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2014-10-28T02:30:38+05:30

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Let us take the case of concave mirror.   See diagram.

Let P be the pole (geometrical center) of the mirror.  Let us draw the axis of the mirror.  Let us draw the focal point F and the center C at distances f and 2f from the pole on the principal axis.

Let us take a point object at A which is at distance u from P.  To find its image draw a ray AM on to the mirror at angle y to the axis.  Draw the normal to the mirror at M which passes through the center C.  Now draw the reflected ray making the same angle x with the normal.  Let it intersect the axis at A'.  Draw another ray from object A along the axis which gets reflected back along the same path.  It intersects A'M at A'.   Hence A' is the image of A.

We want to find a relation between u and v (image distance from P),

Z = exterior angle of Triangle AMA' and hence  z = y + 2 x  or,  z - 2x = y
z-x is the exterior angle of triangle AMC.

Tan y = MP' / AP'           Tan z =  MP' / A'P'         tan z-x = MP' / CP'

When y, x, z are all small angles.  Tangent is nearly same as the angle.  Hence

         y = MP' / AP'            z =  MP' / A'P'          z-x = MP' / CP'

So we  get  x = MP'/AP' - MP'/CP'

Since y = z - 2x ,    MP' / AP' = MP' / A' P' - 2 [  MP'/AP' - MP'/CP' ] 

   Cancel MP'.

             1 /AP' = 1 / A' P' - 2 / A'P' + 2 /CP'

Substitute CP' = 2f and A'P' = v   and u = AP'.   as  PP' is very small and is negligible for small aperture mirror and small angles x, y and z.

           1 / u = - 1 / v + 2 / 2f

            1/u + 1/v = 1/f   -- equation 1

We introduce a cartesian sign convention. The distances in the direction of the light ray are taken as positive and those in the opposite direction are taken as negative.

So u is -ve,  v is negative and f is also negative.  Then  equation 1 is

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

The same equation can be proved for convex mirror also.

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2014-10-28T13:57:32+05:30

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Certified answers contain reliable, trustworthy information vouched for by a hand-picked team of experts. Brainly has millions of high quality answers, all of them carefully moderated by our most trusted community members, but certified answers are the finest of the finest.
Mirror formula is the relationship between object distance (u),  image distance (v)  and focal length (f).
A ray of  light starting from A and incident on the mirror along AD  parallel to the principal axis of the mirror and gets reflected from the mirror and passes through F.
Another ray of light incident along AP is reflected along PA’ such that angle APB=i= angle BPA’= r.
The two reflected ray meet at A’ which is a real image.
Therefore A’B’ is real, inverted image of AB formed by reflection from the mirror.

 


In Triangles BAC and A’B’C’

AB/A’B’ = CB/CB’   ……………. (i)
Now  Triangle ABP and A’B’P’ are similar,
AB/A’B’ = PB/PB’ …………………. (ii)
From (i) & (ii) …  
CB/CB’= PB/PB’   ……………….. (iii)
Measuring all distances from  P,
We have CB= PB-PC CB’= PC-PB’

Therefore from (iii),
(PB-PC)/(PC-PB’) = PB/PB’  …………… (iv)
Using new cartesian sign convections,
PB= -u (distance of object) PC= -R PB’ = -v (distance of image) We get from (iv), (-u+R)/(-R+v) = -u/-v
Or,           

+uR – uv = uv – vR
Or,            
uR + vR = 2uv  

Dividing both sides by uvR,
We get, 1/v + 1/u = 2/R As    
R=2f

Therefore, 1/v + 1/u = 2/R = 2/2f = 1/f This is the required formula.
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