InterviewSolution
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InterviewSolution
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Use the mirror equation to deduce that : (a) an object placed between f and 2f of a concave mirror produces a real image beyond 2f. (b) a convex mirror always produces a virtual image independent of the locatioin of the object. (c ) the virtual image produced by a convex mirror is always diminished in size and is located between the focus and the pole. (d) an object placed between the pole and focus of a concave mirror procuces a virtual and enlarged image. |
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Answer» Solution :(a) As per MIRROR formula: `1/v + 1/u = 1/f` `therefore 1/v = 1/f -1/u` and for concave mirror u and f both are NEGATIVE i.e. `f lt 0` and `MU lt 0` If `2f lt u lt f`, then `1/(2f) GT 1/u gt 1/f` or `-1/(2f) lt -1/u lt -1/f` or `1/(2f) lt 1/v lt 0` Thus, `v gt 2f`with a negative sign i.e., the image is real and lies beyond 2f. (b) For a concave mirror `1/v =1/f -1/u` but f si positive and u is negative i.e. `1/v = 1/f -1/(-u) =1/f + 1/u = +ve` Therefore, irrespective of the value of u, value of v is always +ve. It means that the image formed by convex mirror is always virtual independent of the location of the object. ( c) As for a convex mirror f is +ve and u is -ve, hence `1/v =1/f -1/(-u) = 1/f + 1/u` It means that `1/v gt 1/u` or `|v| lt |u|`. Thus, `m =h^(.)/h =|v|/|u| lt 1` i.e, the image is diminished one . Moreover, from above, it is clear that `1/v gt 1/f` or `|v| lt |f|` moreover from the above, it is clear that `1/v gt 1/f` or `|v| lt |f|` Hence, the image is located between the POLE and principal focus of convex mirror. (d) As for a concave mirror f and u both are negative, hence `1/v =1/(-f) -1/(-u) =1/u -1/f` if `|v| lt |f|`, then `1/u gt 1/f` and hence, `1/v gt 0` i.e. is `+ve`. Moreover `1/v lt 1/u` or `v gt u`.It means that for an object placed between the pole and principal focus of a concave mirror the image formed is virtual and magnification`|m| |v|/|u| gt 1` |
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