Question

A converging beam of light passes through a diverging lens of focal length 0.2 metres and comes at a point 0.3 metres behind the lens. find the position of the point at which the beam would converge in the absense of lens

Answer 1

Answer:

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Explanation:

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Answer 2

Answer:

Given-

Lens is concave (diverging lens))

focal length of lens is 0.2 m

and image comes at point 0.3 m

To find-

the position at which the beam is converge and forms a image ?

Explanation:

Firstly we have to change 0.2 metres into centimetres .

so, $0.2 m \times 100=20cm$

so the focal length of the lens is 20 cm.

and the object distance is 30 cm.

According to the lens formula-

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

$\frac{1}{-20} =\frac{1}{v} - \frac{1}{30}$

• here, object distance (u) is 30 cm.
• and focal length (f) is -20

$\frac{1}{-20} +\frac{1}{30} = \frac{1}{v}$

$\frac{+3+2}{ 60} = \frac{1}{v}$

$v=12cm$

The position at which the beam is converge and forms a image is 12 cm.