Central Maximum Shift Due to Glass Slit in Double Slit Experiment

In summary, the conversation discusses the effect of placing a very thin piece of glass in one slit of a double-slit experiment. By calculating the initial phase delay and using the equations for the distances traveled by the light waves, the central maximum is found to move by approximately 0.00113 meters. Another approach to solving this problem is using the equation λ/4 = d sin θ, which leads to a simpler solution.
  • #1
MathIsFun

Homework Statement



Light of wavelength 600 nm passes though two slits separated by 0.20 mm and is observed on a screen 1.5 m behind the slits. The location of the central maximum is marked on the screen and labeled y = 0.

A very thin piece of glass is then placed in one slit. Because light travels slower in glass than in air, the wave passing through the glass is delayed by 5.0 ×10−16s in comparison to the wave going through the other slit.

Part E

By how far does the central maximum move?

Homework Equations



[tex]\Delta \phi=2\pi m[/tex]

The Attempt at a Solution



I calculated that the initial phase delay due to the glass is [itex]\frac{\pi}{2}[/itex]. So in order to have constructive interference, the light wave that travels through the slit with no glass must travel an additional distance of [itex]\frac{\lambda}{4}[/itex]. I called the distance that the light through the glass travels [itex]x_{1}[/itex] and the distance that the light through the normal slit travels [itex]x_{2}[/itex]. I know that [tex]x_{1}=\sqrt{L^{2}+(\frac{d}{2}-h)^{2}}[/tex] and [tex]x_{2}=\sqrt{L^{2}+(\frac{d}{2}+h)^{2}}[/tex]So then I have [tex]\sqrt{L^{2}+(\frac{d}{2}+h)^{2}}-\sqrt{L^{2}+(\frac{d}{2}-h)^{2}}=\frac{\lambda}{4}[/tex]
I don't know how to solve this analytically, so I plugged it into Wolfram Alpha and got [itex]h\approx 0.00113[/itex] meters, which is correct, but I feel that there should be another way to solve for [itex]h[/itex] that doesn't require a computer algebra system.

Is there another way to approach this problem that makes finding [itex]h[/itex] simpler?

Thanks
 
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  • #2
I would expand each radical in Taylor series. You can write the radicals as ##L\sqrt{1+\epsilon_{\pm}^2}## where ##\epsilon_{\pm}=\frac{d/2 \pm h}{L}## and expand about zero for small values of ##\epsilon_{\pm}##.

On edit: I just did it and the answer is very simple.
 
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  • #4
Yes, I can see how that works now.

Thank you for your help.
 

What is the double slit equation?

The double slit equation, also known as the Young's double slit equation, is a mathematical formula that describes the interference pattern created by light passing through two parallel slits.

How is the double slit equation solved?

The double slit equation can be solved by using the following formula: d*sinθ = m*λ, where d is the distance between the two slits, θ is the angle of the incident light, m is the order of the interference pattern, and λ is the wavelength of the light.

What is the significance of solving the double slit equation?

Solving the double slit equation allows scientists to understand and predict the behavior of light as it passes through two parallel slits. It also helps in studying the properties of light and its wave-like nature.

Can the double slit equation be applied to other types of waves?

Yes, the double slit equation can be applied to any type of wave, including sound waves and water waves. It describes the interference pattern that occurs when waves pass through two parallel slits.

What are some real-world applications of the double slit equation?

The double slit equation has various applications in fields such as optics, acoustics, and fluid dynamics. It is used in designing diffraction gratings for optical instruments, studying the acoustic properties of musical instruments, and understanding the behavior of water waves in oceans and lakes.

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