Question

An object is placed at 20 cm from a convex lens of focal length 10 cm and height of an object is 5 cm what is the position size and nature of an image find magnification and power of lens

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Answer 1

Answer:

Given:u=−10cm               f=20cm        (convex lens)

To find: v and image's nature.

Solution: From lens formula

v1−u1=f1

v1+101=201

v1=20−1

v=−20cm

And magnification = uv=2

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

Explanation:

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Answer 2

Answer:

Given: u = −10cm

f = 20cm

(convex lens)

To find: v and image's nature.

Solution: From lens formula

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

$\frac{1}{v} + \frac{1}{10} = \frac{1}{20}$

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

v = −20cm

And magnification

$\frac{v}{u} = 2$

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