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Serhud [2]
2 years ago
15

Plane-polarized light is incident on a single polarizing disk with the direction of E0 parallel to the direction of the transmis

sion axis. Through what angle should the disk be rotated so that the intensity in the transmitted beam is reduced by a factor of:
Physics
1 answer:
DENIUS [597]2 years ago
6 0

Answer: Plane-polarized light is incident on a single polarizing disk with the direction of E0 parallel to the direction of the transmission axis. Through angle \alpha=cos^-1(\sqrt{\frac{1}{n} }) should the disk be rotated so that the intensity in the transmitted beam is reduced by a factor of n .

Explanation: To find the correct answer, we need more clarifications about the Malu's law.

<h3>What is Malu's law?</h3>
  • When a beam of plane polarized light is passed through an analyzer, the intensity I of the transmitted light varies directly as the square of the angle between the direction of polarizer and the analyzer.

                                 I=I_0cos^2\alpha

where, I_0 is the maximum intensity of transmitted light.

  • Intensity of the light transmitted through the polarizer is half of the incident intensity.
<h3>How to solve the problem?</h3>
  • Here in the question ,let's take the factor at which the intensity of transmitted light is reduced by as n.
  • we have the expression

                                            I=I_0cos^2\alpha

  • we have to find the value of angle alpha, so that the intensity in the transmitted beam is reduced by a factor of n.

                                 \alpha =cos^-1\sqrt{(\frac{I}{I_0}} )=cos^-1\sqrt{\frac{I}{nI} } \\\alpha =cos^-1\sqrt{\frac{1}{n} }

  • If we know the value of n, then we can easily find the angle.

Thus, we can conclude that, the value of angle alpha, so that the intensity in the transmitted beam is reduced by a factor of n will be,

                                 \alpha=cos^-1(\sqrt{\frac{1}{n} }) .

Learn more about Malu's law here: brainly.com/question/28045350

#SPJ4

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Answer:

Vo = 4.5 [m/s]

Explanation:

In order to solve this problem, we must use the following equation of kinematics.

v_{f}=v_{o}+a*t

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Now replacing:

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Horizontal component of force is given by

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Let \alpha be the angular acceleration of the end of the rod

Taking torque about the link, we have:

\tau = W \times OM\\ or\\ \tau = mg\times \left(\dfrac{L}{2}\right)\sin 55^\circ ....(i)

Torque is also given in terms of moment of inertia of the rod and its angular acceleration i.e.

\tau = I_{rod}\ \alpha......(ii)

From equations (i) and (ii) we have:

mg\times \left(\frac{L}{2}\right)\sin 55^\circ = \left(\frac{mL^2}{3}\right)\alpha\ \ \ \ \ \ \ (\because I_{rod} = \dfrac{mL^2}{3}) \\ \\ \alpha = \left(\frac{3g}{2L}\right)\sin 55^\circ\\ \\ \alpha = \left(\frac{3\times 9.8}{2\times 2.4}\right)\sin 55^\circ\\ \\ \boxed{\alpha = 5.02\ rad/s^2} }

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7 0
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The ratio of the drag forces is \mathbf{\frac{F_1}{F_2}}=\mathbf{3.55}

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