Question

A concave lens produces an image of height 8cm of an object of height 16cm.The power of the lens is

10D.Calculate the position of the object and image from the lens.​

Answer 1

QuestioN :

A concave lens produces an image of height 8cm of an object of height 16cm.The power of the lens is  10D.Calculate the position of the object and image from the lens.​

ANswer :

The image height is 1.42cm.

Given :

In concave lens,

Object height = 5 cm

Object distance = - 20 cm

Focal length = -8 cm

To find :

• The position and height of the iimage

SolutioN :

Using lens formula,

A formula which gives the relationship between image distance, object distance and focal length of a lens is known as the lens formula.

The lens formula can be written as :

$\frac{1}{V} -\frac{1}{U} =\frac{1}{F}$

where,

• v denotes image distance

• u denotes object distance

• f denotes focal length

• by substituting all the given values in the formula,

by substituting all the given values in the formula,

$=\frac{1}{V} -\frac{1}{U} =\frac{1}{F} \\\\=\frac{1}{V} =\frac{1}{-20} =-\frac{1}{8} \\\\=\frac{1}{V} +\frac{1}{20} =-\frac{1}{8} \\\\=\frac{1}{V} =-\frac{1}{8} -\frac{1}{20} \\\\=\frac{1}{V} =\frac{-5-2}{40} \\\\=\frac{1}{V} =-\frac{7}{40} \\\\=V=-\frac{40}{7} \\\\=V=-5.7\:cm$

thus, position of image is -5.7 cm.

Now, using Magnification that is

$\left[\begin{array}{ccc}m=\frac{h'}{h} =\frac{U}{V}\end{array}\right]$

here,

• h' = image height

• h = object height

• m = magnification

• u = object distance

• v = image distance

by substituting all the given values,

$=\frac{h'}{h} =\frac{V}{U} \\\\=\frac{h'}{5} =\frac{5.7*5}{20} \\\\h'=\frac{28.5}{20} \\\\h'=1.42\:cm$

thus, the image height is 1.42cm.

Answer 2

QuestioN :

A concave lens produces an image of height 8cm of an object of height 16cm.The power of the lens is  10D.Calculate the position of the object and image from the lens.​

ANswer :

The image height is 1.42cm.

Given :

In concave lens,

Object height = 5 cm

Object distance = - 20 cm

Focal length = -8 cm

To find :

• The position and height of the iimage

SolutioN :

Using lens formula,

A formula which gives the relationship between image distance, object distance and focal length of a lens is known as the lens formula.

The lens formula can be written as :

$\frac{1}{V} -\frac{1}{U} =\frac{1}{F}$

where,

• v denotes image distance

• u denotes object distance

• f denotes focal length

• by substituting all the given values in the formula,

by substituting all the given values in the formula,

$=\frac{1}{V} -\frac{1}{U} =\frac{1}{F} \\\\=\frac{1}{V} =\frac{1}{-20} =-\frac{1}{8} \\\\=\frac{1}{V} +\frac{1}{20} =-\frac{1}{8} \\\\=\frac{1}{V} =-\frac{1}{8} -\frac{1}{20} \\\\=\frac{1}{V} =\frac{-5-2}{40} \\\\=\frac{1}{V} =-\frac{7}{40} \\\\=V=-\frac{40}{7} \\\\=V=-5.7\:cm$

thus, position of image is -5.7 cm.

Now, using Magnification that is

$\left[\begin{array}{ccc}m=\frac{h'}{h} =\frac{U}{V}\end{array}\right]$

here,

• h' = image height

• h = object height

• m = magnification

• u = object distance

• v = image distance

by substituting all the given values,

$=\frac{h'}{h} =\frac{V}{U} \\\\=\frac{h'}{5} =\frac{5.7*5}{20} \\\\h'=\frac{28.5}{20} \\\\h'=1.42\:cm$

thus, the image height is 1.42cm.