Central Maximum Shift Due to Glass Slit in Double Slit Experiment

[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

\Delta \phi=2\pi m

The Attempt at a Solution

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

Is there another way to approach this problem that makes finding h simpler?

Thanks

 

[kuruman]

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.

 

[J Hann]

A simpler way is
λ/4 = d sin theta
(Why does that work?)

Reference https://www.physicsforums.com/threads/double-slit-equation-solving.929825/

 

[MathIsFun]

Yes, I can see how that works now.

Thank you for your help.