Physics, asked by Anonymous, 4 months ago

\huge\purple{\mid{\fbox{\tt{Question}}\mid}}

How far should an object be from the pole of a concave mirror of focal length 20 cm to form a real image whose size is 1/5th of the size of the object....

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Answers

Answered by BornCxnfused
4

\huge\purple{{\fbox{\tt{Question}}}}

How far should an object be kept from the concave mirror of focal length 20 cm to form a real image one-fifth of its size?

\huge\pink{{\fbox{\tt{answer}}}}

m =  \frac{ - 1}{5}  =   \frac{ - v}{u}  \\  \\  \frac{v}{u}  =  \frac{1}{5} \\ now \: by \: applying  \: the \:  \\ mirror \: formula \\  \\  →\frac{1}{u}  +  \frac{1}{v} =  \frac{1}{f}   \\  \\  →\frac{1}{u}  +  \frac{5}{u} =  \frac{ - 1}{20}   \\  \\ → \frac{6}{u}  =  \frac{ - 1}{20}  \\ =  \frac{ - 1}{20} cm

hope it will help u buddy ❤️

bdi mehnat kia yar is answer pai, brainliest mark kardena plz ...xd

Answered by Anonymous
19

Answer:

How far should an object be kept from the concave mirror of focal length 20 cm to form a real image one-fifth of its size?

\huge\pink{{\fbox{\tt{answer}}}}

answer

\begin{gathered}m = \frac{ - 1}{5} = \frac{ - v}{u} \\ \\ \frac{v}{u} = \frac{1}{5} \\ now \: by \: applying \: the \: \\ mirror \: formula \\ \\ →\frac{1}{u} + \frac{1}{v} = \frac{1}{f} \\ \\ →\frac{1}{u} + \frac{5}{u} = \frac{ - 1}{20} \\ \\ → \frac{6}{u} = \frac{ - 1}{20} \\ = \frac{ - 1}{20} cm\end{gathered}

m=

5

−1

=

u

−v

u

v

=

5

1

nowbyapplyingthe

mirrorformula

u

1

+

v

1

=

f

1

u

1

+

u

5

=

20

−1

u

6

=

20

−1

=

20

−1

cm

hope it will help u buddy ❤️

bdi mehnat kia yar is answer pai, brainliest mark kardena plz ..

Explanation:

mark as brainlist answer

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