm object is placed perpendicular to principal axis of convex lense of focal length 20 cm the distance of object from the lens is 15 cm. find the nature, position and size of image

Answer 1

Height of object ho = 4 cm

Focal lenght of convex lens = 20 cm

Object distance u = -15 cm

Image distance v = ?

Image height hi = ?

Lens formula: 1/v  -  1/u  =  1/f

                        1/v  = 1/f  +  1/u

                               =  1/20  +  1/(-15)

                               = -1/60

⇒  v = - 60 cm

The image distance is 60 cm on the same side of the lens as the object.

Magnification m = v/u = (-60)/(-15) = 4 = hi/ho = hi/4

⇒   hi = 4 x 4 = 16 cm

The image is enlarged and erect.

So a virtual, erect, enlarged image is formed at 60 cm from the optic centre on the same side of the lens as the object is positioned.

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Answer 2

Height of object ho = 4 cm

Focal lenght of convex

lens = 20 cm Object

distance u = -15 cm

Image distance v = ?

Image height hi = ?

Lens formula: 1/v - 1/u = 1/f

1/v = 1/f + 1/u

= 1/20 + 1/(-15)

= -1/60 ⇒ v = - 60 cm

The image distance is 60 cm on the same side of the lens as the object.


Magnification m = v/u = (-60)/(-15) = 4 = hi/ho = hi/4 ⇒ hi = 4 x 4 = 16 cm


The image is enlarged and erect. So a virtual, erect, enlarged image is formed at 60 cm from the optic centre on the same side of the lens as the object is positioned.

I hope it is helped you