Question

a convex lence has a focal length of 30cm. calculate at what distance from the lence,should an object be placed so that it forms an image at of 60cm on the other side of of the lence. find the magnification produced by the lence in this case

Answer 1

Answer:

Given information –

• Convex Lens
• Focal length = + 30 cm
• Image distance = + 60 cm

To Find –

• Object distance ?
• Magnification ?

Solution –

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

➸ 1/60 - 1/u = 1/30

➸ - 1/u = 1/30 - 1/60

➸ - 1/u = (60 - 30)/1800

➸ - 1/u = 30/1800

➸ - 1/u = 3/180

➸ - 1/u = 1/60

➸ u = - 60 cm

• Position of Object : At F1.
• Position of Image : At F2.
• Size of Image : Same size.
• Nature of Image : Real and Inverted.

Magnification : h'/h = v/u.

➸ +60/-60

➸ -1

Answer 2

Answer :

The object should be placed 60cm away from the lens to obtain required position of image.

The magnification produced is -1

Given :

• The focal length of a convex lens is 30cm
• The image forms at a distance of 60cm on the other side of the lens.

To Find :

• The distance where object should be placed (object distance)
• The magnification produced by the lens.

Formulae to be used :

Lens formula :

$\star \tt \: \: \dfrac{1}{v}- \dfrac{1}{u} = \dfrac{1}{f}$

Magnification for spherical lens is given by :

$\tt \star \: \: m = \dfrac{h_{2}}{h_{1}}= \dfrac{v}{u}$

Solution :

Given ,

focal length , f = 30cm

image distance , v = 60cm

From lens formula :

$\tt \implies \dfrac{1}{60}- \dfrac{1}{u} = \dfrac{1}{30} \\\\ \tt \implies \dfrac{1}{u} - \dfrac{1}{60} = -\dfrac{1}{30} \\\\ \tt\implies \dfrac{1}{u} = \dfrac{1}{60}-\dfrac{1}{30} \\\\ \tt \implies \dfrac{1}{u} = - \dfrac{1}{60} \\\\ \tt \implies u = -60cm$

The object distance is -60cm . Thus the object should be placed 60cm apart from the lens

Now magnification ,

$\tt \implies m = \dfrac{v}{u} \\\\ \tt \implies m = \dfrac{60}{-60} \\\\ \tt \implies m = -1$

Magnification is -1 . The negative sign represents that the image form is real and inverted. The magnitude is 1 i.e. object size and image size remains same.