Question

A concave lens has focal length of 10cm. At what distance should the object from the lens be placed so that it forms an image at 8 cm from the lens. Also , find the magnification produced by the lens.​

Answer 1

Given :-

• Focal length of lens (f) = 10cm
• The distance of image (v) = 8cm

To Find :-

• The magnification produced by the lens = ?

Solution :-

• To calculate the magnification produced by the lens at first we have to find out the distance of object (u). Then calculate the magnification of lens by applying formula. As you know the distance of image is virtual so the value of (v) = - 8 and a concave focal length (f) = - 10cm.

As per the mirror formula :-

⇒ 1/v - 1/u = 1/f

• V denotes = distance of image
• U denotes = distance of object
• F denotes = focal length of lens

⇒ -1/u = 1/f - 1/v

• f = -10cm v = -8cm

⇒ -1/u = -1/10 - (-1/8)

⇒ -1/u = -1/10 + 1/8

⇒ -1/u = (-4 + 5)/40

⇒ -1/u = 1/40

⇒ u = -40cm

• Now calculate magnification of lens :-]

⇒ Magnification of lens = -v/u

⇒ Magnification of lens = -8/-40

⇒ Magnification of lens = 1/5

Hence,

• Magnification produced by the lens = 1/5

Answer 2

Answer:

Given :-

• A concave lens has focal length of 10 cm.
• The lens be placed so that it forms an image at 8 cm from the lens.

To Find :-

• What is the distance of object.
• What is the magnification produced by the lens.

Formula Used :-

$\clubsuit$ Mirror Formula :

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

where,

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

$\clubsuit$ Magnification Formula :

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

where,

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

Solution :-

First, we have to find the object distance :

Given :

• Focal Length (f) = - 10 cm
• Image Distance (v) = - 8 cm

According to the question by using the formula we get,

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

$\implies \sf \dfrac{1}{- 10} =\: \dfrac{1}{- 8} - \dfrac{1}{u}$

$\implies \sf \dfrac{1}{u} =\: \dfrac{1}{10} - \dfrac{1}{8}$

$\implies \sf \dfrac{1}{u} =\: \dfrac{4 - 5}{40}$

$\implies \sf \dfrac{1}{u} =\: \dfrac{- 1}{40}$

By doing cross multiplication we get,

$\implies \sf - 1 \times u =\: 40(1)$

$\implies \sf - 1u =\: 40 \times 1$

$\implies \sf - 1u =\: 40$

$\implies \sf u =\: \dfrac{40}{- 1}$

$\implies \sf\bold{\purple{u =\: - 40\: cm}}$

${\small{\bold{\underline{\therefore\: The\: object\: distance\: is\: 40\: cm\: .}}}}$

Now, we have to find the magnification produced by the lens :

Given :

• Image Distance (v) = 8 cm
• Object Distance (u) = - 40 cm

According to the question by using the formula we get,

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

$\longrightarrow \sf m =\: \dfrac{- 8}{- 40}$

$\longrightarrow \sf m =\: \dfrac{\cancel{-} 8}{\cancel{-} 40}$

$\longrightarrow \sf m =\: \dfrac{8}{40}$

By doing cross multiplication we get,

$\longrightarrow \sf m \times 40 =\: 8$

$\longrightarrow \sf 40m =\: 8$

$\longrightarrow \sf m =\: \dfrac{\cancel{8}}{\cancel{40}}$

$\longrightarrow \sf m =\: \dfrac{1}{5}$

$\longrightarrow \sf\bold{\red{m =\: + 0.2}}$

${\small{\bold{\underline{\therefore\: The\: magnification\: produced\: by\: the\: lens\: is\: + 0.2\: .}}}}$

[Note :- The positive sign shows that the image is erect and virtual. ]