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2d sin(theta)=lambda
Molecules in many solids are arranged in a crystal lattice with distinct patterns and layers. These layers reflect and scatter light rays according to Bragg's law. The Bragg equation relates the distance, 
d, between layers of molecules to the angle, 
theta, of incoming light rays with wavelength 
lambda. Which of the following is the correct equation for the distance in terms of the angle and the wavelength?
Choose 1 answer:
(A) 
d=(lambda)/(2sin(theta))
(B) 
d=(2lambda)/(sin(theta))
(C) 
d=(2sin(theta))/(lambda)
(D) 
d=(sin(theta))/(2lambda)

2dsin(θ)=λ2d \sin(\theta) = \lambda\newlineMolecules in many solids are arranged in a crystal lattice with distinct patterns and layers. These layers reflect and scatter light rays according to Bragg's law. The Bragg equation relates the distance, \newlinedd, between layers of molecules to the angle, \newlineθ\theta, of incoming light rays with wavelength \newlineλ\lambda. Which of the following is the correct equation for the distance in terms of the angle and the wavelength?\newlineChoose 11 answer:\newline(A) d=λ2sin(θ)d = \frac{\lambda}{2\sin(\theta)}\newline(B) d=2λsin(θ)d = \frac{2\lambda}{\sin(\theta)}\newline(C) d=2sin(θ)λd = \frac{2\sin(\theta)}{\lambda}\newline(D) d=sin(θ)2λd = \frac{\sin(\theta)}{2\lambda}

Full solution

Q. 2dsin(θ)=λ2d \sin(\theta) = \lambda\newlineMolecules in many solids are arranged in a crystal lattice with distinct patterns and layers. These layers reflect and scatter light rays according to Bragg's law. The Bragg equation relates the distance, \newlinedd, between layers of molecules to the angle, \newlineθ\theta, of incoming light rays with wavelength \newlineλ\lambda. Which of the following is the correct equation for the distance in terms of the angle and the wavelength?\newlineChoose 11 answer:\newline(A) d=λ2sin(θ)d = \frac{\lambda}{2\sin(\theta)}\newline(B) d=2λsin(θ)d = \frac{2\lambda}{\sin(\theta)}\newline(C) d=2sin(θ)λd = \frac{2\sin(\theta)}{\lambda}\newline(D) d=sin(θ)2λd = \frac{\sin(\theta)}{2\lambda}
  1. Isolate dd: The given equation is 2dsin(θ)=λ2d \sin(\theta) = \lambda. To find the distance dd in terms of θ\theta and λ\lambda, we need to solve for dd.
  2. Divide by 2sin(θ)2\sin(\theta): Divide both sides of the equation by 2sin(θ)2\sin(\theta) to isolate dd on one side:\newlined=λ2sin(θ)d = \frac{\lambda}{2\sin(\theta)}
  3. Final equation for d: The correct equation for the distance dd in terms of the angle θ\theta and the wavelength λ\lambda is therefore d=λ2sin(θ)d = \frac{\lambda}{2\sin(\theta)}, which corresponds to option (A).

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