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Show that a convexmirror always forms a virtual image of diminished size as compared to the object. |
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Answer» Solution :According to the equation of the SPHERICAL mirror, `(1)/(v)+(1)/(u) = (1)/(f) or, (1)/(v) = (1)/(f) - (1)/(u)` `or, "" v=(uf)/(u-f)` Now for a convex mirror f is considered as positive. Again, as the object is real, u is taken as negative. Following this sigh convention, form equation (1) we get, `v=((-u)xxf)/(-u-f) = (uf)/(u+f)` As v is positive, so for any position of the object in front of a convex mirror, the image will be formed behind the mirror i.e., Again, magnification, `m=(v)/(u)` [considering the magnitude of m only] `=(f)/(u+f)` [from equation (2)] Clearly, `u+fgtf.so, m lt 1` i.e, the image formed by a convex mirror is diminished in size as compared to object. |
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