Question #19902

In a Young's double slit experiment a laser of wavelength 600 nm is used to obtain a pattern with a distance between the center bright fringe and the first bright fring of 0.5 mm. A thin plate of glass (thickness of 100 micrometers and refractive index of 1.5) is then placed over one of the slits. What is the lateral displacement of the central fringe on the screen?

Expert's answer

Question: In a Young's double slit experiment a laser of wavelength 600nm600\mathrm{nm} is used to obtain a pattern with a distance between the center bright fringe and the first bright fring of 0.5mm0.5\mathrm{mm}. A thin plate of glass (thickness of 100 micrometers and refractive index of 1.5) is then placed over one of the slits. What is the lateral displacement of the central fringe on the screen?

Answer:

We are given:


λ=600 nm\lambda = 600\ \mathrm{nm}w1=0.5 mmw_1 = 0.5\ \mathrm{mm}Δp=100 μm\Delta p = 100\ \mu\mathrm{m}n1=1.5n_1 = 1.5


In a Young's double slit experiment the spacing of the fringes at a distance zz from the slits is given by


w=zλd, n=0,1,2, …w = \frac{z\lambda}{d},\ n = 0,1,2,\ \dots


where λ\lambda is the wavelength of the light.

Thus, for first case:


w1=0.5 mmw_1 = 0.5\ \mathrm{mm}


so:


d=zλw1d = \frac{z\lambda}{w_1}


In the second case optical path between slits and plane is increased:


p2=p−Δp+Δp∗n1=dθ+Δp(n1−1)p_2 = p - \Delta p + \Delta p * n_1 = d\theta + \Delta p(n_1 - 1)


The interference fringe maxima occur at angles


dsin⁡θ+Δp(n1−1)=nλ, n=0,1,2,3d \sin \theta + \Delta p(n_1 - 1) = n\lambda,\ n = 0,1,2,3


For small angles:


sin⁡θ=θ=tan⁡θ\sin \theta = \theta = \tan \theta


For central fringe n=0n = 0, thus:


θ=−Δp(n1−1)d\theta = - \frac{\Delta p(n_1 - 1)}{d}


And


tan⁡θ=yz≈θ\tan \theta = \frac{y}{z} \approx \theta


So:


y1z=−Δp(n1−1)d=−Δp(n1−1)zλw1\frac{y_1}{z} = - \frac{\Delta p(n_1 - 1)}{d} = - \frac{\Delta p(n_1 - 1)}{\frac{z\lambda}{w_1}}y1z=−Δp(n1−1)w1zλ\frac{y_1}{z} = -\frac{\Delta p(n_1 - 1) w_1}{z \lambda}y1=−Δp(n1−1)w1λy_1 = -\frac{\Delta p(n_1 - 1) w_1}{\lambda}


Calculating:


y1=−Δp(n1−1)w1λ=100∗10−6∗(1.5−1)∗0.5∗10−3600∗10−9=0.0416 m=4.16 cmy_1 = -\frac{\Delta p(n_1 - 1) w_1}{\lambda} = \frac{100 * 10^{-6} * (1.5 - 1) * 0.5 * 10^{-3}}{600 * 10^{-9}} = 0.0416 \, \text{m} = 4.16 \, \text{cm}


See: http://hyperphysics.phy-astr.gsu.edu/hbase/phyopt/slits.html

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