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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Answer:

Correct Answer: 1. (c)

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

Optics 6 Question 1

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Optics 6 Question 1

Answer:

Engineering Preparation

Medical Preparation

NCERT Books & Solutions

Important Resources

Other Resources

Syllabus

Test Platform

Sample Mock Test

Mentorship

NCERT Exercise Video Solution

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