Question

From the given data set determine the angular dispersion by the prism and dispersive power of its material fo
extreme colors ng = 1.62, n = 1.66, 8g = 3.1°.
​

Answer 1

Answer:

The angular dispersion is $0.2^\circ$

The dispersive power of the prism is $0.0625$

Explanation:

Angle of deviation for red colour

$\delta_R=3.1^\circ$

Given, the refractive index dor red colour

$\mu_R=1.62$

We know that

$\delta_R=(\mu_R-1)A$

$3.1^\circ=(1.62-1)A$

$\implies A=\frac{3.1^\circ}{0.62}$

$\implies A=5^\circ$

Therefore, the angle of prism is $5^\circ$

The angle of deviation for the violet colour

$\delta_V=(\mu_V-1)A$

$\implies \delta_V=(1.66-1)\times 5^\circ$

$\implies \delta_V=3.3^\circ$

The angle of deviation for the yellod colour (mean angle of deviation)

$\delta_Y=\frac{\delta_R+\delta_V}{2}$

$\implies \delta_Y=\frac{3.1^\circ+3.3^\circ}{2}$

$\implies \delta_Y=3.2^\circ$

The angular dispersion by the prism

$\theta=\delta_V-\delta_R$

$\theta=3.3^\circ-3.1^\circ$

$\theta=0.2^\circ$

The dispersive power of the prism

$\omega=\frac{\theta}{\delta_Y}$

$\implies \omega=\frac{0.2}{3.2}$

$\implies \omega=0.0625$

Hope this helps.

Answer 2

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.