(a) a convex mirror of suitable focal length.

(b) a concave mirror of suitable focal length

(c) a convex lens of focal length less than 0.25 m

(d) a concave lens of suitable local length.

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(a) a convex mirror of suitable focal length.

(b) a concave mirror of suitable focal length

(c) a convex lens of focal length less than 0.25 m

(d) a concave lens of suitable local length.

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Image can be formed on screen if it is real .

Real image of reduced size can be formed by a concave mirror or a convex lens.

Let $ u= 2f+x$

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

=> $ \large\frac{1}{2f+x}+\frac{1}{v}=\frac{1}{f}$

=> $\large\frac{1}{v} =\frac{1}{f}-\frac{1}{2f (x)}=\frac{f+x}{f(2+x)}$

=> $ v= f \large\frac{(2f+x)}{f+x}$

It is given that $u+v=1 m$

$2f +x +f \large\frac{2+x)}{f+x}$$=(2f+x) \bigg[ 1+\large\frac{f}{f+m}\bigg] $$ < 1m$

or $\large\frac{f(2f+x)^2}{f+x}$$ < 1m\; or \; (2f+x)^2 < (f+x)^2 < (f+x)$

This will true only when $f < 0.25\;m$

Hence c is the correct answer.

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