In the case of a concave mirror of focal length f, when an object is kept between f and 2f, show that its image is formed beyond 2f.

 

Answers (1)
S safeer

Mirror equation is given by:

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

\frac{1}{v}= \frac{1}{f}-\frac{1}{u}

For a concave mirror, f is negative, i.e, f<0.

For a real object, i.e., which is on the left side of the mirror.

For 'u' between f and 2f 

\frac{1}{2f}> \frac{1}{u}> \frac{1}{f} 

(as u, f are negative) 

\frac{1}{-2f}< \frac{1}{-u}< \frac{1}{-f}

Add 1/f every side

\frac{1}{f}- \frac{1}{2f}<\frac{1}{f}- \frac{1}{u}<0

i.e.

 \frac{1}{2f}<\frac{1}{v}<0

i.e.

 \frac{1}{v} is negative 

This implies that 'v' is negative and greater than 2f. Therefore, image is beyond 2f and it is real.

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