Question

an object is placed at a distance of 20 CM in front of a concave lens of focal length 20 cm find the position of image and the magnification of image

Answer 1

Answer:

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Explanation:

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Answer 2

Answer:

The position of the image is -10cm and the magnification of the image is 0.5.

Explanation:

We will use the given formulas to solve this question,

$\frac{1}{f} =\frac{1}{v} -\frac{1}{u}$      (1)

$m=\frac{v}{u}$       (2)

Where,

f=focal length of the lens

v=image distance from the lens

u=object distance from the lens

m=magnification of the image

From the question we have,

u=-20cm

f=-20cm

By substituting the value of f and u in equation (1) we get;

$\frac{1}{-20} =\frac{1}{v} -\frac{1}{-20}$

$\frac{1}{v} =\frac{1}{-20} +\frac{1}{-20}$

$\frac{1}{v}=\frac{-2}{20}$

$\frac{1}{v}=\frac{-1}{10}$

$v=-10cm$

And by substituting the required values in the equation (2) we get;

$m=\frac{-10}{-20}=\frac{1}{2}$

$m=0.5$

Hence, the position of the image is -10cm and the magnification of the image is 0.5.