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A plano-convex lens of refractive index 1.5 and radius of curvature 30 cm is silvered at the curved surface. Now, this lens has been used to form the image of an object. At what distance from this lens, an object be placed in order to have a real image of the size of the object. |
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Answer» 20 cm  `therefore(1)/(f_e)=-(n/f_1""-m/f_m)` .....(i) where `f_m` = focal length of mirror after silvered. `implies` From lens maker.s formula, `(1)/(f_1)=((n_e-n_1)/(n_1))((1)/(R_1)-(1)/(R_2))` `(1)/(f_1)=((1.5-1.0)/(1.0))((1)/(infty)-(1)/(-30))` `=1/2(+(1)/(30))` `=(1)/(60)` `therefore f_1=60` cm From EQUATION (1),`(1)/(f_e)=-((n)/(f_1)-(m)/(f_m))` [No. of lens =2,no. of nirrror=1] `(1)/(f_e)=-((2)/(60)-(1)/((-15)))` `(1)/(f_e)=-((1)/(30)+(1)/(15))` `(1)/(f_e)=-(3)/(30)=-(1)/(10)` `therefore f_e=-10` cm `therefore` MAGNITUDE `f_e=10` cm If image is to be obtained of same size of OBJEC then it should be placed at centre of curvature `therefore` Object distance `u = 2f_e = 2 xx 10 = 20` cm |
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