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PHXII09:RAY OPTICS AND OPTICAL INSTRUMENTS

365041 A prism of refractive index $μ$ and angle $A$ is placed in the minimum deviation position. If the angle of minimum deviation is equal to $A$ , then the value of $A$ in terms of $μ$ is

1

2 \cos^{- 1} (\frac{μ}{2})

2

\sin^{- 1} (\frac{μ}{2})

3

\cos^{- 1} (\frac{μ}{2})

4

\sin^{- 1} \sqrt{\frac{μ - 1}{2}}

Explanation:

Given $δ_{m} = A,$ as $μ = \frac{\sin (\frac{A + δ_{m}}{2})}{\sin (\frac{A}{2})}$
$\Rightarrow μ = \frac{\sin (\frac{A + δ_{m}}{2})}{\sin \frac{A}{2}} = 2 \cos \frac{A}{2}$
$\Rightarrow A = 2 \cos^{- 1} (\frac{μ}{2})$

PHXII09:RAY OPTICS AND OPTICAL INSTRUMENTS

365042 Consider the following statements $A$ and $B$ and identify the correct choice
A) For a given value of angle of incidence, the angle of deviation produced by a prism increases with increasing of refractive index of the prism.
B) Angle of deviation produced by a thin prism is independent of angle of incidence.

1 A is true and B is false

2 A is false and B is true

3 A and B are true

4 A and B are false

PHXII09:RAY OPTICS AND OPTICAL INSTRUMENTS

365043 As angle of incidence on the first face of a prism increases, then the angle of deviation of emergent ray

1 Increases

2 Decreases

3 First increases then decreases

4 First decreases then increases

PHXII09:RAY OPTICS AND OPTICAL INSTRUMENTS

365044 For a prism of refractive index 1.732, the angle of minimum deviation is equal to the angle of the prism. The angle of the prism is

1

80^{\circ}

2

70^{\circ}

3

60^{\circ}

4

50^{\circ}

Explanation:

Given, refractive index of prism
$μ = 1.732, δ_{m} = A$
$∴ 1.732 = \frac{\sin (\frac{A + δ_{m}}{2})}{\sin \frac{A}{2}}$
$\sqrt{3} = \frac{\sin (\frac{2 A}{2})}{\sin \frac{A}{2}} = \frac{\sin A}{\sin \frac{A}{2}}$
$\Rightarrow \sqrt{3} = \frac{2 \sin \frac{A}{2} \cos \frac{A}{2}}{\sin \frac{A}{2}} \Rightarrow \sqrt{3} = 2 \times \cos \frac{A}{2}$
$\Rightarrow \cos \frac{A}{2} = \frac{\sqrt{3}}{2}$
$\frac{A}{2} = \cos^{- 1} (\frac{\sqrt{3}}{2}) = 30^{\circ}$
$A = 30^{\circ} \times 2 = 60^{\circ}$ .
So correct option is (3)

PHXII09:RAY OPTICS AND OPTICAL INSTRUMENTS

365045 When light of wavelength $λ$ is incident on an equilateral prism kept in its minimum deviation position, it is found that the angle of deviation equals the angle of the prism itself. The refractive index of the material of the prism with respect to the wavelength $λ$ is

1

\sqrt{3}

2

\frac{\sqrt{3}}{2}

3 2

4

\sqrt{2}

Explanation:

According to the relation
$μ = \frac{\sin (\frac{A + δ_{m}}{2})}{\sin \frac{A}{2}} = \frac{\sin (\frac{60^{\circ} + 60^{\circ}}{2})}{\sin \frac{60^{\circ}}{2}}$
$μ = \frac{\sin 60^{\circ}}{\sin 30^{\circ}} = \frac{\sqrt{3}}{2} \times \frac{2}{1} = \sqrt{3}$
$μ = \sqrt{3}$ .

PHXII09:RAY OPTICS AND OPTICAL INSTRUMENTS

365041 A prism of refractive index $μ$ and angle $A$ is placed in the minimum deviation position. If the angle of minimum deviation is equal to $A$ , then the value of $A$ in terms of $μ$ is

1

2 \cos^{- 1} (\frac{μ}{2})

2

\sin^{- 1} (\frac{μ}{2})

3

\cos^{- 1} (\frac{μ}{2})

4

\sin^{- 1} \sqrt{\frac{μ - 1}{2}}

Explanation:

Given $δ_{m} = A,$ as $μ = \frac{\sin (\frac{A + δ_{m}}{2})}{\sin (\frac{A}{2})}$
$\Rightarrow μ = \frac{\sin (\frac{A + δ_{m}}{2})}{\sin \frac{A}{2}} = 2 \cos \frac{A}{2}$
$\Rightarrow A = 2 \cos^{- 1} (\frac{μ}{2})$

PHXII09:RAY OPTICS AND OPTICAL INSTRUMENTS

365042 Consider the following statements $A$ and $B$ and identify the correct choice
A) For a given value of angle of incidence, the angle of deviation produced by a prism increases with increasing of refractive index of the prism.
B) Angle of deviation produced by a thin prism is independent of angle of incidence.

