Question

consider an experiment using a diffraction grating with 7000 lines/cm, a screen 2.50 m away, and a 440nm wavelength beam of light. (a) how many side maxima will be observed on one side of the central maximum? (b) how far apart on the screen are the first-order and second-order maxima on the same side of the central maximum?

Answer 1

(a) There are three side maxima on one side of the central maximum.

(b) The distance between the first-order and second-order maxima on the same side of the central maximum is

(a) To find the number of side maxima on one side of the central maximum, we can use the diffraction grating equation:

`\[ m \lambda = d \sin(\theta) \]`

Given that:
`\(d = \frac{1}{7000 \, \text{lines/cm}}\) and \(\lambda = 440 \, \text{nm}\)`, let's find the angles for m = 1, 2, 3.

For m = 1:

`\[ \lambda = d \sin(\theta_1) \]`

`\[ \sin(\theta_1) = (\lambda)/(d) \]`

`\[ \sin(\theta_1) = \frac{440 \, \text{nm}}{\frac{1}{7000 \, \text{cm}}} \]`

`\[ \sin(\theta_1) \approx 440 * 7000 \]`

`\[ \theta_1 \approx \sin^(-1)(440 * 7000) \]`

`\[ \theta_1 \approx 0.155 \, \text{rad} \]`

Now, for m = 2:

`\[ 2\lambda = d \sin(\theta_2) \]`

`\[ \sin(\theta_2) = (2\lambda)/(d) \]`

`\[ \sin(\theta_2) = \frac{2 * 440 \, \text{nm}}{\frac{1}{7000 \, \text{cm}}} \]`

`\[ \sin(\theta_2) \approx 2 * 440 * 7000 \]`

`\[ \theta_2 \approx \sin^(-1)(2 * 440 * 7000) \]`

`\[ \theta_2 \approx 0.310 \, \text{rad} \]`

Now, for m = 3:

`\[ 3\lambda = d \sin(\theta_3) \]\\`

`\[ \sin(\theta_3) = (3\lambda)/(d) \]\\`

`\[ \sin(\theta_3) = \frac{3 * 440 \, \text{nm}}{\frac{1}{7000 \, \text{cm}}} \]\\`

`\[ \sin(\theta_3) \approx 3 * 440 * 7000 \]\\`

`\[ \theta_3 \approx \sin^(-1)(3 * 440 * 7000) \]\\`

`\[ \theta_3 \approx 0.464 \, \text{rad} \]`

Now, we'll find the values of m for which
`\(\sin(\theta_m) \leq 1\)`:

`\[ m = 1: \quad \theta_1 \approx 0.155 \, \text{rad} \]`

`\[ m = 2: \quad \theta_2 \approx 0.310 \, \text{rad} \]`

`\[ m = 3: \quad \theta_3 \approx 0.464 \, \text{rad} \]`

So, there are three side maxima on one side of the central maximum.

(b) Now, let's find the distance between adjacent maxima
`(\(\Delta y\))` on the screen using the formula:

`\[ \Delta y = L \tan(\theta) \]`

For the first-order maximum m = 1:

`\[ \Delta y_1 = L \tan(\theta_1) \]`

`\[ \Delta y_1 = 2.50 \, \text{m} * \tan(0.155) \]`

`\[ \Delta y_1 \approx 2.50 \, \text{m} * 0.155 \]`

`\[ \Delta y_1 \approx 0.3875 \, \text{m} \]`

For the second-order maximum m = 2:

`\[ \Delta y_2 = L \tan(\theta_2) \]`

`\[ \Delta y_2 = 2.50 \, \text{m} * \tan(0.310) \]`

`\[ \Delta y_2 \approx 2.50 \, \text{m} * 0.309 \]`

`\[ \Delta y_2 \approx 0.7725 \, \text{m} \]`

Therefore, the distance between the first-order and second-order maxima on the same side of the central maximum is approximately

0.7725 m - 0.3875 m = 0.385 m.