KVPY SX SB Exam 2016 Question Paper

Question : 37 of 120

Marks: +1, -0

A monochromatic light source S of wavelength 440 nm is placed slightly above a plane mirror M as shown . Image of S in M can be used as a virtual source to produce interference fringes on the screen. The distance of source S from O is 20.0 cm, and the distance of screen from O is 100.0 cm (figure is not to scale). If the angle θ =

0.50 × 10^{–3}

radians, the width of the interference fringes observed on the screen is –

2.20 mm
2.64 mm
1.10 mm
0.55 mm

Validate

Solution:

S and

S_1

are source of YDSE

θ = 0.5 × 10^{–3}

radian (very small)
D = SO cos θ + 100 = 20 × 1 + 100 = 120 cm
d = 2 × SO sin θ
= 2 × 20 × θ

= 40 × 0.5 × 10^{–3}

=2 × 10^{–2} \cm

β={λD}/d

={440×10^{-6}×120×10^2}/{2×10^{-2}×10^2}

=264×10^{-2}=2.64\mm

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