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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!!