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.

Explanation

In plane mirror, imgae is formed behind it as the object is in front of it. So,
`d = 2 mm = 0.2 cm`
`lambda = 6000 Å = 6000 xx 10^(-8) cm`
`D = 120 cm`
Therefore, fringe width,
` beta = (lambda D)/( d) = (6 xx 10^(-5) xx 120)/(0.2) = 0.036 cm`
The width of fringe is AB. From figure
`tan theta_(1) = (0.1)/(5)` and `tan theta_(2) = (0.1)/(10)`
In right-angled triangles `AM_(1) O` and `BM_(2) O`,
`tan theta_(1) = (0.1)/(5) = (OA)/(M_(1) O) implies OA = 115 xx (0.1)/(5) cm`,
`tan theta_(2) = (0.1)/(10) = (OB)/(OM_(2)) impliesOB = 110 xx (0.1)/(10) cm`
Hence, width of fringe pattern is
`OA - OB = (115 xx 0.1)/(5) - (110 xx 0.1)/(10) = 1.2 cm`