- #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
\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