Question

size of the image of an object by a concave lens of focal length 20cm is observed to be reduced to one third of its size find the distance of object from the lens

Answer 1

Answer: The distance of the object is 40 cm from the lens.

Explanation:

Given that,

Focal length f = -20 cm

Magnification $m =\dfrac{1}{3}$

We know that,

The magnification is

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

$\dfrac{1}{3}=\dfrac{v}{u}$

$u = 3v$......(I)

Using lens formula

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

$\dfrac{1}{-20}=\dfrac{1}{v}-\dfrac{1}{3v}$

$v = \dfrac{-40}{3}$

Put the value of v in equation (I)

$u = -40\ cm$

Hence, The distance of the object is 40 cm from the lens.

Answer 2

The distance of object from the lens is 40 cm

Explanation:

Given that,

Focal length of the concave lens, f = -20 cm

Image is reduced to one third of its size, $m=\dfrac{1}{3}$

We need to find the distance of object from the lens. It can be calculated using the definition of magnification. It is given by :

$m=\dfrac{h'}{h}=\dfrac{v}{u}$

Where

h' is the size of image

h is the size of object

v is the image distance

u is the object distance

$\dfrac{1}{3}=\dfrac{v}{u}$

$u=3v$............(1)

Using lens formula :

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

$\dfrac{1}{20}=\dfrac{1}{v}-\dfrac{1}{3v}$

$v=\dfrac{-40}{3}\ cm$

Put v in equation (1) as :

$u=3\times \dfrac{-40}{3}$

u = -40 cm

So, the distance of object from the lens is 40 cm. Hence, this is the required solution.

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Lens formula

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