Question

Example 35.5.1 Sample Problem Michelson's interferometer Light of wavelength i = 589.0 nm is used in a Michelson's interferometer. First, if mirror Mis moved in either direction) by distance d=2.000 jam, we want to determine the phase shift. Second, if a glass sheet with index of refraction - 1.500 is inserted into the path of M (perpendicular to the path) and the pattem shifts by 5.10 fringes, we want to determine the thickness L of the sheet.

will give 2 thumbs up

Answer 1

1. Since wavelength of light in a medium of refractive index n decrease by factor n

So , wavelength in glass = wavelenth in air / n

As light travels 2 times through glass sheet , so total distance moved = 2L

No of wavelenth in glass ( Nglass) = distance travelled /wavelenth in medium

Nglass = 2L/ wavelength in glass

Nglass = 2Ln/wavelength in air

factor in reduces by a When a light of wavelength N. (in alr) . passes through a medium of refractive index to h then the wavelength room In the medium . medium = dair wavelength n glass [Because, speed of light reduces by factor t Now, in this experiment, due to insertion of slab, of length į Light passes through the glass sheet 2 times. So 2L path is covered by light . - Distance trandled in medium No of wavelength = est Wavelength in medium (Ngloss) - 2L .- 210 = Tair. 2Ln 2L = tail if 2L distance travelled was in air instead of glass stad, sheet, then Number of wavelength = Distance travelled in air - Wandlayth in air O Nale = 2 anos ② N glass - Naik = 24 [n-l]. 7 I fringe sl fringes corresponds to corresponds to I wandength sil wavelength.

© Nglans - Naie = sil $.1= 22 [n--] L = sil daie dair L = 5.1 x 589 2[1.5-1] L = 3003.9 mm [L = 30035 tim L = 3.039 um.