Question

prove that mirror formula

Answer 1

hey!!!

from figure
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in ∆ABP and ∆A'B'P are similar

so.. according to theles theorem

AB/A'B'=-u/-v -----1)

and again

in ∆ A'B'C and ∆ ABC are similar to each other

now again according to theles theorem

AB/A'B'=BC/B'C but BC =-u -(-R)

BC=-u+R {we can write }

and B'C=-R-(-v)

=>-R+v

•°•AB/A'B'°-u+R/-R+v

=> u/v=-u+R/-R+v

=> u/v=-u+R/-R+v [•°•by equation 1) ]

=> -uR+uv=-uv+vR

=> -uR/uvR +uv/uvR =-uv/uvR+vR/uvR

[dividing by uvR on both side ]

=> - 1/v+1/R=-1/ R+1/u

=> -1/v-1/u=-1/R-1/R

=>-(1/v+1u)=-( 1+ 1/R)

=>1/v+1/u=2/R

=>1/v+1/u=2/2F

=>1/v+1/u=1/F

hence here Prooved ...

⚛ 1/v+1/u =1/F ........

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hope it helps you !!!

#Rajukumar 111

Answer 2

Hey there !

Mirror Formula -

It tells the relationship between Object distance ( u ) , image distance ( v). , and focal length ( f )

so we get ,

$\frac{1}{f} \: = \frac{ \: 1}{v } \: + \frac{1}{u}$

Thanks !
Hope this helps you !