1where's your result ..
Prove rigorously that for an isoceles prism a ray of light falling on one side will suffer minimum deviation when angle of incidence= angle of emergence ..
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5 Answers
1let me try to solve it by calculas
let d=delta
so
we know that d=i+e-A
on differentiating WRT i we get
dd/di=1+de/di
for min deviation dd=0
so de/di=-1
also r1+r2=A
diff WRT r1
dr2/dr1=-1
so
de/di=dr2/dr1 1
also for first face
n=sini/sinr1
diff
cosi*di=ncosr1*dr1
similarly cose*de=ncosr2*dr2
so cose*de/cosi*di=cosr2*dr2/cosr1*dr1
from 1 cose/cosi=cosr2/cosr1
1sry went to see score
now squaring &rearranging
1-sine(pow 2)/1-sini(pow 2)=1-sinr2(pow 2)/1-sinr1(pow 2) .................2
sinr1=sini/n & sinr2=sine/n
put in 2 & rearrange to get
1-sine(pow 2)/1-sini(pow 2)=n(pow 2)-sine(pow 2)/n(pow 2)-sini(pow 2)
cross multiplying & simplyfing
i=e
9for every i and e if u interchange them the δ remains same (VOA)
so wen u draw graph for i and δ it repeats itself after δm in a symmetrical fashion
at δm therefore i=e