Fringes of equal thickness are observed in a thin glass wedge of RI = 1.52. Calculate the angle of wedge in arc-sec if for a light of $\lambda =5893 \mathring{A}$ the fringe spacing is found to be 1mm

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written 6.9 years ago by

neeta.vanage • 870

modified 2.8 years ago by

pedsangini276 • 4.8k

Subject: Applied Physics 2

Topic: Interference And Diffraction

Difficulty: Medium

engineering physics

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1 Answer

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written 6.8 years ago by

neeta.vanage • 870

modified 6.8 years ago by

mananbothra • 100

μ =1.52, λ = 5893 x10$^{-8}$ cm, D$_x$ =0.1 cm

$D_x = \frac{λ}{2μϴ}$

$ θ = \frac{λ}{(2μ D_x)} \ radians \\[2ex] θ = \frac{5893 \times10^{-8}}{2 \times1.52 \times 0.1} = 1.93 \times 10^{-4} \ radians \\[2ex] θ = \frac{1.93 \times 10^{-4} \times 3600 \times 180}{\pi} $

θ = 40 seconds of an arc

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