An object of height 1.5 cm is placed on the principal axis of a concave lens at
a distance of 40 cm from it. If the focal length of the lens is also 40 cm, find the
location of image, height of image, magnification and draw the ray diagram.
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Explanation:
We have lens equation as, (1/v) -(1/u) =1/f
where v is lens to image distance , u is lens to object distance and f is focal length of concave lens
By Cartesian sign convention u is -ve and focal length of concave lens is also -ve
Hence, (1/v) +(1/40) = -(1/40)
By solving above equation we get , v = -20cm, I. e., image is formed left side of concave lens at a distance 20cm from lens .
Magnification, m = -v/u = 20/40 =0.5
Image is diminished to half , hence image size of 1.5cm object is 0.75 cm. Image is erect aand virtual .
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