Question

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

Answer 1

f= -15

v= -10

u= ?

lens formula 1/f= 1/v - 1/u

1/u = 1/v - 1/f

1/u = 1/ -10 - 1/-15

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

1/u = -3+2/30 (taking LCM)

1/u = -1/30
u = -30
thus, object should be placed at a distance of 30 cm...
magnification = v/u
m = -10/-30
m = 1/3
m= 0.33

Answer 2

$\huge\bold{Question:-}$

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

$\huge\bold{Answer:-}$

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,

$\sf \dfrac{1}{v} - \dfrac{1}{u} = \dfrac{1}{f}$

or,

$\sf \dfrac{1}{u} = \dfrac{1}{v} - \dfrac{1}{f}$

$\sf \dfrac{1}{u} = \dfrac{1}{ - 10} - \dfrac{1}{( - 15)} = - \dfrac{1}{10} + \dfrac{1}{15}$

$\sf \dfrac{1}{u} = \dfrac{ - 3 + 2}{30} = \dfrac{1}{ - 30}$

$\sf or \: u = - 30 \: cm$

Thus the object-distance is 30 cm

Magnification m=v/u

$\sf m = \dfrac{ - 10 \: cm}{ - 30 \: cm} = \dfrac{1}{3} = + 0.33$

The positive sign shows that the image is erect and virtual. The image is one-third of the size of the object.

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