Concepts Of Physics MCQ Edition [Volume 1]
- Exam
- Concepts Of Physics MCQ Edition [Volume 1]
- Subject
- Physics
- Chapter
- Dispersion and Spectra
Concepts Of Physics MCQ Edition [Volume 1] · Physics · Dispersion and Spectra
A thin prism of angle 6.0^ ^ ^ , ω = 0.07 and μ_y = 1.50 is...
A thin prism of angle 6.0^{}^{{}^{\circ}}, and is combined with another thin prism having and . The combination produces no deviation in the mean ray. Determine: (i) the angle of the second prism, (ii) the net angular dispersion produced by the combination when a beam of white light passes through it, (iii) the deviation in the mean ray if the prisms are similarly directed, and (iv) the angular dispersion in the situation where they are similarly directed.
Options
- A(i) 5^{}^{{}^{\circ}}, (ii) 0.03^{}^{{}^{\circ}}, (iii) 6^{}^{{}^{\circ}}, (iv) 0.45^{}^{{}^{\circ}}
- B(i) 6^{}^{{}^{\circ}}, (ii) 0.04^{}^{{}^{\circ}}, (iii) 7^{}^{{}^{\circ}}, (iv) 0.55^{}^{{}^{\circ}}
- C(i) 5^{}^{{}^{\circ}}, (ii) 0.06^{}^{{}^{\circ}}, (iii) 6^{}^{{}^{\circ}}, (iv) 0.40^{}^{{}^{\circ}}
- D(i) 4^{}^{{}^{\circ}}, (ii) 0.02^{}^{{}^{\circ}}, (iii) 5^{}^{{}^{\circ}}, (iv) 0.35^{}^{{}^{\circ}}
Correct answer
A. (i) 5^{}^{{}^{\circ}}, (ii) 0.03^{}^{{}^{\circ}}, (iii) 6^{}^{{}^{\circ}}, (iv) 0.45^{}^{{}^{\circ}}
Step-by-step solution
For the first prism, the deviation of the mean ray is given by .
Substituting the given values, \delta_1 = (1.50 - 1) \times 6.0^{}^{{}^{\circ}} = 3.0^{}^{{}^{\circ}}.
For the combination to produce no deviation in the mean ray, the prisms must be oppositely directed and their deviations must be equal in magnitude, so \delta_2 = \delta_1 = 3.0^{}^{{}^{\circ}}.
The deviation for the second prism is .
Substituting the values, 3.0^{}^{{}^{\circ}} = (1.60 - 1)A_2 \Rightarrow 0.60 A_2 = 3.0^{}^{{}^{\circ}} \Rightarrow A_2 = 5.0^{}^{{}^{\circ}}.
The angular dispersion produced by a prism is .
For the first prism, \theta_1 = 0.07 \times 3.0^{}^{{}^{\circ}} = 0.21^{}^{{}^{\circ}}.
For the second prism, \theta_2 = 0.08 \times 3.0^{}^{{}^{\circ}} = 0.24^{}^{{}^{\circ}}.
When the prisms are oppositely directed (for zero net deviation), the net angular dispersion is the difference between their dispersions:
Net dispersion = |\theta_1 - \theta_2| = |0.21^{}^{{}^{\circ}} - 0.24^{}^{{}^{\circ}}| = 0.03^{}^{{}^{\circ}}.
If the prisms are similarly directed, their deviations and dispersions add up.
Net deviation = \delta_1 + \delta_2 = 3.0^{}^{{}^{\circ}} + 3.0^{}^{{}^{\circ}} = 6.0^{}^{{}^{\circ}}.
Net angular dispersion = \theta_1 + \theta_2 = 0.21^{}^{{}^{\circ}} + 0.24^{}^{{}^{\circ}} = 0.45^{}^{{}^{\circ}}.
Substituting the given values, \delta_1 = (1.50 - 1) \times 6.0^{}^{{}^{\circ}} = 3.0^{}^{{}^{\circ}}.
For the combination to produce no deviation in the mean ray, the prisms must be oppositely directed and their deviations must be equal in magnitude, so \delta_2 = \delta_1 = 3.0^{}^{{}^{\circ}}.
The deviation for the second prism is .
Substituting the values, 3.0^{}^{{}^{\circ}} = (1.60 - 1)A_2 \Rightarrow 0.60 A_2 = 3.0^{}^{{}^{\circ}} \Rightarrow A_2 = 5.0^{}^{{}^{\circ}}.
The angular dispersion produced by a prism is .
For the first prism, \theta_1 = 0.07 \times 3.0^{}^{{}^{\circ}} = 0.21^{}^{{}^{\circ}}.
For the second prism, \theta_2 = 0.08 \times 3.0^{}^{{}^{\circ}} = 0.24^{}^{{}^{\circ}}.
When the prisms are oppositely directed (for zero net deviation), the net angular dispersion is the difference between their dispersions:
Net dispersion = |\theta_1 - \theta_2| = |0.21^{}^{{}^{\circ}} - 0.24^{}^{{}^{\circ}}| = 0.03^{}^{{}^{\circ}}.
If the prisms are similarly directed, their deviations and dispersions add up.
Net deviation = \delta_1 + \delta_2 = 3.0^{}^{{}^{\circ}} + 3.0^{}^{{}^{\circ}} = 6.0^{}^{{}^{\circ}}.
Net angular dispersion = \theta_1 + \theta_2 = 0.21^{}^{{}^{\circ}} + 0.24^{}^{{}^{\circ}} = 0.45^{}^{{}^{\circ}}.