SATHEE: Optics 6 Question 1

Optics 6 Question 1

1. A system of three polarisers $P_{1}, P_{2}, P_{3}$ is set up such that the pass axis of $P_{3}$ is crossed with respect to that of $P_{1}$ . The pass axis of $P_{2}$ is inclined at $60^{\circ}$ to the pass axis of $P_{3}$ . When a beam of unpolarised light of intensity $I_{0}$ is incident on $P_{1}$ , the intensity of light transmitted by the three polarisers is $I$ . The ratio $(I_{0} / I)$ equals (nearly)

(2019 Main, 12 April II)

(a) 5.33

(b) 16.00

(d) 1.80

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Solution:

When unpolarised light pass through polaroid $P_{1}$ , intensity obtained is

$I_{1} = \frac{I_{0}}{2}$

where, $I_{0} =$ intensity of incident light.

Now, this transmitted light is polarised and it pass through polariser $P_{2}$ . So, intensity $I_{2}$ transmitted is obtained by Malus law.

$\Rightarrow I_{2} = I_{1} \cos^{2} θ$

As angle of pass axis of $P_{1}$ and $P_{3}$ is $90^{\circ}$ and angle of pass axis of $P_{2}$ and $P_{3}$ is $60^{\circ}$ , so angle between pass axis of $P_{1}$ and $P_{2}$ is $(90^{\circ} - 60^{\circ}) = 30^{\circ}$ . So,

When this light pass through third polariser $P_{3}$ , intensity $I$ transmitted is again obtained by Malus law. So, $I = I_{2} \cos^{2} 60^{\circ} = \frac{3}{8} I_{0} \cos^{2} 60^{\circ}$

$= \frac{3}{8} I_{0} \times {\frac{1}{2}}^{2} = \frac{3}{32} I_{0}$

So, ratio $\frac{I_{0}}{I} = \frac{32}{3} = 10.67$

2 As we know,

Path difference introduced by thin film,

$Δ = (μ - 1) t$

and if fringe pattern shifts by one frings width, then path difference,

$Δ = 1 \times λ = λ$

So, from Eqs. (i) and (ii), we get

$(μ - 1) t = λ \Rightarrow t = \frac{λ}{μ - 1}$

Alternate Solution

Path difference introduced by the thin film of thickness $t$ and refractive index $μ$ is given by

$Δ = (μ - 1) t$

$∴$ Position of the fringe is

$x = \frac{Δ D}{d} = \frac{(μ - 1) t D}{d}$

Fringe width of one fringe is given by

$β = \frac{λ D}{d}$

Given that $x = β$ , so from Eqs. (i) and (ii), we get

$\begin{array}{ll} \Rightarrow & \frac{(μ - 1) t D}{d} = \frac{λ D}{d} \\ \Rightarrow & (μ - 1) t = λ or t = \frac{λ}{(μ - 1)} \end{array}$

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Optics 6 Question 1

Solution:

Alternate Solution

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एनसीईआरटी व्यायाम वीडियो समाधान

Optics 6 Question 1

Solution:

Alternate 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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