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PHXII09:RAY OPTICS AND OPTICAL INSTRUMENTS

365054 Angle of minimum deviation for a prism of refractive index 1.5 is equal to the angle of prism. Then the angle of prism is $(\sin 48^{0} 36^{'} = 0.75)$

1

80^{\circ}

2

41^{\circ} 24^{'}

3

82^{\circ} 48^{'}

4

A = 60^{\circ}

Explanation:

$μ = \frac{\sin \frac{(A + δ)}{2}}{\sin \frac{A}{2}}$
Given that $δ = A$
$\frac{3}{2} = \frac{2 \sin \frac{A}{2} \cos \frac{A}{2}}{\sin \frac{A}{2}} \Rightarrow \cos \frac{A}{2} = \frac{3}{4}$
$A = 2 \cos^{- 1} (\frac{3}{4}) = 82^{\circ} 42^{'}$

PHXII09:RAY OPTICS AND OPTICAL INSTRUMENTS

365055 The refracting angle of a prism is A and refractive index of material of prism is $\cot \frac{A}{2}$ . The angle of minimum deviation is

1

90^{o} - A

2

180^{o} - 2 A

3

180^{o} - 3 A

4

180^{o} + 2 A

Explanation:

Given, angle of prism =A, $μ = \cot \frac{A}{2}$
$∴ μ = \cot \frac{A}{2} = \frac{\sin (\frac{A + δ_{m}}{2})}{\sin (\frac{A}{2})}$
$\Rightarrow \cos (\frac{A}{2}) = \sin (\frac{A + δ_{m}}{2})$
$\Rightarrow \sin (\frac{π}{2} - \frac{A}{2}) = \sin (\frac{A + δ_{m}}{2})$
$\Rightarrow \frac{π}{2} - \frac{A}{2} = \frac{A}{2} + \frac{δ_{m}}{2}$
$\Rightarrow δ_{m} = π - 2 A = 180^{o} - 2 A$

PHXII09:RAY OPTICS AND OPTICAL INSTRUMENTS

365056 When a ray of light is incident normally on one refracting surface of an equilateral prism (Refractive index of the material of the prism $= 1.5)$

1 Emerging ray is deviated by

45^{\circ}

2 The ray undergoes total internal reflection at the second refracting surface

3 Emerging ray is deviated by

30^{\circ}

4 Emerging ray just grazes the second refracting surface

Explanation:

Critical angle for the material of prism $C = \sin^{- 1} (\frac{1}{μ}) = \sin^{- 1} (\frac{1}{1.5}) = 42^{\circ}$ since angle of incidence at surface $A B (60^{\circ})$ is more than the critical angle $(42^{\circ})$ so total internal reflection takes place.
supporting img

PHXII09:RAY OPTICS AND OPTICAL INSTRUMENTS

365057 The curve of angle of incidence versus angle of deviation shown has been plotted for prism. The value of refractive index of the prism used is :
supporting img

1

\sqrt{3}

2

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

3

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

4

\sqrt{2}

Explanation:

At minimum deviation
$δ_{min} = 2 i - A$
$60^{\circ} = 2 (60^{\circ}) - A$
$A = 60^{\circ}$
The refractive index is
$1 \sin i = μ \sin r$
$\sin 60^{\circ} = μ \sin (\frac{A}{2}) = μ \sin 30^{\circ}$
$μ = \sqrt{3}$

PHXII09:RAY OPTICS AND OPTICAL INSTRUMENTS

365058 The angle of minimum deviation for a prism is 40° and the angle of the prism is 60° The angle of incidence in the position will be

1 60°

2 30°

3 100°

4 50°

Explanation:

$i = \frac{A + δ_{m}}{2} = 50^{0}$

PHXII09:RAY OPTICS AND OPTICAL INSTRUMENTS

365054 Angle of minimum deviation for a prism of refractive index 1.5 is equal to the angle of prism. Then the angle of prism is $(\sin 48^{0} 36^{'} = 0.75)$

1

80^{\circ}

2

41^{\circ} 24^{'}

3

82^{\circ} 48^{'}

4

A = 60^{\circ}

Explanation:

$μ = \frac{\sin \frac{(A + δ)}{2}}{\sin \frac{A}{2}}$
Given that $δ = A$
$\frac{3}{2} = \frac{2 \sin \frac{A}{2} \cos \frac{A}{2}}{\sin \frac{A}{2}} \Rightarrow \cos \frac{A}{2} = \frac{3}{4}$
$A = 2 \cos^{- 1} (\frac{3}{4}) = 82^{\circ} 42^{'}$

