Question

an object of height 8cm is placed at the distance of 40cm infront of a convex lens of focal length 20cm .The size of images is a)12cm b) 4cm c)-8cm d)16cm​

Answer 1

Answer:

u=-40

f=20

ho=8

1/v-1/u=1/f

v=40

m=v/u=hi/ho

hi=-8 cm

obj is placed on curvature

Answer 2

Answer:

The size of the image is -8cm i.e.option(c).

Explanation:

First, let's find the image distance through the lens formula which is given as,

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

Where,

f=focal length of the lens

v=image distance from the lens

u=object distance from the lens

From the question we have,

The object distance from the lens=-40cm

The focal length of the lens=20cm

The height of the object=8cm

Placing all the required values in equation (1) we get;

$\frac{1}{20}= \frac{1}{v}- \frac{1}{-40}$

$\frac{1}{v}=\frac{1}{20}-\frac{1}{40}$

$\frac{1}{v}=\frac{2-1}{40}$

$\frac{1}{v}=\frac{1}{40}$

$v=40cm$      (2)

And we know that the magnification is calculated is two ways i.e.

$m=\frac{h_2}{h_1}$      (3)

$m=\frac{v}{u}$       (4)

h₂=height of image

h₁=height of object

Equation (3) and equation (4) together give us;

$\frac{h_2}{h_1}=\frac{v}{u}$     (5)

When we enter the values into equation (5), we obtain;

$\frac{h_2}{8}=\frac{40}{-40}$

$h_2=-8\ cm$

The size of the image is -8cm i.e.option(c).

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