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A diminished image of an object is to be obtained on a screen 1.0 \mathrm{~m} away from it. This can be achieved by approximately placing:

Option: 1

a convex mirror of suitable focal length


Option: 2

a concave mirror of suitable focal length


Option: 3

a convex lens of focal length less than 0.25 \mathrm{~m}


Option: 4

 a concave lens of suitable focal length.


Answers (1)

best_answer

mage can be formed on the screen if it is real. Real image of reduced size can be formed by a concave mirror or a convex lens.

Let \mathrm{u=2 f+x}, then
\frac{1}{\mathrm{u}}+\frac{1}{\mathrm{v}}=\frac{1}{\mathrm{f}}
\Rightarrow \frac{1}{2 f+x}+\frac{1}{v}=\frac{1}{f}
\Rightarrow \frac{1}{v}= \frac{1}{f}-\frac{1}{2 f+x}=\frac{f+x}{f(2 f+x)}
\Rightarrow v=\frac{f(2 f+x)}{f+x}

It is given that \mathrm{u+v=1 m}.

\mathrm{2 f+x+\frac{f(2 f+x)}{f+x}=(2 f+x) \hat{e}_{\hat{e}}^{e} 1+\frac{f}{f+x} \stackrel{u}{u} d i m}
\mathrm{or, \quad \frac{f(2 f+x)^{2}}{f+x}<1 m}
\mathrm{or, \quad(2 f+x)^{2}<(f+x)}

This will be true only when \mathrm{\mathrm{f}<0.25 \mathrm{~m}}.

Posted by

himanshu.meshram

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