PHXII09:RAY OPTICS AND OPTICAL INSTRUMENTS

365055 The refracting angle of a prism is A and refractive index of material of prism is $\cot \frac{A}{2}$ . The angle of minimum deviation is

1

90^{o} - A

2

180^{o} - 2 A

3

180^{o} - 3 A

4

180^{o} + 2 A

Explanation:

Given, angle of prism =A, $μ = \cot \frac{A}{2}$
$∴ μ = \cot \frac{A}{2} = \frac{\sin (\frac{A + δ_{m}}{2})}{\sin (\frac{A}{2})}$
$\Rightarrow \cos (\frac{A}{2}) = \sin (\frac{A + δ_{m}}{2})$
$\Rightarrow \sin (\frac{π}{2} - \frac{A}{2}) = \sin (\frac{A + δ_{m}}{2})$
$\Rightarrow \frac{π}{2} - \frac{A}{2} = \frac{A}{2} + \frac{δ_{m}}{2}$
$\Rightarrow δ_{m} = π - 2 A = 180^{o} - 2 A$

PHXII09:RAY OPTICS AND OPTICAL INSTRUMENTS

365056 When a ray of light is incident normally on one refracting surface of an equilateral prism (Refractive index of the material of the prism $= 1.5)$

1 Emerging ray is deviated by

45^{\circ}

2 The ray undergoes total internal reflection at the second refracting surface

3 Emerging ray is deviated by

30^{\circ}

4 Emerging ray just grazes the second refracting surface

Explanation:

Critical angle for the material of prism $C = \sin^{- 1} (\frac{1}{μ}) = \sin^{- 1} (\frac{1}{1.5}) = 42^{\circ}$ since angle of incidence at surface $A B (60^{\circ})$ is more than the critical angle $(42^{\circ})$ so total internal reflection takes place.
supporting img

PHXII09:RAY OPTICS AND OPTICAL INSTRUMENTS

365057 The curve of angle of incidence versus angle of deviation shown has been plotted for prism. The value of refractive index of the prism used is :
supporting img

1

\sqrt{3}

2

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

3

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

4

\sqrt{2}

Explanation:

At minimum deviation
$δ_{min} = 2 i - A$
$60^{\circ} = 2 (60^{\circ}) - A$
$A = 60^{\circ}$
The refractive index is
$1 \sin i = μ \sin r$
$\sin 60^{\circ} = μ \sin (\frac{A}{2}) = μ \sin 30^{\circ}$
$μ = \sqrt{3}$

PHXII09:RAY OPTICS AND OPTICAL INSTRUMENTS

365058 The angle of minimum deviation for a prism is 40° and the angle of the prism is 60° The angle of incidence in the position will be

1 60°

2 30°

3 100°

4 50°

Explanation:

$i = \frac{A + δ_{m}}{2} = 50^{0}$

PHXII09:RAY OPTICS AND OPTICAL INSTRUMENTS

365054 Angle of minimum deviation for a prism of refractive index 1.5 is equal to the angle of prism. Then the angle of prism is $(\sin 48^{0} 36^{'} = 0.75)$

1

80^{\circ}

2

41^{\circ} 24^{'}

3

82^{\circ} 48^{'}

4

A = 60^{\circ}

Explanation:

$μ = \frac{\sin \frac{(A + δ)}{2}}{\sin \frac{A}{2}}$
Given that $δ = A$
$\frac{3}{2} = \frac{2 \sin \frac{A}{2} \cos \frac{A}{2}}{\sin \frac{A}{2}} \Rightarrow \cos \frac{A}{2} = \frac{3}{4}$
$A = 2 \cos^{- 1} (\frac{3}{4}) = 82^{\circ} 42^{'}$

PHXII09:RAY OPTICS AND OPTICAL INSTRUMENTS

365055 The refracting angle of a prism is A and refractive index of material of prism is $\cot \frac{A}{2}$ . The angle of minimum deviation is

1

90^{o} - A

2

180^{o} - 2 A

3

180^{o} - 3 A

4

180^{o} + 2 A

Explanation:

Given, angle of prism =A, $μ = \cot \frac{A}{2}$
$∴ μ = \cot \frac{A}{2} = \frac{\sin (\frac{A + δ_{m}}{2})}{\sin (\frac{A}{2})}$
$\Rightarrow \cos (\frac{A}{2}) = \sin (\frac{A + δ_{m}}{2})$
$\Rightarrow \sin (\frac{π}{2} - \frac{A}{2}) = \sin (\frac{A + δ_{m}}{2})$
$\Rightarrow \frac{π}{2} - \frac{A}{2} = \frac{A}{2} + \frac{δ_{m}}{2}$
$\Rightarrow δ_{m} = π - 2 A = 180^{o} - 2 A$

