Question

ident wavefronts barrier 1.0pm. A light beam of wavelength 546 A two-slit barrier has a slit separation distance of dorig- nm passes through the slits. Do not use the small angle approximation for this problem. a. Calculate the angle for the first-order bright fringe. b Suppose the barrier was such that one of the slits oscillates vertically so that the slit separation varies with time according to d (t)-dorig + Asin(at). Calculate ?? for the first C. At what instant after t = 0 does 0,have its maximum value for the first time? what is this d. At what instant after (# 0 does ei have its minimum value for the first time? what is this e. Which of the parameters A,u, ?, dorie affect the values of these angles? order bright fringe as a function of time for A = 0.25um and a?100s-1 maximum value. minimum value?

Answer 1

SOLUTION

Part (a)

According to the following expression:

$d\sin(\theta)=m\lambda\: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: (1)$

Isolating $\theta$ from equation (1) we have:

$\theta=\sin^{-1}\left ( \frac{m \lambda}{d} \right )\: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: (2)$

Now, for the first-order bright fringe we have:

$m=1$

Therefore equation (1) can be written as:

$\theta_{1}=\sin^{-1}\left ( \frac{ \lambda}{d} \right )\: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: (3)$

Replacing values:

$\theta_{1}=\sin^{-1}\left ( \frac{ 546\times 10^{-9}\,m}{1.0\times 10^{-6}\,m} \right )\: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: (4)$

Solving we obtain:

${\color{Blue} \theta_{1}=33.1^{\circ}}$

Part (b)

The new slit separation is:

$d(t)=d_{orig}+A\sin(\omega t)\: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: (5)$

Inserting (5) into (3) we have:

$\theta_{1}=\sin^{-1}\left [ \frac{ \lambda}{d_{orig}+A\sin(\omega t)} \right ]\: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: (6)$

replacing values we obtain:

$\theta_{1}=\sin^{-1}\left [ \frac{ 546 \times 10^{-9}\,m}{1.0\times 10^{-6}\,m+\left (0.25\times 10^{-6}\,m \right )\sin(100 t)} \right ]\: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: (7)$

or

${\color{Blue} \theta_{1}=\sin^{-1}\left [ \frac{ 546 \times 10^{-3}}{1.0+0.25\sin(100 t)} \right ]}\: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: (8)$

Part (c)

The maximum value for $\theta_{1}$ can be obtained when:

$\sin(100.t)=0\, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, (9)$

which is satisfied for:

$t=\frac{\pi}{100}$

replacing (9) into (8) we obtain:

${\color{Blue} \theta_{1,max}=33.1^{\circ}}\: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: (10)$

Part (d)

The minimum value for $\theta_{1}$ can be obtained when:

$\sin(100.t)=1\, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, (11)$

which is satisfied for

$t=\frac{\pi}{200}$

Inserting (11) into (8) we obtain:

${\color{Blue} \theta_{1,min}=25.9^{\circ}}\: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: (12)$

Part (e)

The main parameter of , $\omega$ , $\lambda$ , $d_{orig}$ that affects the value of these angles is $d_{orig}$ .