1 A is true and B is false

2 A is false and B is true

3 A and B are true

4 A and B are false

PHXII09:RAY OPTICS AND OPTICAL INSTRUMENTS

365043 As angle of incidence on the first face of a prism increases, then the angle of deviation of emergent ray

1 Increases

2 Decreases

3 First increases then decreases

4 First decreases then increases

PHXII09:RAY OPTICS AND OPTICAL INSTRUMENTS

365044 For a prism of refractive index 1.732, the angle of minimum deviation is equal to the angle of the prism. The angle of the prism is

1

80^{\circ}

2

70^{\circ}

3

60^{\circ}

4

50^{\circ}

Explanation:

Given, refractive index of prism
$μ = 1.732, δ_{m} = A$
$∴ 1.732 = \frac{\sin (\frac{A + δ_{m}}{2})}{\sin \frac{A}{2}}$
$\sqrt{3} = \frac{\sin (\frac{2 A}{2})}{\sin \frac{A}{2}} = \frac{\sin A}{\sin \frac{A}{2}}$
$\Rightarrow \sqrt{3} = \frac{2 \sin \frac{A}{2} \cos \frac{A}{2}}{\sin \frac{A}{2}} \Rightarrow \sqrt{3} = 2 \times \cos \frac{A}{2}$
$\Rightarrow \cos \frac{A}{2} = \frac{\sqrt{3}}{2}$
$\frac{A}{2} = \cos^{- 1} (\frac{\sqrt{3}}{2}) = 30^{\circ}$
$A = 30^{\circ} \times 2 = 60^{\circ}$ .
So correct option is (3)

PHXII09:RAY OPTICS AND OPTICAL INSTRUMENTS

365045 When light of wavelength $λ$ is incident on an equilateral prism kept in its minimum deviation position, it is found that the angle of deviation equals the angle of the prism itself. The refractive index of the material of the prism with respect to the wavelength $λ$ is

1

\sqrt{3}

2

\frac{\sqrt{3}}{2}

3 2

4

\sqrt{2}

Explanation:

According to the relation
$μ = \frac{\sin (\frac{A + δ_{m}}{2})}{\sin \frac{A}{2}} = \frac{\sin (\frac{60^{\circ} + 60^{\circ}}{2})}{\sin \frac{60^{\circ}}{2}}$
$μ = \frac{\sin 60^{\circ}}{\sin 30^{\circ}} = \frac{\sqrt{3}}{2} \times \frac{2}{1} = \sqrt{3}$
$μ = \sqrt{3}$ .

PHXII09:RAY OPTICS AND OPTICAL INSTRUMENTS

365041 A prism of refractive index $μ$ and angle $A$ is placed in the minimum deviation position. If the angle of minimum deviation is equal to $A$ , then the value of $A$ in terms of $μ$ is

1

2 \cos^{- 1} (\frac{μ}{2})

2

\sin^{- 1} (\frac{μ}{2})

3

\cos^{- 1} (\frac{μ}{2})

4

\sin^{- 1} \sqrt{\frac{μ - 1}{2}}

Explanation:

Given $δ_{m} = A,$ as $μ = \frac{\sin (\frac{A + δ_{m}}{2})}{\sin (\frac{A}{2})}$
$\Rightarrow μ = \frac{\sin (\frac{A + δ_{m}}{2})}{\sin \frac{A}{2}} = 2 \cos \frac{A}{2}$
$\Rightarrow A = 2 \cos^{- 1} (\frac{μ}{2})$

PHXII09:RAY OPTICS AND OPTICAL INSTRUMENTS

365042 Consider the following statements $A$ and $B$ and identify the correct choice
A) For a given value of angle of incidence, the angle of deviation produced by a prism increases with increasing of refractive index of the prism.
B) Angle of deviation produced by a thin prism is independent of angle of incidence.

