# Pendulum Lab. Interference

#### Pendulum Laboratory

The equations of pendulum are

$T= C * \sqrt{L/g}$    (1)

$T= 2\pi * \sqrt{L/g} * [1+(1/2)^2*\sin^2(x/2)+(3/4)^2*\sin^4(x/2)]$   (2)

1. For which angle is the period given by pendulum equation (2) diferrent with 1% from that given by the simplest equation (1)? You can consider just the first and second term in the sum of equation (2).

2. If the pendulum length is 1 meter, and you have a difference of 1% in $T$, what is the error in $L$?

1) 1% error of eq 2 reported wo eq 1 means that

$1+(1/2)^2*sin^2(x/2) +(3/4)^2*sin^4(x/2) = 1.01$

$1/4*sin^2(x/2)+9/16*sin^4(x/2) =0.01$

$1/4*y +9/16*y^2 =0.01$

$0.25*y +0.5625*y^2 =0.01$   or $y= 0.03693$

where we used the equation solver

$sin(x/2) =0.1922$   or $x/2=11.079$ degree  or $x =22.159$ degree

2) $T =C*\sqrt{L/g}$

$f(x) = C*\sqrt{x/L}$

$df = C*dx*(1/2)*1/\sqrt{x*g}$ where $C=2*\pi$

By differentiating to the left with respect to $T$ and to the right with resoect to $L$ we get obtain

$dT = C*dL *\sqrt{1/Lg}*(1/2)$

$dT/T = dL*(1/2)*\sqrt{1/Lg}*\sqrt{g/L} =(dL/L)*(1/2)$

1% error in $T$ means $dT/T =0.01$

$0.01 =(dL/L)*(1/2)$

$0.02 =dL/L$

for $L =1 m$ it means $dL =0.0200 m =2.00 cm$

Answer: The difference in $L$ is 2.00 cm for an error of $T$ of 1%.

#### Interference laboratory

Compute the intensity of light $I$ on a screen at a distance $L >>$a of two very long (infinite) slits of width $a$, a distance $d (>a)$ apart.

At one point on the screen we have the electric field intensity is the sum of the two intensities given by each slit.

Let the electric filed of the slit 1 be $E1 = E0*\sin(\omega*t)$ and the electric field of slit 2 be $E2 =E0*\sin(\omega t +\phi)$ where $\phi$ is the phase difference between the phase shift.

Total electric field is the sum :

$E = E1+E2 =E0*[\sin(\omega t) +\sin(\omega t+\phi)] = 2*E0*\cos(\phi/2)*sin(\omega t +\phi/2)$

because $sin(a) +sin(b) = 2*\sin((a+b)/2)*\sin((a-b)/2)$

The intensity of ligt is the time average of the square of electric field.

$I =<E^2> = 4E0^2*<\cos^2(\phi/2)>*<\sin^2(\omega t+\phi /2)> = 2*E0^2*\cos^2(\phi/2)$

since $<sin^2(\omega t+\phi/2)> =1/2$

Now phase $\phi$ as function of path difference ($\Delta$) can be written as

$\Delta/\lambda = \phi/(2*\pi)$

and

$\Delta =d*\sin(\theta)$

where $\theta$ is the angle between the horizontal and the ray of light emerging from one slit.

$tan(\theta) = D/L$

($D$ is the distance from central brightest frindge to the interference point)

Therefore

$I =I0*\cos^2(\pi*d*\sin(\theta)/\lambda)$

Observation: for $L>>a$ (slit opening width) it only counts the distance between slits $d$, not the slit opening width. See the figure below.