Question

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?

Answer 1

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.

Answer 2

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:

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

Thickness = 670.10 nm