Question

an object placed 40 cm from a lens produces a virtual image at a distance of 10 cm in Front of thr the lens . what is the focal length of lens ? is it converging or diverging lens ?

Answer 1

$\underline{solution}$

given :-

$\underline{in \: question \: already \: given}$

u = - 40 cm

v = -10 cm

f = ?

$\underline{according \: to \: lens \: formula \: }$

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

put the values of and solving we get.

$\frac{ - 1}{10} - ( \frac{1}{40} ) \: = \frac{1}{f}$

or,

$\frac{1}{f} = \frac{1}{40} - \frac{1}{10}$

$\frac{1}{f} = \frac{1 - 5 }{40} - \frac{-3 }{40}$

= $\frac{-3}{40}$


or ,

f = - 13.33


since focal length is negative ,. the lens must be a diverging lens or concave lens.

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