Question

Proof for angle of deviation is equal to angle of incidence +angle of emergence-angle of prism

Answer 1

Here i1 is incident angle and i2 is the emergent angle while A is angle of prism

(Consider i2=e and i1=i )

A+α =180

r1+r2+α =180

This gives A=r1+r2

δ=(i−r1)+(e−r2)

This results in δ=i+e−A

Taking derivatives

dδdi=1+dedi

Since for minimum deviation, dδdi=0, since δ is constant.

This leaves us with

−1=dedi

And using the equation , A=r1+r2

dAdr1=1+dr2dr1

Since angle of prism A is constant, dAdr1=0

which leaves us with −1=dr2dr1

Also we know the following relation by snell's law

sin isin r1=μ

sin esin r2=μ

Differentiating both the equations

cosididr1=μcos r1

cosededr2=μcos r2

Using the relation which we got above, the derivative terms get cancelled out

Dividing these two equations, it leaves us with

cosicose=cos r1cos r2

If we simplify this equation, it results in

sin i=sin e

And voila i=e