Question

2
An object is placed 10 cm from a concave lens of focal length 20 cm .
Find the position of the
image and find its magnification .
Pls give appropriate answer ​

Answer 1

Solution :-

As per the given data ,

• Image distance (u) = - 10 cm ( distance is always measured from the optical center of the lens  )

• Focal length (f) = - 20 cm ( focal length of a concave lens is negative )

As per the lens formula ,

➝ 1  / f = 1 / v - 1 / u

➝ 1 / v = 1 / f + 1 / u

➝ 1 / v = - 1 / 20 - 1 / 10

➝ 1 / v = -1 - 2 / 20

➝ 1 / v = -3 / 20

➝ v = - 20 / 3

➝ v = - 6.6 cm

The image is formed at a distance of 6.66 cm from the lens  on the same side as that of the object

Magnification ,

➝ m = v / u

➝ m = -6.6 / - 10 = 0.66

The magnification of the object is 0.66  

Answer 2

$\sf \huge \underline \green{Solution}$

Given:

$\sf u = - 10cm. \\ \sf f = 20cm.$

So,

From Lens Formula:

$\sf \frac{1}{v} - \frac{1}{u} = \frac{1}{f} \\ \\ \sf \frac{1}{v} + \frac{1}{10} = \frac{1}{20} \\ \\ \sf \frac{1}{v} = \frac{ - 1}{20} .$

$\sf\small\green{v = - 20}$

$\sf{And \: \: Magnification: \: \frac{v}{u} = 2.}$

Hence image formed is virtual, erect and magnified two times.

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Hope it helps you!