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Assume that the limits of the visible spectrum are arbitrarily chosen as 430and680nm. Calculate the number of rulings per millimeter of a grating that will spread the first-order spectrum through an angle of 20.0°20.0°

Short Answer

Expert verified

The number of ruling is1.09×103 ruling per mm

Step by step solution

01

Introduction of diffraction

Diffraction is a phenomena in which the interference or the bending of waves occurs around the corners of the opening or obstacle through or on which it strikes.

The diffraction formula is following,

²Ôλ=2»å²õ¾±²Ôθλ=wavelengthd=spacingθ=bragg anglen=order of diffraction

02

Step 2: Identification of the given data

The shortest Wavelengthλ1=430nm

The longest Wavelengthλ2=680nm

03

Determination of the number of rulings per milimeter

The diffraction formula,

λ1=dsinθλ2=dsinθ+Δθsinθ+Δθ=sinθcosΔθ+cosθsinΔθ

From above equation,

sinθ=λ1d

And,

cosθ=1-λ1d2

After replacing thesinθ andcosθ in λ2equation,

λ1cosΔθ+d2-λ12sinΔθ=λ2

The value of d will be,

d=λ2-λ1cosΔθ2+λ1sinΔθ2sin2Δθ

Substitute all the value in the above equation.

d=680 nm-430 nm×cos20°2+430 nmsin20°2sin220°d=914 nmd=9.14×10-4 mm

1d=19.14×10-4 mm1d=1.09×103 ruling per mm

Hence the number of ruling is, 1.09×103 ruling per mm.

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Most popular questions from this chapter

Visible light is incident perpendicularly on a diffraction grating of 200 rulings/mm. What are the (a) longest, (b) second longest, and (c) third longest wavelengths that can be associated with an intensity maximum at θ = 30.0°?

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