Lloyds mirror experiment

In Lloyd's mirror experiment (Fig.) a light wave emitted directly be the source S (narrow slit) interferes with the wave reflected form a mirror M . As s result, an interference fringe pattern is formed on the screen Sc . the source and the mirror are separated by a distance l = 100 cm .At a certain position of the source the fringe width on the screen was equal to Deltax = 0.25 mm , and after source was moved away from the mirror plane by Delta h = 0.60 mm , the fringe width decreased eta = 1.5 times. Find the wavelength of light.

When a light wave passes from a rarer medium to a denser medium, there will be a phase change of pi radians. This difference brings change in the conditions for constructive and destructive interference. This phenomena also reasons the fromation of interference pattern in thin films like, oily layer, soap film, etc., but has no reason on the shifting of fringes from the central portion outward. The shift is dependent on the refractive index of the material as per the relation, Delta y = (mu - 1) t The condition for constructive interference in Lloyd's single mirror experiment is the path difference which is equal to

The arrangement for a mirror experiment is shown in figure. S is a point source of frequency 6xx10^(14)Hz . D and C represent the two ends of a mirror placed horizontally and LOM represents the screen. Calculate the number of fringes

A Lloyd's mirror or length 5 cm is illuminated with mon ochromatic light of wavelength lambda (= 6000 Å) from a narrow 1 mm slit in its plane and 5 cm plane from its near edge. Find the fringe width on a screen 120 cm from the slit and width of interference pattern on the screen.

In Lloyd's experiment , the slit and its image have a separation of 4.32 mm and the obsevations made at a plane 2.0 m away fringes of separation 0.26 mm. Find the wavelength of light used.