Question

prove that 1÷f =1÷u+1÷v

Answer 1

Mirror formula is the relationship between object distance (u), image distance (v) and focal length.

1/v + 1/u = 1/f

In Triangle ABC  and Triangle A’B’C

<A = <A’ = 900

<C =<C ( vert. opp. <s]

Triangle ABC ~Triangle A’B’C [AA similarity]              => AB /A’B’ = AC/A’C ----(I)
Similarly,

In Triangle ABC  and A’B’C

<A = <A’ = 900

<C =<C ( vert. opp. <s]

Also, in Triangle ABC ~Triangle A’B’C [AA similarity]

AB /A’B’ = AC/A’C ----(1)

Similarly, In DFPE  ~ A’B’F

EP /A’B’ = PF/A’F

AB /A’B’ = PF/A’F   [ AB=EP]  ----(II)

From(i) &(ii)

AC/A’C =  PF/A’F   

=>  A’C/AC = A’F/PF

=> (CP-A’P)/(AP- CP)   = (A’P – PF)/PF

Now, PF = -f ;  CP = 2PF = -2f ; AP = -u ; and A’P = -v 

Put these value in above relation:

[(-2f) –(-v)] /(-u)-(-2f) = {(-v) –(-f) }/(-f)

=> uv = fv +uf

=>  1/f = 1/u + 1/v

ya here is ur answer


Mirror formula is the relationship between object distance (u), image distance (v) and focal length.

1/v + 1/u = 1/f

In Triangle ABC  and Triangle A’B’C

<A = <A’ = 900

<C =<C ( vert. opp. <s]

Triangle ABC ~Triangle A’B’C [AA similarity]              => AB /A’B’ = AC/A’C ----(I)
Similarly,

In Triangle ABC  and A’B’C

<A = <A’ = 900

<C =<C ( vert. opp. <s]

Also, in Triangle ABC ~Triangle A’B’C [AA similarity]

AB /A’B’ = AC/A’C ----(1)

Similarly, In DFPE  ~ A’B’F

EP /A’B’ = PF/A’F

AB /A’B’ = PF/A’F   [ AB=EP]  ----(II)

From(i) &(ii)

AC/A’C =  PF/A’F   

=>  A’C/AC = A’F/PF

=> (CP-A’P)/(AP- CP)   = (A’P – PF)/PF

Now, PF = -f ;  CP = 2PF = -2f ; AP = -u ; and A’P = -v 

Put these value in above relation:

[(-2f) –(-v)] /(-u)-(-2f) = {(-v) –(-f) }/(-f)

=> uv = fv +uf

=>  1/f = 1/u + 1/v

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