when the distance of an object from a concave mirror is decreased from 15 cm to 9 cm, image gets magnified 3 times than that in first case . calculate the focal length of the mirror.
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FOR THE FIRST CASE, LET OBJECT DISTANCE FOR THE FIRST CASE BE U & FOR THE SECOND CASE LET OBJECT DISTANCE BE U'. SIMILARLY,LET THE IMAGE DISTANCE FOR THE FIRST CASE BE V & FOR THE SECOND CASE BE V'. NOW, U= -15CM (THE OBJECT DISTANCE IS ALWAYS TAKEN AS NEGATIVE AS PER THE SIGN CONVENTION) U'= -9CM
LET THE MAGNIFICATION BE X FOR THE FIRST CASE AND 3(X) FOR THE SECOND CASE. FOR THE FIRST CASE, -V/U=M -V/-15=x V=15x.......... (i) FOR THE SECOND CASE, -V'/U'=M' -V'/-9=3(x) V'=27x............ (ii) BY APPLYING THE MIRROR FORMULA IN FIRST CASE, 1/V + 1/U =1/F 1/15x + 1/-15 =1/F (1-x)/15x=1/F 15x/(1-x)= F.............. (iii) BY APPLYING THE MIRROR FORMULA IN SECOND CASE, 1/V' + 1/U'=1/F 1/27x + 1/-9=1/F (1-3x)/27x =1/F 27x/(1-3x)=F.............. (iv) FROM EQUATION (iii) & (iv), 15x/(1-x)=27x/(1-3x)=F 15x(1-3x)=27x(1-x) x= -1 SUBSTITUTING THE VALUE OF x IN EQUATION (iii) OR (iv), 15(-1)/(1-(-1)) = -15/2CM=F SO, FOCAL LENGTH (F)= -7.5CM
LET THE MAGNIFICATION BE X FOR THE FIRST CASE AND 3(X) FOR THE SECOND CASE. FOR THE FIRST CASE, -V/U=M -V/-15=x V=15x.......... (i) FOR THE SECOND CASE, -V'/U'=M' -V'/-9=3(x) V'=27x............ (ii) BY APPLYING THE MIRROR FORMULA IN FIRST CASE, 1/V + 1/U =1/F 1/15x + 1/-15 =1/F (1-x)/15x=1/F 15x/(1-x)= F.............. (iii) BY APPLYING THE MIRROR FORMULA IN SECOND CASE, 1/V' + 1/U'=1/F 1/27x + 1/-9=1/F (1-3x)/27x =1/F 27x/(1-3x)=F.............. (iv) FROM EQUATION (iii) & (iv), 15x/(1-x)=27x/(1-3x)=F 15x(1-3x)=27x(1-x) x= -1 SUBSTITUTING THE VALUE OF x IN EQUATION (iii) OR (iv), 15(-1)/(1-(-1)) = -15/2CM=F SO, FOCAL LENGTH (F)= -7.5CM
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