derive the mirror formula in science
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The Mirror Formula can be defined as 1/f=1/v +1/u
Here I am attaching the image to solve the mirror formula :
https://drive.google.com/file/d/1xShmwoayseuSn-J8SNSbojdXHKrJWeHb/view
Proof:
ACB= A1CB1;
Similarly,ABC=A1B1C;
Now Since two right angles of triangle ACB and A1CB1 are equal and hence the third angle is given by;
BAC=B1A1C;and
AB/A1B1 = BC/B1C............1
Similarly the triangle of FED and FA1B1 are also equal and Similar,So;
ED/A1B1 = EF/FB1;
Also since ED is equal to AB so we have;
AB/A1B1 = EF/FB1;............2
Combining 1 and 2 ,we have;
BC/B1C = EF/FB1
Consider that the point D is very close to P and hence EF = PF,So;
(PC-PB)/PB1 - PC = PF/(PB1 - PF);
Now substituting the values of the above segments along with the sign,we have;
PC= -R
PB=u;
PB1=-v;
PF=-f
so the above equation becomes;
{-R -(-u)} / {-v - (-R)} = -f / {-v -(-f)}
(u - R) / (R-v) = -f / (f-v) ;
(u - R) / (R-v) = f/(v-f);
By solving we get;
uv - uf -Rv + Rf = Rf - vf;
uv - uf - Rv + vf =0;
since R = 2f;
uv - uf - 2fv +vf = 0;
uv - uf - vf = 0;
dividing with uv on both sides,then we get;
1/f = (1/u) + (1/v)
Hope it helps!!
Thank you!!
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