Question

a 2cm tall object is placed perpendicular to the principal axis of a convex lens of focal length 10cm . distance of object is 15 cm .find it's position ,nature, and size​

Answer 1

Type of mirror = convex

h (Height of object) = 2 cm

f = 10 cm

u = -15 cm

v =  ?

h' (Height of image) = ?

m = ?

Solution

By using mirror formula,

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

$\frac{1}{-15}+\frac{1}{v}=\frac{1}{10}$

$\frac{1}{v}=\frac{1}{10}+\frac{1}{15}$

$\frac{1}{v}=\frac{3+2}{30}$

$\frac{1}{v}=\frac{5}{30}$

$\frac{1}{v}=\frac{1}{6}$

Reciprocating both sides

$v=6 cm$

$m=\frac{-v}{u}$

$m=\frac{-6}{-15}$

$m=\frac{2}{5}$

$m=+0.4$

$m=\frec{h'}{h}$

$0.4=\frec{h'}{2}$

$h'= 0.8 cm$

Position = 6 cm behind the mirror (Between P and F)

Size = 0.8 cm

Nature = Virtual, Erect and Diminished

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Answer 2

h=2cm

f=10cm

u= -15cm

v=?; h'=?

1/v-1/u=1/f

1/v-(1/-15)=1/10

1/v=1/10-1/15

1/v=3-2/30

1/v=30

v=30cm

m=v/u=h'/h

30/(-15)=h'/2

-2*2=h'

h'= -4cm

The position of the image=30cm

The nature of the image is real and inverted

The size of the image = 4cm

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