Double-Slit Experiment with Light

You have a double slit setup with a screen that is only \( 15 \, \text{cm} \) wide. The double slit and the screen are attached to each other at a distance of \( 3 \, \text{m} \), and the slit spacing is \( 0.15 \, \text{mm} \). On the screen, you want to create an interference pattern, and you want to see at least \( 15 \) bright stripes there!

What wavelength \( \lambda \) do you need to use for this?

Solution to Exercise #1

From the condition for interference maxima: $$\Delta s ~=~ m \, \lambda $$ and the sketch of the double slit (with the approximation that the screen is far away from the double slit): $$ \frac{x}{a} ~=~ \frac{\Delta s}{g}$$ you get the wavelength: $$\lambda ~=~ \frac{ x \, g }{ a \, m }$$

Substituting the given values yields: \[ \lambda ~=~ \frac{ 0.15\cdot10^{-3}\text{m} ~\cdot~ 0.15\text{m} }{ 3\text{m} ~\cdot~ 15 } ~=~ 5\cdot10^{-7} \, \text{m} \]

So, you need to use light with at least \( 500 \, \text{nm} \) wavelength (red light) to generate at least 15 stripes on a \( 15 \, \text{cm} \) wide screen (at a distance of \(3 \, \text{m} \) from the double slit).

Monochromatic light strikes the double slit in parallel, with both slits spaced \( g ~=~ 0.8 \, \mathrm{mm} \) apart. You observe interference stripes on the screen, which is \( a ~=~ 5 \, \mathrm{m} \) away from the double slit. Two bright fringes are \( x ~=~ 3.5 \, \mathrm{mm} \) apart.

What wavelength \( \lambda \) does the light have?

Solution to Exercise #2

Since the monochromatic light strikes the double slit in parallel, the light waves arrive at both slits in phase. Because the waves are in phase, they satisfy the condition for constructive interference as follows: $$\Delta s ~=~ m \, \lambda$$

Here, \( \Delta s \) is the path difference. We have taken the condition for constructive and not destructive interference because the task specifies the distance between two bright rather than dark fringes.

Additionally, the distance from the slit to the screen \( a ~=~ 3 \, \text{m} \) is much larger than the slit spacing \( g ~=~ 0.3 \, \text{mm} \). This means you can use the following approximation: $$\tan(\phi) ~\approx~ \sin(\phi) ~=~ \frac{x}{a}$$

From the right triangle, where the opposite side is the path difference \( \Delta s \), you gain the information: $$\sin(\phi) ~=~ \frac{\Delta s}{g}$$

Set this equation equal to the above approximation: $$\frac{x}{a} ~=~ \frac{\Delta s}{g}$$

Also, use the condition for constructive interference for the path difference: $$\frac{x}{a} ~=~ \frac{m \, \lambda}{g}$$

Now, you have derived a relationship that contains only quantities given in the exercise. Just rearrange the last equation to solve for the desired wavelength: $$\lambda ~=~ \frac{x \, g}{m \, a}$$

Assuming that the bright fringes always have the same spacing and nothing further is mentioned in the exercise, you can consider the distance from the 0th to the 1st maximum: \( m ~=~ 1 \). Substituting the given values yields: $$\begin{align}\lambda ~&=~ \frac{ 3.5 \cdot 10^{-3}\text{m} ~\cdot~ 0.8 \cdot 10^{-3}\text{m} }{ 1 ~\cdot~ 5\text{m} } \\ ~&=~ 5.6 \cdot 10^{-7} \, \text{m}\end{align}$$

This path difference is equivalent to a wavelength of \( 560 \text{nm} \), corresponding to the typical wavelength of green light. $$\lambda ~=~ \frac{ 3.5 \cdot 10^{-3}\text{m} ~\cdot~ 0.8 \cdot 10^{-3}\text{m} }{ 1 ~\cdot~ 5\text{m} } ~=~ 5.6 \cdot 10^{-7} \, \text{m}$$

A double slit is illuminated with red light of wavelength \( \lambda ~=~ 650 \, \text{nm} \) in parallel. At a distance \( a ~=~ 3 \, \text{m} \) from the double slit, there is a screen on which a sharp interference pattern is visible. Since it is not easy to determine the distance from the main maximum to the 1st maximum, instead, the distance from the 5th minimum to the opposite 5th minimum is measured, which is \( \Delta x ~=~ 6 \, \text{cm} \).

What is the spacing \( g \) between the two slits through which the light passed?

Solution to Exercise #3

Since the red light in parallel strikes the double slit, the light waves arrive at both slits in phase. And, because the waves are in phase, the condition for destructive interference applies as follows: \[ \Delta s ~=~ \left( m ~-~ \frac{1}{2} \right) \, \lambda \]

Here, \( \Delta s \) is the path difference, and \( m ~=~ 1,2,3... \) indicates the order of the minima. We have taken the condition for destructive and not constructive interference because the task specifies the distance between two minima. Minima are the points on the screen that are dark. The light waves have canceled out at these points.

Regarding the slit spacing: It is unknown. But what you can say about it just by looking at it is that it is very small... (I calculated it, it IS small *cough*). The distance from the slit to the screen \( a ~=~ 3 \, \text{m} \) is much larger than the unknown slit spacing \( g \). This means you can use the following approximation: \[ \tan(\phi) ~\approx~ \sin(\phi) ~=~ \frac{x}{a} \]

The position \( x \) on the screen (measured from the center) is only indirectly known. The distance between the 5th minima was measured. Since the interference pattern is symmetrical, the distance from the main maximum to the 5th minimum is just half of the measured value. This is also the sought position \( x \) on the screen: \( x ~=~ \frac{\Delta x}{2} \). Substitute this position into the above equation: \[ \sin(\phi) ~=~ \frac{\Delta x}{2a} \]

From the right triangle, where the opposite side is the path difference \( \Delta s \), you can read off: \[ \sin(\phi) ~=~ \frac{\Delta s}{g} \]

Set these last two formulas equal: \[ \frac{\Delta x}{2a} ~=~ \frac{\Delta s}{g} \]

Since you want to look at the minima, also set the condition for destructive interference: \[ \frac{x}{a} ~=~ \frac{ \left( m ~-~ \frac{1}{2} \right) \, \lambda }{g} \]

Now you have derived a relationship that contains only quantities given in the task. Just rearrange it with respect to the desired slit spacing \( g \): \[ g ~=~ \frac{ 2a \, \left( m ~-~ \frac{1}{2} \right) \, \lambda }{ \Delta x } \]

Substituting the given values yields: \[ g ~=~ \frac{ 2 \cdot 3\text{m} ~\cdot~ \left( 5 ~-~ \frac{1}{2} \right) ~\cdot~ 650 \cdot 10^{-9}\text{m} }{ 0.06\text{m} } ~=~ 2.925 \cdot 10^{-4} \, \text{m} \]

This corresponds to a slit spacing of approximately \( 0.3 \text{mm} \), which is hardly measurable with a ruler... but fortunately, it works with the double slit experiment!