PHXII09:RAY OPTICS AND OPTICAL INSTRUMENTS

365056 When a ray of light is incident normally on one refracting surface of an equilateral prism (Refractive index of the material of the prism $= 1.5)$

1 Emerging ray is deviated by

45^{\circ}

2 The ray undergoes total internal reflection at the second refracting surface

3 Emerging ray is deviated by

30^{\circ}

4 Emerging ray just grazes the second refracting surface

Explanation:

Critical angle for the material of prism $C = \sin^{- 1} (\frac{1}{μ}) = \sin^{- 1} (\frac{1}{1.5}) = 42^{\circ}$ since angle of incidence at surface $A B (60^{\circ})$ is more than the critical angle $(42^{\circ})$ so total internal reflection takes place.
supporting img

PHXII09:RAY OPTICS AND OPTICAL INSTRUMENTS

365057 The curve of angle of incidence versus angle of deviation shown has been plotted for prism. The value of refractive index of the prism used is :
supporting img

1

\sqrt{3}

2

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

3

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

4

\sqrt{2}

Explanation:

At minimum deviation
$δ_{min} = 2 i - A$
$60^{\circ} = 2 (60^{\circ}) - A$
$A = 60^{\circ}$
The refractive index is
$1 \sin i = μ \sin r$
$\sin 60^{\circ} = μ \sin (\frac{A}{2}) = μ \sin 30^{\circ}$
$μ = \sqrt{3}$

PHXII09:RAY OPTICS AND OPTICAL INSTRUMENTS

365058 The angle of minimum deviation for a prism is 40° and the angle of the prism is 60° The angle of incidence in the position will be

1 60°

2 30°

3 100°

4 50°

Explanation:

$i = \frac{A + δ_{m}}{2} = 50^{0}$

PHXII09:RAY OPTICS AND OPTICAL INSTRUMENTS

365054 Angle of minimum deviation for a prism of refractive index 1.5 is equal to the angle of prism. Then the angle of prism is $(\sin 48^{0} 36^{'} = 0.75)$

1

80^{\circ}

2

41^{\circ} 24^{'}

3

82^{\circ} 48^{'}

4

A = 60^{\circ}

Explanation:

$μ = \frac{\sin \frac{(A + δ)}{2}}{\sin \frac{A}{2}}$
Given that $δ = A$
$\frac{3}{2} = \frac{2 \sin \frac{A}{2} \cos \frac{A}{2}}{\sin \frac{A}{2}} \Rightarrow \cos \frac{A}{2} = \frac{3}{4}$
$A = 2 \cos^{- 1} (\frac{3}{4}) = 82^{\circ} 42^{'}$

PHXII09:RAY OPTICS AND OPTICAL INSTRUMENTS

365055 The refracting angle of a prism is A and refractive index of material of prism is $\cot \frac{A}{2}$ . The angle of minimum deviation is

1

90^{o} - A

2

180^{o} - 2 A

3

180^{o} - 3 A

4

180^{o} + 2 A

Explanation:

Given, angle of prism =A, $μ = \cot \frac{A}{2}$
$∴ μ = \cot \frac{A}{2} = \frac{\sin (\frac{A + δ_{m}}{2})}{\sin (\frac{A}{2})}$
$\Rightarrow \cos (\frac{A}{2}) = \sin (\frac{A + δ_{m}}{2})$
$\Rightarrow \sin (\frac{π}{2} - \frac{A}{2}) = \sin (\frac{A + δ_{m}}{2})$
$\Rightarrow \frac{π}{2} - \frac{A}{2} = \frac{A}{2} + \frac{δ_{m}}{2}$
$\Rightarrow δ_{m} = π - 2 A = 180^{o} - 2 A$

PHXII09:RAY OPTICS AND OPTICAL INSTRUMENTS

365056 When a ray of light is incident normally on one refracting surface of an equilateral prism (Refractive index of the material of the prism $= 1.5)$

1 Emerging ray is deviated by

45^{\circ}

2 The ray undergoes total internal reflection at the second refracting surface

3 Emerging ray is deviated by

30^{\circ}

4 Emerging ray just grazes the second refracting surface

Explanation:

Critical angle for the material of prism $C = \sin^{- 1} (\frac{1}{μ}) = \sin^{- 1} (\frac{1}{1.5}) = 42^{\circ}$ since angle of incidence at surface $A B (60^{\circ})$ is more than the critical angle $(42^{\circ})$ so total internal reflection takes place.
supporting img

