In your research lab, a very thin, flat piece of glass with refractive index 1.5
ID: 1353538 • Letter: I
Question
In your research lab, a very thin, flat piece of glass with refractive index 1.50 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 lambda 0 in vacuum at normal incidence onto the surface of the glass. When lambda 0 = 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. Use these measurements to calculate the thickness of the glass. Express your answer with the appropriate units. What is the longest wavelength in vacuum for which there is constructive interference for the reflected light?Explanation / Answer
Refractive index of Glass = 1.5
For Constructive interference,
2nt = (m + 1/2)
In 1st Case,
2nt = (m + 1/2) 496 nm
For 2nd Case,
2nt = (m +1+ 1/2) 386 nm
2nt = (m+3/2) * 386 nm
(m + 1/2) 496 nm = (m+3/2) * 386 nm
m = 3
Putting value of m in 1.
2nt = (m + 1/2) 496 nm
2*1.5t = (3 + 1/2) * 496 nm
t = ((3 + 1/2) * 496 nm)/ 3
t = 578.6 nm
Thickness of the glass, t = 578.6 nm
b)
We know, For constructive interfernce
2nt = (m + 1/2)
= 2nt / ((m + 1/2))
For Longest Wavelength, m = 0
= 2*1.5*578.6/ (1/2)
= 3472 nm