1 A is true and B is false

2 A is false and B is true

3 A and B are true

4 A and B are false

PHXII09:RAY OPTICS AND OPTICAL INSTRUMENTS

365043 As angle of incidence on the first face of a prism increases, then the angle of deviation of emergent ray

1 Increases

2 Decreases

3 First increases then decreases

4 First decreases then increases

PHXII09:RAY OPTICS AND OPTICAL INSTRUMENTS

365044 For a prism of refractive index 1.732, the angle of minimum deviation is equal to the angle of the prism. The angle of the prism is

1

80^{\circ}

2

70^{\circ}

3

60^{\circ}

4

50^{\circ}

Explanation:

Given, refractive index of prism
$μ = 1.732, δ_{m} = A$
$∴ 1.732 = \frac{\sin (\frac{A + δ_{m}}{2})}{\sin \frac{A}{2}}$
$\sqrt{3} = \frac{\sin (\frac{2 A}{2})}{\sin \frac{A}{2}} = \frac{\sin A}{\sin \frac{A}{2}}$
$\Rightarrow \sqrt{3} = \frac{2 \sin \frac{A}{2} \cos \frac{A}{2}}{\sin \frac{A}{2}} \Rightarrow \sqrt{3} = 2 \times \cos \frac{A}{2}$
$\Rightarrow \cos \frac{A}{2} = \frac{\sqrt{3}}{2}$
$\frac{A}{2} = \cos^{- 1} (\frac{\sqrt{3}}{2}) = 30^{\circ}$
$A = 30^{\circ} \times 2 = 60^{\circ}$ .
So correct option is (3)

PHXII09:RAY OPTICS AND OPTICAL INSTRUMENTS

365045 When light of wavelength $λ$ is incident on an equilateral prism kept in its minimum deviation position, it is found that the angle of deviation equals the angle of the prism itself. The refractive index of the material of the prism with respect to the wavelength $λ$ is

1

\sqrt{3}

2

\frac{\sqrt{3}}{2}

3 2

4

\sqrt{2}

Explanation:

According to the relation
$μ = \frac{\sin (\frac{A + δ_{m}}{2})}{\sin \frac{A}{2}} = \frac{\sin (\frac{60^{\circ} + 60^{\circ}}{2})}{\sin \frac{60^{\circ}}{2}}$
$μ = \frac{\sin 60^{\circ}}{\sin 30^{\circ}} = \frac{\sqrt{3}}{2} \times \frac{2}{1} = \sqrt{3}$
$μ = \sqrt{3}$ .

PHXII09:RAY OPTICS AND OPTICAL INSTRUMENTS

365041 A prism of refractive index $μ$ and angle $A$ is placed in the minimum deviation position. If the angle of minimum deviation is equal to $A$ , then the value of $A$ in terms of $μ$ is

1

2 \cos^{- 1} (\frac{μ}{2})

2

\sin^{- 1} (\frac{μ}{2})

3

\cos^{- 1} (\frac{μ}{2})

4

\sin^{- 1} \sqrt{\frac{μ - 1}{2}}

Explanation:

Given $δ_{m} = A,$ as $μ = \frac{\sin (\frac{A + δ_{m}}{2})}{\sin (\frac{A}{2})}$
$\Rightarrow μ = \frac{\sin (\frac{A + δ_{m}}{2})}{\sin \frac{A}{2}} = 2 \cos \frac{A}{2}$
$\Rightarrow A = 2 \cos^{- 1} (\frac{μ}{2})$

PHXII09:RAY OPTICS AND OPTICAL INSTRUMENTS

365042 Consider the following statements $A$ and $B$ and identify the correct choice
A) For a given value of angle of incidence, the angle of deviation produced by a prism increases with increasing of refractive index of the prism.
B) Angle of deviation produced by a thin prism is independent of angle of incidence.

1 A is true and B is false

2 A is false and B is true

3 A and B are true

4 A and B are false

PHXII09:RAY OPTICS AND OPTICAL INSTRUMENTS

365043 As angle of incidence on the first face of a prism increases, then the angle of deviation of emergent ray

1 Increases

2 Decreases

3 First increases then decreases

4 First decreases then increases

PHXII09:RAY OPTICS AND OPTICAL INSTRUMENTS

365044 For a prism of refractive index 1.732, the angle of minimum deviation is equal to the angle of the prism. The angle of the prism is

1

80^{\circ}

2

70^{\circ}

3

60^{\circ}

4

50^{\circ}

Explanation:

Given, refractive index of prism
$μ = 1.732, δ_{m} = A$
$∴ 1.732 = \frac{\sin (\frac{A + δ_{m}}{2})}{\sin \frac{A}{2}}$
$\sqrt{3} = \frac{\sin (\frac{2 A}{2})}{\sin \frac{A}{2}} = \frac{\sin A}{\sin \frac{A}{2}}$
$\Rightarrow \sqrt{3} = \frac{2 \sin \frac{A}{2} \cos \frac{A}{2}}{\sin \frac{A}{2}} \Rightarrow \sqrt{3} = 2 \times \cos \frac{A}{2}$
$\Rightarrow \cos \frac{A}{2} = \frac{\sqrt{3}}{2}$
$\frac{A}{2} = \cos^{- 1} (\frac{\sqrt{3}}{2}) = 30^{\circ}$
$A = 30^{\circ} \times 2 = 60^{\circ}$ .
So correct option is (3)

