Question

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Answer 1

Solution :-

First, we have to find the image distance :

\implies \sf \dfrac{- v}{u} =\: \dfrac{3\cancel{h_o}}{\cancel{h_o}}⟹

u

−v

=

h

o

3

h

o

\implies \sf \dfrac{- v}{u} =\: 3⟹

u

−v

=3

By doing cross multiplication we get,

\implies \sf - v =\: 3u⟹−v=3u

\implies \sf \bold{\purple{v =\: - 3u}}⟹v=−3u

Now, we have to find the object distance :

Given :

Image Distance (v) = - 3u

Focal Length (f) = - 20 cm

\longrightarrow \sf \dfrac{1}{- 20} =\: \dfrac{1}{- 3u} + \dfrac{1}{u}⟶

−20

1

=

−3u

1

+

u

1

\longrightarrow \sf \dfrac{1}{- 20} =\: \dfrac{- 1 + 3}{3u}⟶

−20

1

=

3u

−1+3

\longrightarrow \sf \dfrac{1}{- 20} =\: \dfrac{2}{3u}⟶

−20

1

=

3u

2