Central Maximum Shift Due to Glass Slit in Double Slit Experiment

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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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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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Yes, I can see how that works now.

Thank you for your help.