Question

A concave lens has focal length of 15cm. At what distance should object be placed so that image at 10cm.

Answer 1

Answer:

-30

Explanation:

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

$\frac{1}{15} = \frac{1}{10} + \frac{1}{u}$

$\frac{1}{15} - \frac{1}{10} = \frac{1}{u}$

$\frac{1}{ - 30} = \frac{1}{u}$

$- 30 = u$

Answer 2

Explanation:

HLW MATE....HERE IS YOUR CORRECT QUESTION :

A concave lens has focal length of 15cm. At what distance should object be placed so that image at 10cm from the lens? Also , find the magnification produced by the lens.

YOUR CORRECT ANSWER IS:

SOLUTION :

A concave lens always forms a virtual, erect image on the same side of the object.

Image distance, V = -10 cm

Focal length, f = -15 cm

Object distance, u = ?

Since 1/v - 1/u =1/f

OR , 1/u = 1/v - 1/f

1/u = 1/-10 - 1/(-15) = -1/10 + 1/15

1/u = -3+2/30 = 1/-30

OR, u= -30 cm

THUS THE OBJECT DISTANCE IS 30 cm.

MAGNIFICATION :

m = -10 cm /-20 cm=1/3 = +0.33

THE POSITIVE SIGN SHOWS THAT THE IMAGE IS ERECT AND VIRTUAL. THE IMAGE IS ONE-THIRD OF THE SIZE OF THE OBJECT.