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3 years ago

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Vertically polarized light with an intensity of 36.8 lux passes through a polarizer whose transmission axis is an angle of 51.0°

with the vertical. What is the intensity and direction of transmitted light? If the second polarizer whose transmission axis is an angle of 88.0° with the vertical is placed after first polarizer, what is the intensity of transmitted light?

1 answer:

Annette [7]3 years ago

5 0

Answer:

0.01774 lux

Explanation:

$I_0$ = Polarized light intensity = 36.8 lux

$\theta$ = Angle

Through first filter

$I_1=I_0cos^2\theta\\\Rightarrow I_1=36.8cos^2{51}\\\Rightarrow I_1=14.57\ lux$

Through second filter

$I_2=I_1cos^2\theta\\\Rightarrow I_2=14.57cos^2{88}\\\Rightarrow I_2=0.01774\ lux$

intensity of transmitted light is 0.01774 lux

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Readme [11.4K]

Question 1.

mass = 4500 kg
potential energy (p.e) = 67500 J

now, we know :

=》

$p.e = mgh$

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$67500 = 4500 \times 10 \times h$

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$67500 = 45000 \times h$

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Question 2.

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$k.e = \dfrac{1}{2} mv {}^{2}$

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$63000 = \dfrac{1}{2} \times 4500 \times {v}^{2}$

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${v}^{2} = \dfrac{63000 \times 2}{4500}$

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2 years ago

A photoelectric effect experiment finds a stopping potential of 1.93 V when light of wavelength 200 nm is used to illuminate the

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The stopping potential (V) is the potential needed to stop the photoelectrons with maximum kinetic energy: so, the corresponding electric potential energy must be equal to the maximum kinetic energy,

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So we can rewrite (1) as

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Converting into electronvolts,

$E=\frac{9.95\cdot 10^{-19}J}{1.6\cdot 10^{-19} J/eV}=6.22 eV$

And now we can solve eq.(1) to find the work function of the metal:

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The intensity of the incident light is proportional to the number of photons hitting the surface of the metal. However, the energy of the photons depends only on their frequency, so it does not depend on the intensity of the light. This means that the term E in eq.(1) does not change.

Moreover, the work function of the metal is also constant, since it depends only on the properties of the material: so $\phi$ is also constant in the equation. As a result, the term (eV) must also be constant, and therefore V, the stopping potential, is constant as well.

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3 years ago