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bcjochim07
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Homework Statement
A two slit experiment uses a laser with a wavelength of 633 nm. A very thin piece of glass with an index of refraction of 1.50 is placed over one of the slits. Afterward, the central point on the screen is occupied by what had been the m=10 dark fringe. What is the thickness of the glass?
Homework Equations
The Attempt at a Solution
The intial position of the m=10 dark fringe is (10 + .5)*(6.33e-7)*L/d, where L is the distance to the screen and d is the distance between the slits.
The central maximum must move up the same distance, so that its new position is at
y=(10.5)(6.33e-7)*L/d
Then I used Ltan(theta)=(10.5)(6.33e-7)L/d tan(theta) = (10.5)(6.33e-7)/d With the small angle approximation theta is about equal to (6.65e-6)/d
Then thinking about the delay between the waves going through the covered and uncovered slits. n=c/v so 1.5=(3.0e8)/v where v is the velocity through glass. The time it takes the wave to get through the glass is t=b/v, where b is thickness of the glass
The distance that the waves going the uncovered slit travels while the wave is still traveling through the glass is given by (b/v)*c.
In order to produce the constructive interference, the waves must get back in phase with each other, so the value of dsin(theta) should equal (b/v)*c. Sin(theta is about equal to theta, so substituting:
(b/v)*c=d*(6.65e-6)/d
1.5b= 6.65e-6
b= 4.43e-6 m = 4.43 micrometers
What am I doing wrong here?