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A short object of length L is placed along the principal axis of a concave mirror away from focus. The object distance is u. If the mirror has a focal length f, what will be the length of the image ? You may take L lt lt |v- f|. |
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Answer» SOLUTION :Distance of closect end of an object (of length L ) from the pole of concave mirror, `u_1 = u + 1/2` `RARR` Similarly distance of farthest end of above object from the pole of concave mirror, `mu_2 = u +L/2` `therefore u_1 - u_2 = - L` `therefore | mu_1 - mu_2| = L` If image distance for above two ends are repectively `v_1` and `v_2` then length of image will be `L. = |v_1 - v_2|` `rArr` From mirror formula `1/f = 1/u + 1/v` we have `1/v = 1/f - 1/u` `therefore 1/v = (u-f)/(fu)` `rArr` From above equation `therefore= (fu)/(u-f)` `v_ = (f(u-L/2))/(u-L/2-f) and v_2 = (f(u+L/2))/(u+L/2-f)` `therefore ` Length of image `L. = | v_1 - v_2|` `= (fu-(fL)/(2))/(u-f-L/2) - (fu+(fL)/(2))/(u-f + L/2)` ` =1/((u-f)^(2) - L^2/4)xx [ fu^2-f^2u + (FUL)/(2)-(fuL)/(2) + (f^2L)/(2)-(fL^2)/(4)-fu^2 + f^2u + (fuL)/(2) - (fuL)/(2) + (f^2L)/(2) + (fL^2)/(4)]` `= (f^2L)/((u-f^2)-L^2/4)` `rArr` Hence we have `(L^2)/(4) lt lt (u- f)^2` andneglecting `(L^2)/(4)` from the denominator, `therefore L. = |v_1 - v_2| = (f^2)/((u-f)^2)L` |
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