Question

A diffraction grating is used to resolve two yellow lines 5770 Å and 5790 Å in the 3rd order. If the grating element () is 0.35 mm, calculate the angular location of 3rd order yellow line with respect to the normal. Determine the total width of the ruling/ruled surface.

Answer 1

Answer on Question #49309, Engineering, Other

A diffraction grating is used to resolve two yellow lines 5770 Å and 5790 Å in the 3rd order. If the grating element d is 0.35 mm, calculate the angular location of 3rd order yellow line with respect to the normal. Determine the total width of the ruling/ruled surface.

Solution:

When light is normally incident on the grating, the diffracted light will have maxima at angles $\theta$ given by:

$$d \sin \theta = m \lambda$$

Thus,

$$\sin \theta = \frac{m \lambda}{d} = \frac{3 * 5770 * 10^{-10}}{0.35 * 10^{-3}} = 0.0049457.$$

The angular location of 3rd order yellow line

$$\theta = \sin^{-1} 0.0049457 \approx 0.28^{\circ}$$

The expression for the resolving power for a plane transmission grating is

$$\frac{\lambda}{d\lambda} = mN$$

where $\lambda = 5770\,\text{\AA}$, $\lambda + d\lambda = 5790\,\text{\AA}$, $d\lambda = 20\,\text{\AA}$, $m = 3$

The number of lines

$$N = \frac{\lambda}{m d \lambda} = \frac{5770}{3 * 20} = 96 \text{ lines}$$

The total width of the ruling/ruled surface is

$$L = N * d = 96 * 0.35 * 10^{-3} = 0.0336\,\text{m} = 33.6\,\text{mm}$$

Answer: $\theta = 0.28^{\circ}, L = 33.6\,\text{mm}$