Double Slit Problem: Part d-i Homework Statement Solutions

[unscientific]

Homework Statement

Part(d): Find an expression for y

Part (e): Sketch the interference pattern on screen and locate maximas

Part (f): One slit is covered by a film that reduces amplitude by half. Explain how the interference pattern changes.

Part (g): The film is now replaced by 0.5mm glass. Explain how the interference pattern changes.

Part (h): Explain what happens if source is moved upwards by distance x.

Part (i) Explain what happens if a second independent source is placed distance x above from the first. What will be the minimum value of x at which fringes disappear? Estimate biggest possible inchoherent source in order to see fringes.

Homework Equations

The Attempt at a Solution

I'm not sure if the lens would affect the system, and what's the point of telling us its focal length?

Part (d)

sin(\theta) \approx tan (\theta)
\frac{\lambda}{d} \approx \frac{y}{D}
y = \frac{\lambda D}{d}

Part(e)

For small angles, $sin \theta \approx \theta$:

Maxima: $\theta_{max} = \frac{n\lambda}{d}$

Part(f)

Using Method of Phasors:

Phase difference $\delta = kd sin \theta$.

u_r^2 = (\frac{u_0}{r})^2 + (\frac{u_0}{2r})^2 + 2 (\frac{u_0}{r})^2\frac{1}{2}cos (\delta)
I \propto u_r^2 = I_0\left( \frac{5}{4} + cos (kd sin(\theta))\right)

Maximum Intensity:
kd sin \theta = 2n\pi
\theta_{max} = sin^{-1}(\frac{n\lambda}{d})
\theta_{max} \approx (\frac{n\lambda}{d})

Minimum Intensity:
kd sin \theta = (\frac{2n+1}{2})\pi
\theta_{min} = sin^{-1}\left( (\frac{2n+1}{4d}\lambda \right )
\theta_{min} \approx \left( (\frac{2n+1}{4d}\lambda \right )

Incomplete destructive interference so no regions of zero intensity and regions of maximum intensity now lower than before.

Part(g)

Within the block, wavelength is shorter. Therefore within the block more wavelengths could be squeezed in compared to the same length in air.

Path difference in terms of wavelength:

n_{\lambda} = L(\frac{1}{\lambda'} - \frac{1}{\lambda_0})

Therefore the pattern shifts by $n_{\lambda} \lambda_0$ distance away from the block.

Part(h)

I'm guessing that the pattern moves in the opposite direction?

Part (i)
This remains a complete mystery.

 

[mfb]

The lens influences the directions of the rays, or, as alternative description, the lengths of the light pathes. To see how it works, I think it is easier to work backwards. Start with a specific point at the screen, see how light propagates and which phase difference you get at the slits.

 

[unscientific]

The lens influences the directions of the rays, or, as alternative description, the lengths of the light pathes. To see how it works, I think it is easier to work backwards. Start with a specific point at the screen, see how light propagates and which phase difference you get at the slits.

Can I assume the rays approaching the lens are parallel (i.e. from infinity)?

then the path difference would be $d sin\theta$

 

[mfb]

Can I assume the rays approaching the lens are parallel (i.e. from infinity)?

They come from the slits.

then the path difference would be $d sin\theta$

For proper d, θ and a description where this path difference is supposed to exist.

 

[unscientific]

They come from the slits.

For proper d, θ and a description where this path difference is supposed to exist.

From the lens equation,
1/u + 1/v = 1/f

Treating u = ∞ since we can assume light rays entering are parallel, v = f. Thus the rays will be focused to a distance equal to it's focal length.

Now for part (i)

I have a strong feeling this has to do with additional phase difference before the light enters the slits.

Consider the light entering the slit at angle $\phi$ and leaving at angle $\theta$. I'm supposed to show that the maxima fringe condition is given by:

d(sin \theta + sin\phi) = n\lambda

The fringes disappear when $d(sin \theta + sin\phi) = (n+1/2) \lambda$ for some $\theta$.

This means that originally at some angle $\theta$ where it is maximum, it implies that:

d(sin \theta + sin\phi) = (n+1/2) \lambda

where $d sin \theta = n\lambda$

Thus, subtracting from above:

d sin \phi = 1/2 \lambda

We get an expression for the first minimum angle $\phi$ that makes the fringes disappear. Other angles of $\phi$ that makes the fringes disappear are of $d sin \phi = \frac{2k+1}{2} \lambda$

Does that reasoning sound correct?

 

[unscientific]

bumpp

 

[unscientific]

bumpp

 

[unscientific]

bump - Any input on parts (h) and (i) ?

 

[unscientific]

bumpp

 

[unscientific]

Is there nobody out there that can solve parts (h) and (i)?

 

[unscientific]

bumpp parts (h) and (i) anyone?

 

[TSny]

I think (h) and (i) are related. I believe your qualitative answer to (h) is correct.

To make it more quantitative, you should consider a single source displaced upward by x and derive an expression for the path difference from the source to the two slits as a function of x. It should be OK to make the approximation x << s to help simplify the expression. You can then use the result to get the phase difference of the light at the two slits. Think about what the interference pattern looks like if the phase difference at the slits is $\small \pi$. Then, if you add a second incoherent source at x = 0, what would the interference pattern look like from both sources?