Question

Light of wavelength 680 nm is incident normally on a diffraction grating having 4000 lines/cm. The diffraction angle (in degrees) corresponding to the third-order maximum is ........

Hint:

For diffraction from a grating, use Bragg's law \( n\lambda = d \sin \theta \) to find the diffraction angle.

Answer 1

Correct Answer: 54.67

Step 1: Use the diffraction grating equation.
The diffraction condition for a grating is given by Bragg’s law: \[ n\lambda = d \sin \theta \] where \( n \) is the diffraction order, \( \lambda \) is the wavelength of the light, \( d \) is the grating spacing, and \( \theta \) is the diffraction angle.
Step 2: Calculate the grating spacing.
The number of lines per centimeter is 4000, so the grating spacing \( d \) is: \[ d = \frac{1}{4000} \, \text{cm} = \frac{1}{4000 \times 10^2} \, \text{m} = 2.5 \times 10^{-6} \, \text{m} \]
Step 3: Apply the diffraction equation.
For the third-order maximum (\( n = 3 \)) and wavelength \( \lambda = 680 \, \text{nm} = 680 \times 10^{-9} \, \text{m} \): \[ 3 \times 680 \times 10^{-9} = 2.5 \times 10^{-6} \sin \theta \] Solving for \( \theta \): \[ \sin \theta = \frac{3 \times 680 \times 10^{-9}}{2.5 \times 10^{-6}} = 0.816 \] \[ \theta = \sin^{-1}(0.816) = 21.04^\circ \]
Step 4: Conclusion.
Thus, the diffraction angle corresponding to the third-order maximum is 21.04°.