A plane of spacing 'd' shows first order Bragg diffraction at angle θ. A plane of spacing 2d

Question

A plane of spacing 'd' shows first order Bragg diffraction at angle θ. A plane of spacing 2d (A) shows Bragg diffraction at 2 θ (B) shows Bragg diffraction at (θ/2) (C) shows Bragg diffraction at sin-1((sin θ/2)) (D) shows Bragg diffraction at sin-1((sin 2θ/2))

Tardigrade Admin · Accepted Answer

From Bragg's equation, we have, 2d sin θ= n λ λ=2d sin θ (for first order reflection) d=(λ/2 sin θ) Then for spacing 2d, we have 2 (2d) sinθ=λ sin θ=(λ/4d)=(λ/4(λ 2 sin θ))=(sin θ/2) θ =sin-1((sin θ/2))

Q. A plane of spacing $^{'} d^{'}$ shows first order Bragg diffraction at angle $θ$ . A plane of spacing $2 d$

shows Bragg diffraction at $2 θ$

shows Bragg diffraction at $\frac{θ}{2}$

shows Bragg diffraction at $s i n^{- 1} (\frac{s in θ}{2})$

shows Bragg diffraction at $s i n^{- 1} (\frac{s in 2 θ}{2})$

From Bragg's equation, we have, $2 d s in θ = nλ$
$λ = 2 d s in θ$ (for first order reflection)
$d = \frac{λ}{2 s in θ}$
Then for spacing $2 d$ , we have
$2 (2 d) s in θ = λ$
$s in θ = \frac{λ}{4 d} = \frac{λ}{4 ( λ 2 s in θ )} = \frac{s in θ}{2}$
$θ = s i n^{- 1} (\frac{s in θ}{2})$

Q. A plane of spacing ′d′ shows first order Bragg diffraction at angle θ. A plane of spacing 2d

shows Bragg diffraction at 2θ

shows Bragg diffraction at 2θ​

shows Bragg diffraction at sin−1(2sinθ​)

shows Bragg diffraction at sin−1(2sin2θ​)

From Bragg's equation, we have, 2dsinθ=nλ λ=2dsinθ (for first order reflection) d=2sinθλ​ Then for spacing 2d, we have 2(2d)sinθ=λ sinθ=4dλ​=4(λ2sinθ)λ​=2sinθ​ θ=sin−1(2sinθ​)

Q. A plane of spacing $^{'} d^{'}$ shows first order Bragg diffraction at angle $θ$ . A plane of spacing $2 d$

shows Bragg diffraction at $2 θ$

shows Bragg diffraction at $\frac{θ}{2}$

shows Bragg diffraction at $s i n^{- 1} (\frac{s in θ}{2})$

shows Bragg diffraction at $s i n^{- 1} (\frac{s in 2 θ}{2})$