Question

a 5cm tall object is placed on the principal axis of a convex lens of fl 50 cm at a distance of 40 cm from it .use lens formula to find the nature and position of the image

Answer 1

h=5cm, u = -40cm, f = 50cm

$\frac{1}{v} - \frac{1}{u} = \frac{1}{f} \\ \\ \frac{1}{v} = \frac{1}{u} + \frac{1}{f} =\frac{1}{-40} + \frac{1}{50} \\ \\ \frac{1}{v}= \frac{-5+4}{200} = \frac{-1}{200} \\ \\ v=-200cm$

$m= \frac{h'}{h}= \frac{v}{u} = \frac{-200}{-40} =5 \\ \\ h'=5* 5cm=25cm$

The image is formed at 200cm to the left from lens, erect and virtual.

Answer 2

Explanation:

HERE;...

h=5cm;

h=5cm;u=-40cm,

h=5cm;u=-40cm,f =50cm.

h=5cm;u=-40cm,f =50cm. since we know that.,

1/v -1 /u=1/f

1/v = 1/f +1/u

1/v = 1/50+ 1/-40

1/v= -1/200

1/v= -1/200v = -200cm.

1/v= -1/200v = -200cm.magnification = h' / h. =v /u

-200/-40=5

h'=5*5=25

HENCE THE IMAGE IS FORMED AT 200 CM TO THE LEFT FROM THE LENS AND THE IMAGE IS ERECT AND VIRTUAL.