For an isosceles prism of angle A and refractive index $μ$ , it is found that the angle of minimum deviation is $δ_{m} = A .$ .

Which of the following options is/are correct [2017]

Question

For an isosceles prism of angle A and refractive index $μ$ , it is found that the angle of minimum deviation is $δ_{m} = A .$ .

Which of the following options is/are correct? [2017]

Smart Achievers · Accepted Answer

(1, 3, 4)

For minimum deviation,

$i_{1} = e$

$r_{1} = r_{2} = r (say) = \frac{A}{2}$

$δ_{m} = 2 i_{1} - A$

Here, $δ_{m} = A$ $\Rightarrow 2 i_{1} - A = A$

$∴$ $i_{1} = A \Rightarrow e = A$

$∴$ $r_{1} = \frac{i_{1}}{2}$

Relation between $μ$ and $A$

$μ = \frac{\sin i_{1}}{\sin r_{1}} = \frac{\sin A}{\sin (\frac{A}{2})} = \frac{2 \sin (\frac{A}{2}) \cos (\frac{A}{2})}{\sin (\frac{A}{2})} = 2 \cos (\frac{A}{2})$

When emergent ray is tangential to the surface,

$μ = \frac{\sin 90 °}{\sin r_{2}} = \frac{1}{\sin r_{2}}$ $\Rightarrow r_{2} = \sin^{- 1} (\frac{1}{μ})$

But, $r_{1} + r_{2} = A$ $∴$ $r_{1} = A - r_{2}$

$∴$ $r_{1} = A - \sin^{- 1} (\frac{1}{μ})$

Applying Snell's law at P,

$μ = \frac{\sin i_{1}}{\sin r_{1}}$ $∴$ $i_{1} = \sin^{- 1} [μ \sin (A - \sin^{- 1} (\frac{1}{μ}))]$

For minimum deviation through an isosceles prism, if $∠ B = ∠ C, P Q ∥ B C .$

Solution

Q.
For an isosceles prism of angle A and refractive index $μ$ , it is found that the angle of minimum deviation is $δ_{m} = A .$ .

Which of the following options is/are correct [2017]

1 For the angle of incidence $i_{1} = A$ , the ray inside the prism is parallel to the base of the prism.

2 For this prism, the refractive index $μ$ and the angle of prism A are related as $A = \frac{1}{2} \cos^{- 1} (\frac{μ}{2})$

3 At minimum deviation, the incident angle $i_{1}$ and the refracting angle $r_{1}$ at the first refracting surface are related by $r_{1} = \frac{i_{1}}{2}$

4 For this prism, the emergent ray at the second surface will be tangential to the surface when the angle of incidence at the first surface is $i_{1} = \sin^{- 1} [\sin A \sqrt{4 \cos^{2} (\frac{A}{2}) - 1} - \cos A]$

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Solution

Q. For an isosceles prism of angle A and refractive index μ, it is found that the angle of minimum deviation is δm=A.. Which of the following options is/are correct [2017]

1 For the angle of incidence i1=A, the ray inside the prism is parallel to the base of the prism.

2 For this prism, the refractive index μ and the angle of prism A are related as A=12cos-1(μ2)

3 At minimum deviation, the incident angle i1 and the refracting angle r1 at the first refracting surface are related by r1=i12