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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
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 also -ve.

Hence, (1/v) + (1/40) = -(1/40)

By solving above equation we get , v = - 20 cm , i.e., image is formed left side of concave lens at a distance 20 cm from lens.

Magnification , m = - v/u  = 20/40 = 0.5

Image is diminished to half, hence image size of 1.5 cm object is 0.75 cm . Image is erect and virtual
Answered by Thiyagarajan K | 14 Dec, 2020, 12:16: AM

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