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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 location 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 produces a virtual and enlarged image. |
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Answer» Solution :MIRROR formula, `1/f = 1/u + 1/v` `therefore 1/v = 1/f - 1/u`...... (1) Now, for CONCAVE mirror, `f lt 0` (negative) and as objective is in left to mirror `u lt 0` `therefore 2 f lt u lt f` .. ... (1) Now, for concave mirror `f lt 0` (negative) and as object is in leftto mirror , `u lt 0` `therefore 2 f lt u lt f` `therefore 1/ (2f) gt (1)/(u) gt (1)/(f)` `therefore (1)/f - (1)/(2f) lt (1)/(f) - (1)/(u) lt (1)/(f) - (1)/(f)` [ By adding 1/f in each term] `therefore 1/(2f) lt (1)/(v) lt 0[ therefore ` from equation (1)] This represent that `v lt 0,` hence REAL image is obtained in left of mirror. `2 f gt 0` `therefore| 2f | gt | v| [ because 2f` and v are negative ] Hence, real image is obtained beyond 2f. For convex lens, ` f gt 0` and object is in left , hence, `u lt 0` `therefore`From mirror equation , `1/v = 1/f - 1/u` Now,` f lt 0` and `u lt 0` and it REPRESENTS |
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