Question

In an experiment for determination of refractiveindex of glass of a prism by i – δ, plot, it was foundthat a ray incident at angle 35°, suffers a deviationof 40° and that it emerges at angle 79°. In that casewhich of the following is closest to the maximumpossible value of the refractive index?

Answer 1

Answer on Question #65984, Physics / Optics

In an experiment for determination of refractive index of glass of a prism by $i - \delta$, plot, it was found that a ray incident at angle $35{}^{\circ}$, suffers a deviation of $40{}^{\circ}$ and that it emerges at angle $79{}^{\circ}$. In that case which of the following is closest to the maximum possible value of the refractive index?

Find: $\mu - ?$

Given:

$$i = 35{}^{\circ}$$ $$\delta = 40{}^{\circ}$$ $$\gamma = 79{}^{\circ}$$

Solution:

Refractive index:

$$\mu = \frac{\sin\left(\frac{\delta_{\min} + A}{2}\right)}{\sin\frac{A}{2}} \quad (1)$$ $$\delta = i + \gamma - A \quad (2)$$ $$\text{Of} \quad (2) \Rightarrow A = i + \gamma - \delta \quad (3)$$ $$\text{Of} \quad (3) \Rightarrow A = 74{}^{\circ} \quad (4)$$

For minimum deviation: $i = \frac{35{}^{\circ} + 79{}^{\circ}}{2} = 57{}^{\circ}$ (5)

$$\delta_{\min} = 2i - A \quad (6)$$

(3) and (5) in (6): $\delta_{\min} = 40{}^{\circ}$ (7)

(4) and (7) in (1): $\mu = \frac{\sin 57{}^{\circ}}{\sin 37{}^{\circ}}$ (8)

$$\text{Of} \quad (8) \Rightarrow \mu = 1.4$$

Answer:

1.4

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