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A Lloyd's mirror or length 5 cm is illum...

A Lloyd's mirror or length 5 cm is illuminated with monochromatic 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.

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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`
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