What is the highest-order (m value)maximum for wavelength 450 nm that can be obtained with a grating with 300 lines per mm?
A)8
B)3
C)7
D)4

Question

Quizplus · Accepted Answer

The highest-order maximum (m value) can be calculated using the grating equation $d\sin(\theta) = m\lambda$, where $d$ is the grating spacing, $\lambda$ is the wavelength, and $m$ is the order of the maximum. The grating spacing $d$ is the inverse of the number of lines per mm, so for 300 lines per mm, $d = \frac{1}{300 \text{ mm}^{-1}} = \frac{1}{300000 \text{ m}^{-1}} = 3.33 \times 10^{-6} \text{ m}$. To find the highest possible order, we need the maximum value of $\sin(\theta)$, which is 1 (when $\theta = 90^\circ$). Thus, $m = \frac{d}{\lambda}$. Substituting the given values, $m = \frac{3.33 \times 10^{-6} \text{ m}}{450 \times 10^{-9} \text{ m}} = 7.4$. Since $m$ must be an integer, the highest possible order is 7.

What Is the Highest-Order (M Value)maximum for Wavelength 450 Nm