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Okay, what I meant to ask is as follows: Take a isosceles prism having a refracting angle A . A ray of light incident on one of the sides containing angle A (the incident angle can be varied) and emerges from the other side containing A. We know that the light gets deviated from the original path. Experimentally, it is known that the deviation is minimum when the incident angle is same as emergent angle. We have to prove this fact.
Case 1 : Let us consider for minimum deviation the angle of incidence is greater than angle of emergence.....
that is in the above diagram if 
is the minimum deviation, then according to pur assumption,
i >e................(1)
Now, if v send a ray along ZS, it will retrace its path and will emerge along PA. Now as the angle of deviation is same, by our assumption, ie the angle of incidence is greater than angle of emergence, we ger
e > i....... which is a contradiction to our assumption ie (1).....
Case 2 : similarly if we assume that the angle of incidence is smaller than the angle of emergence , we again get a contradiction.......
Thus v can conclude that for minimum deviation, angle of incidence is equal to the angle of emergence... ie
i = e. Thus proved.
that is also quite easy .. let angle of minimum deviation be d
well we know the relations
A=r1+r2 --- - -(i)
d=i+e -A
further from (i) we get
d(r1)/di = -d(r2)/di
and we know
nsin(r1)=sin(i) -- -(ii)
nsin(r2)=sin(e) -- - (iii)
=>
ncos(r1)dr(1)/d(i)=cosi
ncos(r2)d(r2)/d(i)=cos(e)(de/di)
dividing both equations
-cos(r1)/cos(r2)=(cosi/cose)(di/de)
further from (ii) and (iii)
cos(r1) = (1-(sin(i)/n)2)1/2
and
cos(r2) = (1-(sin(e)/n)2)1/2
further for d to be extremum
d(d)/di=0
thus
d(i+e-A)/di=0
thus
1+de/di=0
but we have already obtained de/di
solving this equation we get
(n2 - sin2i)/cos2i = (n2-sin2e)/cos2e
=>
(n2-sin2i-cos2i)/cos2i = (n2-sin2e-cos2e)/cos2e
thus we get two possibilities
n=1 or i=e
but n=1 is excluded( as n we cant manipulate)
and thus we get i=e for extremum
and we can see that the value of d obtained in this case is less than the value obtained when ray falls perpendicularly
thus IT HAS TO BE MINIMUM
please reply sir ..
(except that I found it quite hard to read). That was what I was looking for.
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i cudnt get the question
do we have to consider rays parallel to base only and different values of angle of prism
and do we have to take an isoceles prism ?
i have proved it when the case is as mentioned above ...(was quite easy so most probably it is not the case)