A diffraction grating has 500 lines/mm. What is the maximum order of diffraction visible for light with a wavelength of 600 nm?

The grating has 500 lines per millimeter.

$$d = \frac{1}{\text{number of lines per unit length}}$$ $$d = \frac{1}{500 \text{ lines/mm}} = \frac{1}{500 \times 10^3 \text{ lines/m}}$$ $$d = \frac{1}{5 \times 10^5 \text{ lines/m}} = 2 \times 10^{-6} \text{ m}$$

Given wavelength $\lambda = 600 \text{ nm}$.

$$\lambda = 600 \times 10^{-9} \text{ m} = 6 \times 10^{-7} \text{ m}$$

The maximum possible value for $\sin \theta$ is 1 (when $\theta = 90^\circ$).
So, for the maximum order, the equation becomes:

$$d \times 1 = m_{max} \lambda$$ $$m_{max} = \frac{d}{\lambda}$$ $$m_{max} = \frac{2 \times 10^{-6} \text{ m}}{6 \times 10^{-7} \text{ m}}$$ $$m_{max} = \frac{20 \times 10^{-7}}{6 \times 10^{-7}} = \frac{20}{6} = \frac{10}{3} \approx 3.33$$

Since the order $m$ must be an integer, the maximum observable integer order of diffraction is $m=3$.

The possible integer orders of diffraction are $m = 0, \pm 1, \pm 2, \pm 3$.

• $m=0$ corresponds to the central image (1 image).
• $m=1, 2, 3$ correspond to images on one side of the central maximum (3 images).
• $m=-1, -2, -3$ correspond to images on the other side of the central maximum (3 images).

Total number of images = (images on one side) + (images on other side) + (central image)
Total number of images = $3 + 3 + 1 = 7$.

The total number of diffraction images seen, including the central image, is 7.

The final answer is $\text{D. 7}$.