I want to prove mirror formula
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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
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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