Question

what is the condition for obtaining the minimum angle of deviation in the prism.show that the refractive index of the material of a prism is given by
mew=sin(A+fm)/2/ sin A/2​

Answer 1

Answer:

Let the refractive index of the prism be μ.

Applying Snell's law, n

air

sini=μ sinr

where n

air

=1

Or sini=μ sinr

1

⟹ μ=

sinr

1

sini

........(1)

Deviation by prism δ=i+e−A

where A=r

1

+r

2

For minimum deviation i=e and r

1

=r

2

∴ A=2r

1

⟹ r

1

=

2

A

.....(b)

Angle of minimum deviation is given by δ

m

=2i−A

⟹ i=

2

A+δ

m

......(c)

Putting (b) and (c) in (1), we get

μ=

sin

2

A

sin(

2

A+δ

m

)