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
v =50 cm
since a same sized, real and inverted image is formed, therefore the object must be placed at the centre of curvature i.e., at 2F1
and since the image is formed at 2F2. hence the diastance b/w 2F1 and O is 50cm.
Thus the needle is placed 50 cm in front of the lens.
R = 50cm
f= 50/2 = 25cm
f= 0.25 m
P = 1/f
P = 1/0.25
P = 100/25
P = 4D
hence the power of the lens is 4D.
size of image = sixe of object And image is real and inverted.
So, magnification = - 1
or v/u = - 1 or v = - u
v = 50cm
50 = - u
or u = - 50
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).
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.