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Exercises · 9.6

Q.A prism is made of glass of unknown refractive index. A parallel beam of light is incident on a face of the prism. The angle of minimum deviation is measured to be 40∘40^\circ. What is the refractive index of the material of the prism? The refracting angle of the prism is 60∘60^\circ. If the prism is placed in water (refractive index 1.331.33), predict the new angle of minimum deviation of a parallel beam of light.

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Using the prism formula n=sin⁡[(A+δm)/2]sin⁡(A/2)n=\frac{\sin[(A+\delta_m)/2]}{\sin(A/2)} with A=60∘A=60^\circ and δm=40∘\delta_m=40^\circ, the refractive index of the glass is ng≈1.53n_g\approx1.53. When the prism is immersed in water (nw=1.33n_w=1.33), the effective (relative) refractive index becomes nrel=ng/nw≈1.15n_{rel}=n_g/n_w\approx1.15, and the new angle of minimum deviation is δm′≈10.3∘\delta_m'\approx10.3^\circ.

Step 1 — refractive index of the glass in air

ng=sin⁡(A+δm2)sin⁡(A2)=sin⁡50∘sin⁡30∘=0.7660.5=1.532n_g = \frac{\sin\left(\frac{A+\delta_m}{2}\right)}{\sin\left(\frac{A}{2}\right)} = \frac{\sin 50^\circ}{\sin 30^\circ} = \frac{0.766}{0.5} = 1.532

Step 2 — relative refractive index in water

The prism formula's nn is always the refractive index of the prism relative to the surrounding medium. In water, that relative index is

nrel=ngnw=1.5321.33≈1.152n_{rel} = \frac{n_g}{n_w} = \frac{1.532}{1.33} \approx 1.152

Step 3 — new angle of minimum deviation

Using the prism formula again, now with nreln_{rel} in place of ngn_g, and the same prism angle A=60∘A=60^\circ:

nrel=sin⁡(A+δm′2)sin⁡(A/2)  ⇒  sin⁡(A+δm′2)=1.152×0.5=0.576n_{rel} = \frac{\sin\left(\frac{A+\delta_m'}{2}\right)}{\sin(A/2)} \;\Rightarrow\; \sin\left(\frac{A+\delta_m'}{2}\right) = 1.152 \times 0.5 = 0.576 …

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