Question

Question 3: Consider the image below for the following problem statement. ODE Suppose the transmission axes of the left and right polarizing disks are perpendicular to each other. Also, let the center disk be rotated on the common axis with and angular speed w, this means 0 = wt. a) Show that if unpolarized light is incident on the left disk with an intensity of Imax, the intensity of the beam emerging from the right disk is: 16 Imax (1 – Cos(4wt)) You will find the following trigonometric identities helpful: Sin (0) Cos? (0) = 3 (1 + Cos(20) (1 - Cos(20)) Cos? (90° – ) = Sin (0) b) Sketch a plot of the emerging intensity as a function of time using w = 0.5 d, and Imax 16 c) Compare the frequency the disk rotates and the frequency of oscillation of you intensity plot. Explain how/why one is 4x the other.

I just need help with b. and c. Thanks!

Answer 1

b.

Using the function:

$I=\frac{I_{max}}{16}(1-\cos(4\omega t))$

Setting and $\omega=0.5\ rad/s$

Therefore, the function boils down to:

$I(t)=1-\cos(2t)$

The plot:

-11,5 -0.5 T1/3 2T1/3 TT 4T1/3 5Tt/3 2T o

To plot the function manually or sketching it note a couple of things:

The function is like 1-cos x. Therefore, everytime the cos function has a value of 1, the intensity becomes 0, whenever the cos function becomes -1, the intensity becomes 2. This is equivalent to shifting the entire cos function up the y axis such that there are no negative values, followed by shifting it to the right by half a period (pi) such that the function becomes 0 at $x=n\pi;\ n=0,1,2,...$

As for the period,

$1-\cos 2t=2\sin^2(2t/2)=2\sin^2(t)$

Looking at this function, the period of the wave is exactly $2\pi$ , which is the exact same period as a regular sinusoidal wave.

c.

Looking at the function:

$\cos(4\omega t)$

Now, setting $\Omega=4\omega$

Therefore,

$\cos(\Omega t)$ which now looks like a regular sinusoidal wave

Therefore, the angular frequency of oscillation:

$\Omega=2\pi / T\\ \implies \frac 1T= \frac{\Omega}{2\pi}$

where T is the time period

$f= \frac{\Omega}{2\pi}\\ \implies f=\frac{4\omega}{2\pi}$    f being the frequency

which is 4 times the frequency of the disc having angular frequency

$\omega,\ f=\frac{\omega}{2\pi}$