A 1 cm tall object is placed perpendicular to the principal axis of a convex lens of focal length 20cm.The distance of the object from the lens is 15cm.Find the nature, position, size and magnification of the image.
Focal length f = 20 cm
Object distance u = -15 cm
Image distance v= ?
The lens formula is 1/f = 1/v - 1/u
1/v = 1/u + 1/f
1/v = 1/-15 + 1/20
1/v = -4+3/60
1/v = -1/60 or v = -60 cm
Here the negative sign indicates that the image is formed on the same side of the lens as the object.
Therefore, the image is virtual.
Now, Magnification m = h'/h = v/u = -60/-15 = +4
Therefore, h' = 4xh = 4x1 = +4 cm
The positive sign of h' shows that the image is erect. Thus, in this case, the image is erect, virtual, 4 cm tall, and it is formed at a distance of 60 cm on the same side of the lens as the object.