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Cloud [144]
3 years ago
14

A pair of closely spaced slits is illuminated with 650.0-nm light in a Young's double-slit experiment. During the experiment, on

e 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?
Physics
2 answers:
Masteriza [31]3 years ago
7 0

Answer:

The minimum thickness of the Lucite plate is 0.670 μm.

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)=\dfrac{\lambda}{2}

t=\dfrac{\lambda}{2(n-1)}

Where, n = Index of refraction

\lambda = wavelength

Put the value into the formula

t=\dfrac{650.0\times10^{-9}}{2(1.485-1)}

t =0.670\ \mu m

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

nata0808 [166]3 years ago
5 0

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\times (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\times (1.485-1)}

<u>Thickness = 670.10 nm</u>

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