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Use the mirror equation to show 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 ) An object placed between the pole and focus of a concave mirror produces a virtual and an enlarged image. |
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Answer» SOLUTION :`(1)/( v) + ( 1)/( u ) = ( 1)/( f)` for CONCAVE mirror, `f lt 0 ` or f = - ve for convex mirror,` f gt 0 ` or `f = + ve ` For concave mirror `:` LET ` f = - c ` Also,Let `u = nf =- NC ` ` ( 1)/( v ) = ( 1)/( f) - ( 1)/( u ) = - ( 1)/ ( c ) + ( 1)/( nc ) = ( - n +1 )/( nc )` `:.` `v = ( nc)/((1-n))` (a) When object is betweeen f and 2f, we have`1 lt n lt 2 ` `:. ` v is -ve ` rArr `real image ( for n = 1 and n = 2 ), magnitude of v becomes `oo` and 2c, respectively `:.` Real image is formed beyong 2F . (b) For convex mirror `:` `f = + d , `Let u = -pd ( p can have any value ) `(1)/(v ) = ( 1)/( d) = ( 1)/( pd) = ((1+p))/( pd)` `v = ( pd)/( ( p+_1))` `:.`v is alsways `+ve` and always less than d. `:.` Convex mirror always produces a virtual imagebetween pole and focus. (c ) Object between pole and F we have `0 lt n lt 1`v is `+ve rArr` ( virtual image ) and `|v| gt c ` `:.` We GET a virtual, and an enlarged image. |
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