At what angle for the first - order diffraction, spacing between two planes respectively is `lambda and (lambda)/(2)`?

Options

A. `0^(@), 90^(@)`
B. `90^(@), 0^(@)`
C. `30^(@), 90^(@)`
D. `90^(@), 30^(@)`

Explanation

To solve the question regarding the angles for first-order diffraction with spacing between two planes being λ and λ/2, we will use Bragg's Law, which is given by the equation:

\[ n\lambda = 2d\sin\theta \]

where:
- \( n \) is the order of diffraction (for first-order diffraction, \( n = 1 \)),
- \( \lambda \) is the wavelength of the incident wave,
- \( d \) is the distance between the planes,
- \( \theta \) is the angle of diffraction.

### Step 1: For the first spacing \( d = \lambda \)

1. Substitute \( n = 1 \) and \( d = \lambda \) into Bragg's equation:
\[ 1 \cdot \lambda = 2\lambda \sin\theta \]

2. Simplify the equation:
\[ \lambda = 2\lambda \sin\theta \]

3. Divide both sides by \( \lambda \) (assuming \( \lambda \neq 0 \)):
\[ 1 = 2\sin\theta \]

4. Solve for \( \sin\theta \):
\[ \sin\theta = \frac{1}{2} \]

5. Find \( \theta \):
\[ \theta = \sin^{-1}\left(\frac{1}{2}\right) = 30^\circ \]

### Step 2: For the second spacing \( d = \frac{\lambda}{2} \)

1. Substitute \( n = 1 \) and \( d = \frac{\lambda}{2} \) into Bragg's equation:
\[ 1 \cdot \lambda = 2\left(\frac{\lambda}{2}\right) \sin\theta \]

2. Simplify the equation:
\[ \lambda = \lambda \sin\theta \]

3. Divide both sides by \( \lambda \) (assuming \( \lambda \neq 0 \)):
\[ 1 = \sin\theta \]

4. Solve for \( \theta \):
\[ \theta = \sin^{-1}(1) = 90^\circ \]

### Final Result

The angles for the first-order diffraction with spacings \( \lambda \) and \( \frac{\lambda}{2} \) are:
- \( 30^\circ \) for \( d = \lambda \)
- \( 90^\circ \) for \( d = \frac{\lambda}{2} \)