![]() ![]() On the axes below, sketch a graph of I vs. (1) At which value(s) of theta is I at a maximum? (2) What is I_max, the maximum value of I, in terms of I_0? Explain. (3) Explain conceptually why it makes sense for I to be at a minimum at these value(s) of theta. (1) At which value(s) of theta is I at a minimum? (2) What is the minimum value of I? Explain. Use this fact to find all theta (where 0 lessthanorequalto theta lessthanorequalto 90 degree) at which I is at a maximum or a minimum. (As always, simplify.) Maximum and minimum values of I occur where dI/d theta = 0. Find an equation for the intensity I of the light after it passes through filter 3, in terms of only I_0 and theta. Finally, the light passes through filter 3, which has its polarization axis at 90 degree relative to the axis of filter 1. ![]() which has its polarization axis at an angle theta relative to the axis of filter 1, where 0 lessthanorequalto theta lessthanorequalto 90 degree. So angle between them equals to 45 plus 30, which is 75 degrees.Incident unpolarized light of intensity I_0 passes first through polarizing filter 1. ![]() Now it is given that p 1 is rotated, 45 degree, clockwise and p. I 0 again falls on polarized v 1 and the output intensity will then be i 0 by 2, and this then falls on polarizer p. In the next part, we have an un polarized light of intensity. Solving this we get the output intensity equals to 3. The angle between them, that is, 30 degrees. That is, i 0 by 2 multiplied by cos square. P 2 is rotated 60 degree clockwise, as both the polarizers are rotated clockwise angle between them equals to 90 minus 60, which is equal to 30 degrees, so the output will be the incident intensity. It is given that polarize, p 1 is rotated 90 degree clockwise and polarizer. Now this intensity falls on the polarizer p 2. Then we know the output intensity will be. Now using this we will solve these parts of the question at first we are given that an un polarized light is incident on the polarizer. Polarized side, then the output is, i 0 cos square theta. I 0 falls on a plane, polarizer, making an angle theta with the plane. Point 1 When unpolarised light is incident on an ideal polariser, the intensity of the transmitted light is exactly half that of the incident unpolarised light, no matter how the polarising axis is oriented. Let us understand the expression of Malus law. I out, is i 0 by 2, when a polarized light, plain polarized light of intensity. The law helps us quantitatively verify the nature of polarised light. Hello students, when an earn, polarized light of intensity, i 0 falls on a plane polarizer. Then, change the angle of the second polarizer to… View More The intensity after the first polarizer is 'I0'. Discuss the polarization after each component for both cases. If the first quarter wave plate makes an angle of 45 degrees with the first polarizer and the second quarter wave plate makes an angle of zero with the first polarizer, then change its angle from zero to 90 degrees. For both cases, explain the type and direction of polarization after each component. If the two quarter wave plates make zero and 90-degree angles with the linear polarizer successively. The two quarter wave plates make a 45-degree angle with the first linear polarizer axis. Draw a figure of the above setup and show the type and direction of light after each optical component. The unpolarized light is passed through a linear polarizer followed by two quarter wave plates (with aligned axes) and followed by a second linear polarizer. If that polarizer is now rotated 90° from its original position, what intensity is now emitted? Plane-polarized light of intensity I is incident on a polarizer that produces an intensity of 0.75 I. If the first one is rotated clockwise to 45° of its original position and the second one rotated anti-clockwise to 308° of its original position, what is the intensity of the final emerging light? The intensity of unpolarized light incident on the second polarizer in line with the first has an axis that is parallel with respect to the first polarizer. ![]()
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