Answer:
3) Transmitted intensity of light if unpolarized light passes through a single polarizing filter = 40 W/m²
- Transmitted intensity of light if an additional polarizer is added perpendicular to the first polarizer in the setup described = 7.5 W/m²
Explanation:
Complete Question
3) What is the transmitted intensity of light if unpolarized light passes through a single polarizing filter and the initial intensity is 80 W/m²?
- What is the transmitted intensity of light if an additional polarizer is added perpendicular to the first polarizer in the setup described in Question 3 (the setup)? Show all work in your answer.
The image of this setup attached to this question as obtained from online is attached to this solution.
Solution
3) When unpolarized light passes through a single polarizer, the intensity of the light is cut in half.
Hence, if the initial intensity of unpolarized light is I₀ = 80 W/m²
The intensity of the light rays thay pass through the first single polarizer = I₁ = (I₀/2) = (80/2) = 40 W/m²
- According to Malus' law, the intensity of transmitted light through a polarizer is related to the intensity of the incident light and the angle at which the polarizer is placed with respect to the major axis of the polarizer before the current polarizer of concern.
I₂ = I₁ cos² θ
where
I₂ = intensity of light that passes through the second polarizer = ?
I₁ = Intensity of light from the first polarizer that is incident upon the second polarizer = 40 W/m²
θ = angle between the major axis of the first and second polarizer = 30°
I₂ = 40 (cos² 30°) = 40 (0.8660)² = 30 W/m²
In the same vein, the intensity of light that passes through the third/additional polarizer is related to the intensity of light that passes through the second polarizer and is incident upon this third/additional polarizer through
I₃ = I₂ cos² θ
I₃ = intensity of light that passes through the third/additional polarizer = ?
I₂ = Intensity of light from the second polarizer that is incident upon the third/additional polarizer = 30 W/m²
θ = angle between the major axis of the second and third/additional polarizer = 60° (although, it is 90° with respect to the first polarizer, it is the angle it makes with the major axis of the second polarizer, 60°, that matters)
I₃ = 30 (cos² 60°) = 30 (0.5)² = 7.5 W/m²
Hope this Helps!!!
Answer:
1.
2.
Explanation:
Polarizes axis can create two possible angles with the vertical.
first we have to find the intensity of first polarizer
which is given as
For a smaller angle for the first polarizer:
According to Malus Law
taking square root on both sides
For a larger angle for the first polarizer:
According to Malus Law
taking square root on both sides
Explanation:
(a) For an isothermal process, work done is represented as follows.
W =
Putting the given values into the above formula as follows.
W =
=
=
=
= 29596.78 J
or, = 29.596 kJ (as 1 kJ = 1000 J)
Therefore, the required work is 29.596 kJ.
(b) For an adiabatic process, work done is as follows.
W =
=
=
= 49.41 kJ
Therefore, work required to produce the same compression in an adiabatic process is 49.41 kJ.
(c) We know that for an isothermal process,
or,
=
= 11 atm
Hence, the required pressure is 11 atm.
(d) For adiabatic process,
or,
=
= 28.7 atm
Therefore, required pressure is 28.7 atm.