At what angles for the first-order diffraction, spacing between two planes X respectively areλ andλ/2 ?a)0°,90°b)90°,0°c)30°,90°d)90°,30°Correct answer is option 'C'. Can you explain this answer?

Question

At what angles for the first-order diffraction, spacing between two planes X respectively are&#955; and&#955;/2 ?a)0&#176;,90&#176;b)90&#176;,0&#176;c)30&#176;,90&#176;d)90&#176;,30&#176;Correct answer is option 'C'. Can you explain this answer?

Answer

(c) By Bragg's equation, spacing between two planes is given by

For first-order diffraction, n =1

∴

If d = λ, then 2 sin θ =1

if d = λ/2, then
sin θ = 1 => θ = 90^o

Answer

(c) By Bragg's equation, spacing between two planes is given by

For first-order diffraction, n =1

∴

If d = λ, then 2 sin θ =1

if d = λ/2, then
sin θ = 1 => θ = 90^o

Answer

First-order diffraction is given by the equation:

nλ = 2d sin(θ)

where n is the order of diffraction, λ is the wavelength of the incident light, d is the spacing between the two planes, and θ is the angle of diffraction.

We are given that the spacing between the two planes is X/2, where X is the wavelength of the incident light.

To find the angles at which first-order diffraction occurs, we can substitute X/2 for d in the equation:

nλ = 2(X/2) sin(θ)

nλ = Xsin(θ)

sin(θ) = (nλ)/X

Taking the inverse sine of both sides, we get:

θ = sin^(-1)((nλ)/X)

Since we are interested in the first-order diffraction, we can set n = 1:

θ = sin^(-1)((λ)/X)

Now, let's substitute the given values:

θ = sin^(-1)((λ)/(X/2))

θ = sin^(-1)((2λ)/X)

The question asks for the angles at which the first-order diffraction occurs for the spacing between two planes X/2. This means that the wavelength of the incident light is X.

θ = sin^(-1)((2X)/X)

θ = sin^(-1)(2)

Since the sine function has a maximum value of 1, the maximum angle for which first-order diffraction occurs is 90 degrees. Therefore, the correct answer is option C: 30, 90.

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