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A pair of closely spaced slits is illuminated with 650.0-nm light in a Young's double-slit experiment. During the experiment, one of the two slits is covered by an ultrathin Lucite plate with index of refraction n = 1.485. What is the minimum thickness of the Lucite plate that produces a dark fringe at the center of the viewing screen?

User Phil Kulak
by
7.9k points

2 Answers

3 votes

Answer:

Thickness = 670.10 nm

Explanation:

Given that:

The refractive index of ultrathin Lucite plate = 1.485

The wavelength of the light = 650 nm

The minimum thickness that produces a dark fringe at the center can be calculated by using the formula shown below as:


Thickness=\frac {\lambda}{2* (n-1)}

Where, n is the refractive index of ultrathin Lucite plate = 1.485


{\lambda} is the wavelength

So, thickness is:


Thickness=\frac {650\ nm}{2* (1.485-1)}

Thickness = 670.10 nm

User Andha
by
8.8k points
0 votes

Answer:

The minimum thickness of the Lucite plate is 0.670 μm.

Step-by-step explanation:

Given that,

Wavelength = 650.0 nm

Index of refraction = 1.485

We need to calculate the minimum thickness of the Lucite plate

Using formula of thickness


t(n-1)=(\lambda)/(2)


t=(\lambda)/(2(n-1))

Where, n = Index of refraction


\lambda = wavelength

Put the value into the formula


t=(650.0*10^(-9))/(2(1.485-1))


t =0.670\ \mu m

Hence, The minimum thickness of the Lucite plate is 0.670 μm.

User Alex Lowe
by
7.4k points
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