Physics, asked by lohakarekalyani1798, 5 months ago

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​

Answers

Answered by prayoshichowdhury
0

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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