a convex lens forms a real and inverted image of a needle at a distance of 50 cm from it. Where is the needle placed in front of the lens if the image size is equal to the object size. Also find the power of the lens

optical centre

Here, 

size of image = sixe of object  And image is real and inverted.

So, magnification = - 1

or  v/u = - 1 or  v = - u

Here,

v = 50cm

therefore,   

50 = - u 

or u = - 50

Here,
object is placed at a distance of 50cm from the lens.

Since  object distance = image distance

The object is at the Centre of curvature.

And the distance of object from centre of curvature is the radius of curvature (R).

Here,

R = 50 cm

we know that R = 2 f

so, 50 = 2 focal length

or  focal length  = 50/2
or focal length = 25 cm

Since it is a convex lens,   f = 25

And  Power(P) = 100 / f (cm)

P = 100 / 25
So,  P = 4 D

Hence Power of the lens is 4 D.
 

@Swayam epic answer

f=25
p=4d

thanks 

Answer:- v = 50 m = 1 1 P = = ? f As the image formed by lens is real v m = u v 1 = u v = u u = − 50 cm ∴ Object distance is = 50 cm 1 1 1 − = v u f 1 1 1 − = 50 − 50 f 1 + 1 1 = 50 f 2 f = 50 f = 25 cm = .25 m 1 100 P = + = + = + 4 dioptre .25 25 The power of convex lens is + 4 dioptre

oh yeah
 

V =50 CM, m= h' /h= -1 (bcz the image is real and inverted) Also m=v/u u=v/m = 50 / -1 = -50 cm Needle is placed at 50 CM in front of the lens Using lens formula -1/u +1/v = 1/f we get 1/ f =1/50 +1/ 50 =2/50 1/ 25 =1/f, f= 25cm P= 100 / f =100 /25= 4.0d

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