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For yellow light, what is the minimum deviation produced by a prism of angle [tex]\(60^{\circ}\)[/tex] and refractive index 2?

Given:
[tex]\[
\mu = 2, \quad A = 60^{\circ}, \quad \delta_{m} = ?
\][/tex]

Substituting in the formula,
[tex]\[
\sin \left(\frac{A + \delta_m}{2}\right) = \mu \sin \left(\frac{A}{2}\right)
\][/tex]

[tex]\[
\sin \left(\frac{60^{\circ} + \delta_m}{2}\right) = 2 \sin 30^{\circ}
\][/tex]

[tex]\[
\therefore \quad \delta_{m} = 36.3^{\circ}
\][/tex]


Sagot :

To find the minimum deviation angle [tex]\(\delta_m\)[/tex] produced by a prism with an angle [tex]\(A = 60^\circ\)[/tex] and a refractive index [tex]\(\mu = 2\)[/tex], we can use the formula for the angle of minimum deviation for a prism:

[tex]\[ \delta_m = 2 \arcsin\left(\mu \cdot \sin\left(\frac{A}{2}\right)\right) - A \][/tex]

Let's solve this step-by-step:

1. Convert the prism angle from degrees to radians:

[tex]\[ A = 60^\circ \][/tex]

We need this angle in radians:

[tex]\[ A_{\text{rad}} = \frac{60 \times \pi}{180} = \frac{\pi}{3} \approx 1.0471975511965976 \, \text{radians} \][/tex]

2. Calculate the half-angle of the prism in radians:

[tex]\[ \frac{A}{2} = \frac{60^\circ}{2} = 30^\circ \][/tex]

In radians:

[tex]\[ \left(\frac{A}{2}\right)_{\text{rad}} = \frac{30 \times \pi}{180} = \frac{\pi}{6} \approx 0.5235987755982988 \, \text{radians} \][/tex]

3. Calculate [tex]\(\sin\left(\frac{A}{2}\right)\)[/tex]:

[tex]\[ \sin\left(\frac{A}{2}\right) = \sin\left(30^\circ\right) = \frac{1}{2} = 0.5 \][/tex]

4. Calculate the term inside the arcsine function:

[tex]\[ \mu \cdot \sin\left(\frac{A}{2}\right) = 2 \cdot 0.5 = 1 \][/tex]

However, due to numerical precision, this value might slightly differ, for example:
[tex]\[ \mu \cdot \sin\left(\frac{A}{2}\right) \approx 0.9999999999999999 \][/tex]

5. Calculate the arcsine of this term:

[tex]\[ \arcsin\left(0.9999999999999999\right) \approx \arcsin(1) = \frac{\pi}{2} \approx 1.5707963118937354 \, \text{radians} \][/tex]

6. Calculate the minimum deviation in radians:

[tex]\[ \delta_{m, \text{rad}} = 2 \cdot \arcsin\left(\mu \cdot \sin\left(\frac{A}{2}\right)\right) - A_{\text{rad}} \][/tex]

Substitute the values we have calculated:

[tex]\[ \delta_{m, \text{rad}} = 2 \times 1.5707963118937354 - 1.0471975511965976 \approx 2.0943950725908733 \, \text{radians} \][/tex]

7. Convert the minimum deviation from radians back to degrees:

[tex]\[ \delta_{m} \approx 2.0943950725908733 \times \frac{180}{\pi} \approx 119.99999829245273^\circ \][/tex]

Thus, the minimum deviation ([tex]\(\delta_m\)[/tex]) for yellow light in this prism is approximately:

[tex]\[ \delta_m \approx 119.99999829245273^\circ \][/tex]