Answer: 2.068*m
Explanation: According to work energy-theorem , the workdone in accelerating the electron equals the energy it would give off in terms of light.
workdone= qV
energy = hc/λ
q=magnitude of an electronic charge= 1.602*
h= planck constant = 6.626*
c= speed of light =2.998*
v= potential difference= 6*
λ= wavelength=unknown
by making λ subject of formulae we have that
λ=
λ = 6.626* * 2.998*
/ 1.602*
* 6*
λ =
by doing the necessary calculations, we have that
λ = 2.068*m
Answer:
0.010 m
Explanation:
So the equation for a pendulum period is: where L is the length of the pendulum. In this case I'll use the approximation of pi as 3.14, and g=9.8 m\s. So given that it oscillates once every 1.99 seconds. you have the equation:
Evaluate the multiplication in front
Divide both sides by 6.28
Square both sides
Multiply both sides by m/s^2 (the s^2 will cancel out)
Now now let's find the length when it's two seconds
Divide both sides by 6.28
Square both sides
Multiply both sides by 9.8 m/s^2 (s^2 will cancel out)
So to find the difference you simply subtract
0.984 - 0.994 = 0.010 m
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!!!
