interference fringes are observed with a by prism of refracting angle 1 Degree and the refractive index 1.5 on a screen hundred CM away from it the wavelength of light used is 5890 if the distance between source and by prism is 20 CM the principal duties
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Answer:
Young's double-slit experiment.
first of all, find deviation,
we know,
where a is the distance between source and biprism. e.g., a=20m
so, d=2×π/180×20=2π/9m
now, use formula of fringe width,
here,
D=100m,d=2π/9m
so,
0.1×10
−
3=λ×
(2π/9)
100
=>10
−
4×2π/9×10
−
2=λ
=>λ=2π/9×10
−
6m
=>λ=0.697×10
−
6=697×10
−
9m
=>λ=697nm
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