Question

The focus distance of a convex lens is 20 cm. At what distance from this lens should an object be placed so that its image is made 2 times larger than the object, while the image is real?​

Answer 1

Solution :-

Focal length of the convex lens f = 20 cm

Given :

The image formed is 2 times larger than the object

Therefore, Magnification m = - 2

[ 2 is taken as negative since the image is real ]

Also, Magnification m = v/u

⇒ - 2 = v/u

⇒ - 2u = v

⇒ v = - 2u

Now, by using lens formula

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

Substituting f = 20 and v = - 2u

$\implies \rm \dfrac{1}{20} = \dfrac{1}{ - 2u} - \dfrac{1}{u}$

$\implies \rm \dfrac{1}{20} = - \dfrac{1}{2u} - \dfrac{1}{u}$

$\implies \rm \dfrac{1}{20} = \dfrac{ - 1 - 2}{2u}$

$\implies \rm \dfrac{1}{20} = \dfrac{ - 3}{2u}$

$\implies \rm 2u = - 3 \times 20$

$\implies \rm u = \dfrac{ - 60}{2}$

$\implies \rm u = - 30$

Therefore the object is placed 30 cm from the convex lens.

Answer 2

Answer:

$\large\boxed{\sf{30\:\;cm}}$

Explanation:

Given that there is a convex lens.

• focal length, f = 20 cm

Since, the image is real, therefore magnification will be negative.

.°. Magnification, m = -2

Let us imagine that

• Object distance = u
• Image distance = v

Also, we know that , $\bold{m=\dfrac{v}{u}}$

Substituting the values, we get

$= > \sf{ \frac{v}{u} = - 2} \\ \\ = > \sf{v = - 2u \: \: \: \: \: \: \: \: \: .............(1)}$

Now , from lens formula, we have

$\sf{ \red{\dfrac{1}{v} - \dfrac{1}{u} = \dfrac{1}{f} }}$

Again, substituting the values, we get

$= > \sf{ \frac{1}{ - 2u} - \frac{1}{u} = \frac{1}{20} \: \: \: \: \: \: \: \: \: \: .............(from \: \: 1)} \\ \\ \sf{= > \frac{ - 1 - 2}{2u} = \frac{1}{20}} \\ \\ = > \sf{ - \frac{3}{2u} = \frac{1}{20} }\\ \\ \sf{= > 2u = - 60 }\\ \\ \sf{= > u = - 30}$

Hence, object should be placed at a distance of 30 cm from the optical center of the convex lens.