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.

Interference laboratory

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.


valentin68