Physics

Physics

Problem

The graph drawn with object distance along abscissa & image as ordinate for a convex lens is a) straight b) circle c) rectangular hyperbola d) parabola
Intuitive explanation

The lens equation 1f=1u+1v\frac{1}{f} = \frac{1}{u} + \frac{1}{v} relates object distance uu and image distance vv for a fixed focal length. When you rearrange this to express vv as a function of uu, you get a relationship where the product (u−f)(v−f)(u - f)(v - f) is constant—this is the defining property of a rectangular hyperbola. The graph won't be a straight line because the relationship is nonlinear, and the asymptotes occur at the focal length.

Intuitive explanation

The lens equation 1f=1u+1v\frac{1}{f} = \frac{1}{u} + \frac{1}{v} relates object distance uu and image distance vv for a fixed focal length. When you rearrange this to express vv as a function of uu, you get a relationship where the product (u−f)(v−f)(u - f)(v - f) is constant—this is the defining property of a rectangular hyperbola. The graph won't be a straight line because the relationship is nonlinear, and the asymptotes occur at the focal length.

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