Question #ea4f0

1 Answer
P dilip_k
Aug 10, 2017

We know the following relation of refractive index (#mu#) of the prism with the angle of prism (A) and minimum deviation (#del_m#)

#mu=sin((del_m+A)/2)/sin(A/2)#

As per given condition we have #del_m=A#

So we have

#mu=sin((A+A)/2)/sin(A/2)#

#=>mu=sinA/sin(A/2)#

#=>mu=(2sin(A/2)cos(A/2))/sin(A/2)#

#=>cos(A/2)=mu/2#

#=>A=2cos^-1(mu/2)#

If we take for glass #mu=1.5#

Then #A=2cos^-1(1.5/2)=82.8^@#

If we think otherwise

This relation reveals that the value of #A# increases with decrease in value of #mu# and A will be maximum when #mu# is minimum.The medium of the material of the prism is optically denser than the surrounding medium. So the minimum value of the refractive (#mu #) of the medium with respect to the surrounding one may be taken as #~~1#.

Imposing this condition we can have the maximum value of #A#.

Hence #" "A_"max"=2cos^-1(1/2)=2xx60^@=120^@#

Related questions
See all questions in Refractions
Impact of this question
1807 views around the world
You can reuse this answer
Creative Commons License