Question

a concave lens of focal length 5 cm with the object is placed at a distance of 10 cm from the lens . find the nature, image, distance and magnification?​

Answer 1

Solution :

As per the given data ,

• Focal length (f) = - 5 cm ( focal length of concave mirror is negative )

• Image distance (u) = - 10 cm ( Distances are always measured from the optical center of the lens )

By applying Lens formula ,

➜ 1 / f = 1 / v - 1 / u

➜ 1 / v = 1 / f + 1 / u

➜ 1 / v = - 1/ 5 - 1 / 10

➜ 1 / v = -2-1 / 10

➜ 1 / v = -3 / 10

➜ v = - 10 / 3

➜ v = - 3.33 cm

The image is formed as a distance of 3.33 cm in front of the concave mirror on same side as that of the object .

Magnification

➜ m =  v / u = - 3.33 / - 10 = 0.33

The image formed is erect , virtual and diminished

Answer 2

Given :

focal length, f = -5cm

object distance, u = -10cm(It is to the left of lens)

Remember : In concave lens focal length is negative

To find :

the nature, image, distance and magnification

Solution :

using lens formula,

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

The lens formula can be written as :

${\leadsto{\bf{\dfrac{1}{v} - \dfrac{1}{u} = \dfrac{1}{f}}}}$

${\leadsto{\bf{\dfrac{1}{v} - \dfrac{1}{ - 10} = \dfrac{1}{ - 5}}}}$

${\leadsto{\bf{\dfrac{1}{v} = - \dfrac{1}{5} - \dfrac{1}{10} }}}$

${\leadsto{\bf{\dfrac{1}{v} = \dfrac{ - 2 - 1}{10}}}}$

${\leadsto{\bf{\dfrac{1}{v} = - \dfrac{ 3}{10}}}}$

${\leadsto{\bf{v = - \dfrac{10}{3}}}}$

${\leadsto{\bf{v = 3.33 \: cm}}}$

thus, image distance = 3.33cm

Magnification :

The Linear magnification produced by a lens is equal to the ratio of image distance to the object distance .i.e.,

${\leadsto{\bf{m= \dfrac{v}{u}}}}$

${\leadsto{\bf{m= \dfrac{ - 3.33}{ - 10}}}}$

${\leadsto{\bf{m=0.33}}}$

thus, magnification = 0.33

Nature of image :

• the image is virtual and erect
• the image is diminished