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Question

# A Lloyd’s mirror of length 5 cm is illuminated with monochromatic light of wavelength λ(=6000∘A) 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.

A
0.036 cm,1.2 cm
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B
0.045 cm,0.96 cm
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C
1.2 cm,0.036 cm
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D
3.6 cm,0.12 cm
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Solution

## The correct option is A 0.036 cm,1.2 cm Fringe width (β)=λDd from the given information, β =6000×10−8×1200.2cmβ=36×10−3 cm Width of interference pattern on the screen: ΔPAO tanθ1=AO115 ...(1) ΔMBO tanθ2=BO110 ...(2) From the origin side, tanθ1=0.15 tanθ2=0.110 putting these values on eq.(1) and eq.(2) respectively. so, tanθ1=AO1150.15=AO115AO=0.1×1155=2.3 and tanθ2=BO1100.110=BO110BO=0.1×11010=1.1 Now, width of interference pattern on the screen = AO−BO 2.3−2.2=1.2 cm

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