Question

A convex lens forms an erect and three times magnified image of an object placed at a distance 10 cm in front of it. Find:
(a) the position of the image,
(b) the focal length of the image.

Answer 1

Answer:

Given Information,

• Object Distance (u) = -10 cm
• Magnification = +3

To Find,

• Image Distance (v)
• Focal Length of the lens

Solution,

We know that, magnification of a lens is given as:

$\implies m = \dfrac{v}{u}\\\\\text{Substituting the known values we get:}\\\\\implies 3 = \dfrac{v}{-10}\\\\\implies v = -10 \times 3\\\\\implies \boxed{ \bf{ v = -30\:\:cm}}$

Hence the position of the image is formed 30 cm towards the left of the optic center (in diagram).

To calculate the focal length, we use the lens formula. Hence we get:

$\implies \dfrac{1}{f} = \dfrac{1}{v} - \dfrac{1}{u}\\\\\\\implies \dfrac{1}{f} = \dfrac{1}{-30} - \dfrac{1}{(-10)}\\\\\\\implies \dfrac{1}{f} = \dfrac{1}{-30} + \dfrac{1}{10}\\\\\\\text{Taking LCM we get:}\\\\\\\implies \dfrac{1}{f} = \dfrac{-1+3}{30}\\\\\\\implies \dfrac{1}{f} = \dfrac{2}{30} = \dfrac{1}{15}\\\\\\\implies \boxed{ \bf{f = 15\:\:cm}}$

Hence the focal length of the image is 15 cm.

Answer 2

Answer:

Given :-

• A convex lens forms an erect and three times magnified image of an object placed at a distance of 10 cm in front of it.

To Find :-

• What is the position or image distance of the image.
• What is the focal length of the image.

Formula Used :-

$\clubsuit$ Magnification Formula :

$\mapsto \sf\boxed{\bold{\pink{M =\: \dfrac{v}{u}}}}$

where,

• M = Magnification
• v = Image Distance
• u = Object Distance

$\clubsuit$ Lens Formula :

$\mapsto \sf\boxed{\bold{\pink{\dfrac{1}{v} - \dfrac{1}{u} =\: \dfrac{1}{f}}}}$

where,

• v = Image Distance
• u = Object Distance
• f = Focal Length

Solution :-

First, we have to find the image distance or the position of the image :

Given :

• Magnification = 3
• Object Distance = - 10 cm

According to the question by using the formula we get,

$\bigstar\: \: \bf M =\: \dfrac{v}{u}\: \: \bigstar$

$\implies \sf 3 =\: \dfrac{v}{- 10}$

By doing cross multiplication we get,

$\implies \sf v =\: - 10(3)$

$\implies \sf v =\: - 10 \times 3$

$\implies \sf\bold{\purple{v =\: - 30\: cm}}$

${\small{\bold{\underline{\therefore\: The\: position\: of\: the\: image\: is\: 30\: cm\: .}}}}$

[Note : The position of the image is infront of the mirror. ]

Now, we have to find the focal length of the image :

Given :

• Image Distance = - 30 cm
• Object Distance = - 10 cm

According to the question by using the formula we get,

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

$\longrightarrow \sf \dfrac{1}{- 30} - \dfrac{1}{- 10} =\: \dfrac{1}{f}$

$\longrightarrow \sf \dfrac{1}{- 30} - \bigg\{- \dfrac{1}{10}\bigg\} =\: \dfrac{1}{f}$

$\longrightarrow \sf - \dfrac{1}{ 30} + \dfrac{1}{10}$

$\longrightarrow \sf \dfrac{- 1 + 3}{30} =\: \dfrac{1}{f}$

$\longrightarrow \sf \dfrac{2}{30} =\: \dfrac{1}{f}$

By doing cross multiplication we get,

$\longrightarrow \sf 2 \times f =\: 30(1)$

$\longrightarrow \sf 2f =\: 30 \times 1$

$\longrightarrow \sf 2f =\: 30$

$\longrightarrow \sf f =\: \dfrac{\cancel{30}}{\cancel{2}}$

$\longrightarrow \sf f =\: \dfrac{15}{1}$

$\longrightarrow \sf\bold{\red{f =\: 15\: cm}}$

${\small{\bold{\underline{\therefore\: The\: focal\: length\: of\: the\: image\: is\: 15\: cm\: .}}}}$