In your research lab, a very thin, flat piece of glass with refractive index 1.30 and uniform thickness covers the opening of a chamber that holds a gas sample. The refractive indexes of the gases on either side of the glass are very close to unity. To determine the thickness of the glass, you shine coherent light of wavelength λ0 in vacuum at normal incidence onto the surface of the glass. Whenλ0= 496 nm, constructive interference occurs for light that is reflected at the two surfaces of the glass. You find that the next shorter wavelength in vacuum for which there is constructive interference is 386 nm.

1) Use these measurements to calculate the thickness of the glass.

2) What is the longest wavelength in vacuum for which there is constructive interference for the reflected light?

A) The thickness of the glass is 868 nm

B) The wavelength is 3472 nm

Explanation:

We are given;

Refractive index = 1.30

Wavelength = 496 nm

Next wavelength = 386 nm

Now, we need to calculate the thickness of the glass

Using formula for constructive interference which is given as;

2nt = (m + ½)λ

Where;

m is the order of the interference

λ is wavelength

t is thickness

Now, for the first wavelength, we have;

2nt = (m + ½)496 - - - - eq1

for the second wavelength, we have;

2nt = (m + 1 + ½)386

2nt = (m + 3/2)386 - - - - eq2

Thus, combining eq1 and eq2, we have;

(m + ½)496 = (m + 3/2)386

496m + 248 = 386m + 579

496m - 386m = 579 - 248

110m = 331

m = 331/110

m = 3

Put 3 for m in eq 1;

2nt = (3 + ½)496

2nt = 1736

t = 1736/(2 x 1)

t = 868 nm

B) now we need to calculate the longest wavelength.

From earlier, we saw that ;

2nt = (m + ½)λ

Making wavelength λ the subject, we have;

λ = 2nt/(m + ½)

The longest wavelength will be at m = 0

Thus,

λ = (2 x 1 x 868)/(0 + ½)

λ = 3472 nm

r=rho l/a where a is the area, l is the length and rho is resistivity so the greater the area the lesser the resistance thats why the earth pin is thicker bcos current follow the shortest path, so less resistive .