Question

A thin prism of angle 2°30' and of index of refraction 1.58 for sodium light

Answer 1

Answer:

Here

A=$2^{0} 30^{'}$=π/72

n=1.58

OP=0.2m

From small angle A

assuming the prism to be set in minimum deviation position

δ=(μ-1)A

 =(1.58-1)π/72

 =0.025

Again, $sin_{a_{1} } =h/OB$

                                    =h/OP

                                 $h=OP sina_{1}$

                                 $sina_{1} =a_{1}$

                                 $h=OPa_{1}$

                                 $h=OPa_{1}$

                                δ=α1+α2

                                α1=h/OP

                                α2=h/PC

                              h/OP+h/PC=δ

                              h(1/OP+1/PC)=δ

                              OPα1(1/OP+1/PC)=δ

                              α1(1+OP/PC)=δ-----(1)

     Let r be the angle of refraction

we have

2r=A

r=A/2-------(2)

Again i+e=A+δ

  i=α1

 α1+e=A+δ------(3)

sin(e)/sin(r)=μ

Assuming e and r to be small

e/r=μ

e=μr

e=1/2μA

By equation 3

α1+1/2μA=A+δ

α1=A(1-1/2μ)+δ-------(4)

By using equation 1

{A(1-1/2μ)+δ}(1+OP/PC)=δ

{Π/72(1-0.5*1.58)+0.025}{1+0.2P/C}=0.025

PC=0.75m

OC=0.75+0.2=0.95m