Question

An object 4 cm high is placed at a distance of 10cm from a convex lens of focal length 20cm find the position, nature and size of image

Answer 1

Explanation:

H= 4 cm

u = -10 cm

f = 20 cm.

According to lens formula,,

1/v - 1/u = 1/f

1/v = 1/f + 1/u

1/v = 1/20 - 1/10

1/v = -1/20

v = -20

We have, m = v/u

                    = -20/-10 = 2

Also, m= h'/h  ( where h = height of object and h'= height of image)

2=h'/4

h' = 4X2 = 8cm.

Therefore, the image is real and inverted.

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Answer 2

Answer:

The image formed is virtual of height 8cm and is formed at distance 8cm from the pole.

Explanation:

Len's formula :   $\frac{1}{v} -\frac{1}{u} =\frac{1}{f}$

where,

$v$ image distance,

$u$ is the distance if the object,

$f$ is the focal length.

Given :   $u=-10cm; f=20cm; v=?$

$\implies \frac{1}{v} -\frac{1}{-10} =\frac{1}{20}$

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

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

$\impliese v=-20cm$

Magnification = $\frac{v}{u} = \frac{h_i}{h_o}$

where, $h_i, h_o$ are the heights of image and the object respectively.

$\implies \frac{v}{u} = \frac{-20}{-10} = \frac{h_i}{4}$

$\implies h_i=2*4 = 8cm$

So, the image formed is virtual of height 8cm and is formed at distance 8cm from the pole.

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