Determine the angular dispersion by the prism and dispersive by the prism and dispersive power of it's material for extreme colours. nR=1.62 nV=1.66,SR =3.1°
Answers
Answer:
The angular dispersion is 0.2^\circ0.2
∘
The dispersive power of the prism is 0.06250.0625
Explanation:
Angle of deviation for red colour
\delta_R=3.1^\circδ
R
=3.1
∘
Given, the refractive index dor red colour
\mu_R=1.62μ
R
=1.62
We know that
\delta_R=(\mu_R-1)Aδ
R
=(μ
R
−1)A
3.1^\circ=(1.62-1)A3.1
∘
=(1.62−1)A
\implies A=\frac{3.1^\circ}{0.62}⟹A=
0.62
3.1
∘
\implies A=5^\circ⟹A=5
∘
Therefore, the angle of prism is 5^\circ5
∘
The angle of deviation for the violet colour
\delta_V=(\mu_V-1)Aδ
V
=(μ
V
−1)A
\implies \delta_V=(1.66-1)\times 5^\circ⟹δ
V
=(1.66−1)×5
∘
\implies \delta_V=3.3^\circ⟹δ
V
=3.3
∘
The angle of deviation for the yellod colour (mean angle of deviation)
\delta_Y=\frac{\delta_R+\delta_V}{2}δ
Y
=
2
δ
R
+δ
V
\implies \delta_Y=\frac{3.1^\circ+3.3^\circ}{2}⟹δ
Y
=
2
3.1
∘
+3.3
∘
\implies \delta_Y=3.2^\circ⟹δ
Y
=3.2
∘
The angular dispersion by the prism
\theta=\delta_V-\delta_Rθ=δ
V
−δ
R
\theta=3.3^\circ-3.1^\circθ=3.3
∘
−3.1
∘
\theta=0.2^\circθ=0.2
∘
The dispersive power of the prism
\omega=\frac{\theta}{\delta_Y}ω=
δ
Y
θ
\implies \omega=\frac{0.2}{3.2}⟹ω=
3.2
0.2
\implies \omega=0.0625⟹ω=0.0625
Hope this helps.