Question

derive Bragg's equation​

Answer 1

Answer:

bragg 's law can be easily derived by considering the conditions necessary to make the phases of the beams coincide when the incident angle equals to reflecting angle. the rays of the incident beam are always in phase and parallel up to the point at which the top beam strikes the top layer at atom Z

Answer 2

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$\small\underline\mathcal\pink{{Bragg's\:equation:-}}$ W.H.bragg has proposed an equation to explain the relation between inter planar distance (d),wavelength of x-rays (λ)and angle of diffraction of x-rays (θ ) is called Bragg's equation.

$\small\underline\mathbb\green{{From\:The\: Above \:Diagram}}$

$\blue\longmapsto\:$First X-ray is diffracted from a point A in the first plane and second X-ray is diffracted from point 'B' in the second plane.

$\pink\longmapsto\:$Second X-ray wave travels more distance than first X-ray wave.

$\red\longmapsto\:$The extra distance travelled by second x-ray =CB+BD.

$\orange\longmapsto\:$path difference in first & second layer of second x-ray =CB+BD .

$\purple\longmapsto\:$If waves of two rays exists in same phase then,path difference is equal to an integral multiple of wavelength of x-rays

Therefore,CB + BD = nλ ..........eq no.1

where, n = 1, 2, 3,...........

In traingle ABC,

sinθ = CB/AB =CB/d

CB=d sinθ............eq no.2

In triangle ABD

sinθ=BD/AB=BD/d

BD= d sinθ...........eq no.3

when equation 2,3 are substituted in equation 1 we get

d sinθ + d sinθ

This relation is known as bragg's equation

where,

n= order of diffraction

θ= angle of diffraction

d= inter planar distance

λ= Wavelength of x-rays.

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