SATHEE: Optics 7 Question 39

Optics 7 Question 39

40. The refractive indices of the crown glass for blue and red light are 1.51 and 1.49 respectively and those of the flint glass are 1.77 and 1.73 respectively. An isosceles prism of angle $6^{\circ}$ is made of crown glass. A beam of white light is incident at a small angle on this prism. The other flint glass isosceles prism is combined with the crown glass prism such that there is no deviation of the incident light.

(a) Determine the angle of the flint glass prism.

(b) Calculate the net dispersion of the combined system.

$(2001, 5 M)$

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Solution:

(a) When angle of prism is small and angle of incidence is also small, the deviation is given by $δ = (μ - 1) A$ .

Net deviation by the two prisms is zero. So,

$δ_{1} + δ_{2} = 0$

or $(μ_{1} - 1) A_{1} + (μ_{2} - 1) A_{2} = 0$

Here, $μ_{1}$ and $μ_{2}$ are the refractive indices for crown and flint glasses respectively.

Hence,

$\begin{aligned} μ_{1} = \frac{1.51 + 1.49}{2} = 1.5 \\ μ_{2} = \frac{1.77 + 1.73}{2} = 1.75 \end{aligned}$

and

$A_{1} =$ Angle of prism for crown glass $= 6^{\circ}$

Substituting the values in Eq. (i), we get

$(1.5 - 1) (6^{\circ}) + (1.75 - 1) A_{2} = 0$

This gives $A_{2} = - 4^{\circ}$

Hence, angle of flint glass prism is $4^{\circ}$ (Negative sign shows that flint glass prism is inverted with respect to the crown glass prism.)

(b) Net dispersion due to the two prisms is

$\begin{aligned} = (μ_{b_{1}} - μ_{r_{1}}) A_{1} + (μ_{b_{2}} - μ_{r_{2}}) A_{2} \\ = (1.51 - 1.49) (6^{\circ}) + (1.77 - 1.73) (- 4^{\circ}) = - {0.04}^{\circ} \end{aligned}$

$∴$ Net dispersion is $- {0.04}^{\circ}$

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Optics 7 Question 39

Solution:

इंजीनियरिंग की तैयारी

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एनसीईआरटी व्यायाम वीडियो समाधान

Optics 7 Question 39

Solution:

इंजीनियरिंग की तैयारी

चिकित्सा तैयारी

एनसीईआरटी पुस्तकें और समाधान

महत्वपूर्ण संसाधन

अन्य संसाधन

पाठ्यक्रम

परीक्षा मंच

नमूना मॉक टेस्ट

सदस्यता

एनसीईआरटी व्यायाम वीडियो समाधान

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