PHXII09:RAY OPTICS AND OPTICAL INSTRUMENTS

365045 When light of wavelength $λ$ is incident on an equilateral prism kept in its minimum deviation position, it is found that the angle of deviation equals the angle of the prism itself. The refractive index of the material of the prism with respect to the wavelength $λ$ is

1

\sqrt{3}

2

\frac{\sqrt{3}}{2}

3 2

4

\sqrt{2}

Explanation:

According to the relation
$μ = \frac{\sin (\frac{A + δ_{m}}{2})}{\sin \frac{A}{2}} = \frac{\sin (\frac{60^{\circ} + 60^{\circ}}{2})}{\sin \frac{60^{\circ}}{2}}$
$μ = \frac{\sin 60^{\circ}}{\sin 30^{\circ}} = \frac{\sqrt{3}}{2} \times \frac{2}{1} = \sqrt{3}$
$μ = \sqrt{3}$ .

PHXII09:RAY OPTICS AND OPTICAL INSTRUMENTS

365041 A prism of refractive index $μ$ and angle $A$ is placed in the minimum deviation position. If the angle of minimum deviation is equal to $A$ , then the value of $A$ in terms of $μ$ is

1

2 \cos^{- 1} (\frac{μ}{2})

2

\sin^{- 1} (\frac{μ}{2})

3

\cos^{- 1} (\frac{μ}{2})

4

\sin^{- 1} \sqrt{\frac{μ - 1}{2}}

Explanation:

Given $δ_{m} = A,$ as $μ = \frac{\sin (\frac{A + δ_{m}}{2})}{\sin (\frac{A}{2})}$
$\Rightarrow μ = \frac{\sin (\frac{A + δ_{m}}{2})}{\sin \frac{A}{2}} = 2 \cos \frac{A}{2}$
$\Rightarrow A = 2 \cos^{- 1} (\frac{μ}{2})$

PHXII09:RAY OPTICS AND OPTICAL INSTRUMENTS

365042 Consider the following statements $A$ and $B$ and identify the correct choice
A) For a given value of angle of incidence, the angle of deviation produced by a prism increases with increasing of refractive index of the prism.
B) Angle of deviation produced by a thin prism is independent of angle of incidence.

1 A is true and B is false

2 A is false and B is true

3 A and B are true

4 A and B are false

PHXII09:RAY OPTICS AND OPTICAL INSTRUMENTS

365043 As angle of incidence on the first face of a prism increases, then the angle of deviation of emergent ray

1 Increases

2 Decreases

3 First increases then decreases

4 First decreases then increases

PHXII09:RAY OPTICS AND OPTICAL INSTRUMENTS

365044 For a prism of refractive index 1.732, the angle of minimum deviation is equal to the angle of the prism. The angle of the prism is

1

80^{\circ}

2

70^{\circ}

3

60^{\circ}

4

50^{\circ}

Explanation:

Given, refractive index of prism
$μ = 1.732, δ_{m} = A$
$∴ 1.732 = \frac{\sin (\frac{A + δ_{m}}{2})}{\sin \frac{A}{2}}$
$\sqrt{3} = \frac{\sin (\frac{2 A}{2})}{\sin \frac{A}{2}} = \frac{\sin A}{\sin \frac{A}{2}}$
$\Rightarrow \sqrt{3} = \frac{2 \sin \frac{A}{2} \cos \frac{A}{2}}{\sin \frac{A}{2}} \Rightarrow \sqrt{3} = 2 \times \cos \frac{A}{2}$
$\Rightarrow \cos \frac{A}{2} = \frac{\sqrt{3}}{2}$
$\frac{A}{2} = \cos^{- 1} (\frac{\sqrt{3}}{2}) = 30^{\circ}$
$A = 30^{\circ} \times 2 = 60^{\circ}$ .
So correct option is (3)

PHXII09:RAY OPTICS AND OPTICAL INSTRUMENTS

365045 When light of wavelength $λ$ is incident on an equilateral prism kept in its minimum deviation position, it is found that the angle of deviation equals the angle of the prism itself. The refractive index of the material of the prism with respect to the wavelength $λ$ is

1

\sqrt{3}

2

\frac{\sqrt{3}}{2}

3 2

4

\sqrt{2}

Explanation:

According to the relation
$μ = \frac{\sin (\frac{A + δ_{m}}{2})}{\sin \frac{A}{2}} = \frac{\sin (\frac{60^{\circ} + 60^{\circ}}{2})}{\sin \frac{60^{\circ}}{2}}$
$μ = \frac{\sin 60^{\circ}}{\sin 30^{\circ}} = \frac{\sqrt{3}}{2} \times \frac{2}{1} = \sqrt{3}$
$μ = \sqrt{3}$ .