PHXII09:RAY OPTICS AND OPTICAL INSTRUMENTS

365057 The curve of angle of incidence versus angle of deviation shown has been plotted for prism. The value of refractive index of the prism used is :
supporting img

1

\sqrt{3}

2

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

3

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

4

\sqrt{2}

Explanation:

At minimum deviation
$δ_{min} = 2 i - A$
$60^{\circ} = 2 (60^{\circ}) - A$
$A = 60^{\circ}$
The refractive index is
$1 \sin i = μ \sin r$
$\sin 60^{\circ} = μ \sin (\frac{A}{2}) = μ \sin 30^{\circ}$
$μ = \sqrt{3}$

PHXII09:RAY OPTICS AND OPTICAL INSTRUMENTS

365058 The angle of minimum deviation for a prism is 40° and the angle of the prism is 60° The angle of incidence in the position will be

1 60°

2 30°

3 100°

4 50°

Explanation:

$i = \frac{A + δ_{m}}{2} = 50^{0}$

PHXII09:RAY OPTICS AND OPTICAL INSTRUMENTS

365054 Angle of minimum deviation for a prism of refractive index 1.5 is equal to the angle of prism. Then the angle of prism is $(\sin 48^{0} 36^{'} = 0.75)$

1

80^{\circ}

2

41^{\circ} 24^{'}

3

82^{\circ} 48^{'}

4

A = 60^{\circ}

Explanation:

$μ = \frac{\sin \frac{(A + δ)}{2}}{\sin \frac{A}{2}}$
Given that $δ = A$
$\frac{3}{2} = \frac{2 \sin \frac{A}{2} \cos \frac{A}{2}}{\sin \frac{A}{2}} \Rightarrow \cos \frac{A}{2} = \frac{3}{4}$
$A = 2 \cos^{- 1} (\frac{3}{4}) = 82^{\circ} 42^{'}$

PHXII09:RAY OPTICS AND OPTICAL INSTRUMENTS

365055 The refracting angle of a prism is A and refractive index of material of prism is $\cot \frac{A}{2}$ . The angle of minimum deviation is

1

90^{o} - A

2

180^{o} - 2 A

3

180^{o} - 3 A

4

180^{o} + 2 A

Explanation:

Given, angle of prism =A, $μ = \cot \frac{A}{2}$
$∴ μ = \cot \frac{A}{2} = \frac{\sin (\frac{A + δ_{m}}{2})}{\sin (\frac{A}{2})}$
$\Rightarrow \cos (\frac{A}{2}) = \sin (\frac{A + δ_{m}}{2})$
$\Rightarrow \sin (\frac{π}{2} - \frac{A}{2}) = \sin (\frac{A + δ_{m}}{2})$
$\Rightarrow \frac{π}{2} - \frac{A}{2} = \frac{A}{2} + \frac{δ_{m}}{2}$
$\Rightarrow δ_{m} = π - 2 A = 180^{o} - 2 A$

PHXII09:RAY OPTICS AND OPTICAL INSTRUMENTS

365056 When a ray of light is incident normally on one refracting surface of an equilateral prism (Refractive index of the material of the prism $= 1.5)$

1 Emerging ray is deviated by

45^{\circ}

2 The ray undergoes total internal reflection at the second refracting surface

3 Emerging ray is deviated by

30^{\circ}

4 Emerging ray just grazes the second refracting surface

Explanation:

Critical angle for the material of prism $C = \sin^{- 1} (\frac{1}{μ}) = \sin^{- 1} (\frac{1}{1.5}) = 42^{\circ}$ since angle of incidence at surface $A B (60^{\circ})$ is more than the critical angle $(42^{\circ})$ so total internal reflection takes place.
supporting img

PHXII09:RAY OPTICS AND OPTICAL INSTRUMENTS

365057 The curve of angle of incidence versus angle of deviation shown has been plotted for prism. The value of refractive index of the prism used is :
supporting img

1

\sqrt{3}

2

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

3

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

4

\sqrt{2}

Explanation:

At minimum deviation
$δ_{min} = 2 i - A$
$60^{\circ} = 2 (60^{\circ}) - A$
$A = 60^{\circ}$
The refractive index is
$1 \sin i = μ \sin r$
$\sin 60^{\circ} = μ \sin (\frac{A}{2}) = μ \sin 30^{\circ}$
$μ = \sqrt{3}$

PHXII09:RAY OPTICS AND OPTICAL INSTRUMENTS

365058 The angle of minimum deviation for a prism is 40° and the angle of the prism is 60° The angle of incidence in the position will be

1 60°

2 30°

3 100°

4 50°

Explanation:

$i = \frac{A + δ_{m}}{2} = 50^{0}$

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