Question

A convex lens of focal length 20 cm forms a virtual image of an object at 40 cm from the lens. The magnification produced will be
TELL WITH EXPLAINATION

Answer 1

S O L U T I O N :

Given :

• Focal length, (f) = 20 cm
• Distance of object, (u) = -40 cm

Explanation :

As we know that formula of the lens:

• 1/f = 1/v - 1/u

According to the question :

➝ 1/f = 1/v - 1/u

➝ 1/20 = 1/v - 1/(-40)

➝ 1/20 = 1/v + 1/40

➝ 1/v = 1/20 - 1/40

➝ 1/v = 2 - 1/40

➝ 1/v = 1/40

➝ v = 40 cm

Therefore, the height of image will be 13.3 cm .

Now,

As we know that formula of magnification:

➝ Magnification = Height of image/Height of object

➝ Magnification = -v/u

➝ Magnification = -40/-40

➝ Magnification = 1

Thus,

The Magnification produced will be 1 .

Answer 2

$\sf Given \begin{cases} & \sf{f = 20 cm} \\ & \sf{ u = -40} \end{cases}$

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$\sf To \: Find \begin{cases} \bf Magnification \: Power\end{cases}$

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$\sf Formula \: used \begin{cases} & \boxed{\sf\dfrac{1}{f} = \dfrac{1}{v} - \dfrac{1}{u}} \\ \\ & \boxed{\sf m = \dfrac{-v}{u}} \end{cases}$

where

• f is the focal length
• u is the distance of object
• v is height of image
• m is magnification power

Solution :-

• f = 20 cm
• u = -40 cm

Substituting the value in formula :-

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

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➩ $\sf \dfrac{1}{20} = \dfrac{1}{v} - \dfrac{1}{-40}$

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➩ $\sf \dfrac{1}{20} = \dfrac{1}{v} + \dfrac{1}{40}$

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➩ $\sf \dfrac{1}{v} = \dfrac{1}{20} - \dfrac{1}{40}$

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➩ $\sf \dfrac{1}{v} = \dfrac{1}{40}$

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➩ $\boxed{\sf v = 40 cm}$

Height of image = 40 cm

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Now, we have

• v = 40 cm
• u = -40 cm

Substituting the value in formula :-

➩ $\sf m = \dfrac{-v}{u}$

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➩ $\sf m = \dfrac{-40}{-40}$

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➩ $\boxed{\sf m = 1}$

Magnification power = 1