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A polarizer P is placed be- tween a crossed polarizer-analyzer pair. The optic of axis of P makes an angle a with respect to that of the analyzer. If the intensity of incident light is I_{0} , neglecting losses due to reflection and absortion at the polarizers, the intensity of the transmitted light is (a) I_{0} * cos^2 alpha (b) I_{0} * sin^2 2 * alpha (c) 1/8 * I_{0} * cos^2 2 * alpha (d) 1/8 * I_{0} * sin^2 2 * alpha?
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A polarizer P is placed be- tween a crossed polarizer-analyzer pair. T...
Solution:
The intensity of light transmitted through a polarizer is given by Malus' law, which states that the intensity of the transmitted light is proportional to the square of the cosine of the angle between the transmission axis of the polarizer and the polarization direction of the incident light.
Given that the optic axis of the polarizer makes an angle α with respect to that of the analyzer, we can determine the transmitted intensity as follows:
1. Incident Light Intensity:
The intensity of the incident light is given as I₀.
2. Transmission Axis Angle:
The transmission axis of the polarizer makes an angle α with respect to the polarization direction of the incident light.
3. Intensity of Transmitted Light:
According to Malus' law, the intensity of the transmitted light is proportional to the square of the cosine of the angle between the transmission axis of the polarizer and the polarization direction of the incident light. Therefore, the intensity of the transmitted light is given by I = I₀ * cos²(α).
Therefore, the correct answer is (a) I₀ * cos²(α).
The intensity of light transmitted through a polarizer is given by Malus' law, which states that the intensity of the transmitted light is proportional to the square of the cosine of the angle between the transmission axis of the polarizer and the polarization direction of the incident light.
Given that the optic axis of the polarizer makes an angle α with respect to that of the analyzer, we can determine the transmitted intensity as follows:
1. Incident Light Intensity:
The intensity of the incident light is given as I₀.
2. Transmission Axis Angle:
The transmission axis of the polarizer makes an angle α with respect to the polarization direction of the incident light.
3. Intensity of Transmitted Light:
According to Malus' law, the intensity of the transmitted light is proportional to the square of the cosine of the angle between the transmission axis of the polarizer and the polarization direction of the incident light. Therefore, the intensity of the transmitted light is given by I = I₀ * cos²(α).
Therefore, the correct answer is (a) I₀ * cos²(α).
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A polarizer P is placed be- tween a crossed polarizer-analyzer pair. The optic of axis of P makes an angle a with respect to that of the analyzer. If the intensity of incident light is I_{0} , neglecting losses due to reflection and absortion at the polarizers, the intensity of the transmitted light is (a) I_{0} * cos^2 alpha (b) I_{0} * sin^2 2 * alpha (c) 1/8 * I_{0} * cos^2 2 * alpha (d) 1/8 * I_{0} * sin^2 2 * alpha?
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A polarizer P is placed be- tween a crossed polarizer-analyzer pair. The optic of axis of P makes an angle a with respect to that of the analyzer. If the intensity of incident light is I_{0} , neglecting losses due to reflection and absortion at the polarizers, the intensity of the transmitted light is (a) I_{0} * cos^2 alpha (b) I_{0} * sin^2 2 * alpha (c) 1/8 * I_{0} * cos^2 2 * alpha (d) 1/8 * I_{0} * sin^2 2 * alpha? for Physics 2024 is part of Physics preparation. The Question and answers have been prepared according to the Physics exam syllabus. Information about A polarizer P is placed be- tween a crossed polarizer-analyzer pair. The optic of axis of P makes an angle a with respect to that of the analyzer. If the intensity of incident light is I_{0} , neglecting losses due to reflection and absortion at the polarizers, the intensity of the transmitted light is (a) I_{0} * cos^2 alpha (b) I_{0} * sin^2 2 * alpha (c) 1/8 * I_{0} * cos^2 2 * alpha (d) 1/8 * I_{0} * sin^2 2 * alpha? covers all topics & solutions for Physics 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for A polarizer P is placed be- tween a crossed polarizer-analyzer pair. The optic of axis of P makes an angle a with respect to that of the analyzer. If the intensity of incident light is I_{0} , neglecting losses due to reflection and absortion at the polarizers, the intensity of the transmitted light is (a) I_{0} * cos^2 alpha (b) I_{0} * sin^2 2 * alpha (c) 1/8 * I_{0} * cos^2 2 * alpha (d) 1/8 * I_{0} * sin^2 2 * alpha?.
A polarizer P is placed be- tween a crossed polarizer-analyzer pair. The optic of axis of P makes an angle a with respect to that of the analyzer. If the intensity of incident light is I_{0} , neglecting losses due to reflection and absortion at the polarizers, the intensity of the transmitted light is (a) I_{0} * cos^2 alpha (b) I_{0} * sin^2 2 * alpha (c) 1/8 * I_{0} * cos^2 2 * alpha (d) 1/8 * I_{0} * sin^2 2 * alpha? for Physics 2024 is part of Physics preparation. The Question and answers have been prepared according to the Physics exam syllabus. Information about A polarizer P is placed be- tween a crossed polarizer-analyzer pair. The optic of axis of P makes an angle a with respect to that of the analyzer. If the intensity of incident light is I_{0} , neglecting losses due to reflection and absortion at the polarizers, the intensity of the transmitted light is (a) I_{0} * cos^2 alpha (b) I_{0} * sin^2 2 * alpha (c) 1/8 * I_{0} * cos^2 2 * alpha (d) 1/8 * I_{0} * sin^2 2 * alpha? covers all topics & solutions for Physics 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for A polarizer P is placed be- tween a crossed polarizer-analyzer pair. The optic of axis of P makes an angle a with respect to that of the analyzer. If the intensity of incident light is I_{0} , neglecting losses due to reflection and absortion at the polarizers, the intensity of the transmitted light is (a) I_{0} * cos^2 alpha (b) I_{0} * sin^2 2 * alpha (c) 1/8 * I_{0} * cos^2 2 * alpha (d) 1/8 * I_{0} * sin^2 2 * alpha?.
Solutions for A polarizer P is placed be- tween a crossed polarizer-analyzer pair. The optic of axis of P makes an angle a with respect to that of the analyzer. If the intensity of incident light is I_{0} , neglecting losses due to reflection and absortion at the polarizers, the intensity of the transmitted light is (a) I_{0} * cos^2 alpha (b) I_{0} * sin^2 2 * alpha (c) 1/8 * I_{0} * cos^2 2 * alpha (d) 1/8 * I_{0} * sin^2 2 * alpha? in English & in Hindi are available as part of our courses for Physics.
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