Calculate the wavelength of light that has its secondorder maximum at \(45.0^{\circ}\) when falling on a diffraction grating that has 5000 lines per centimeter.

To convert the number of lines per centimeter (5000) to the distance between the lines (d), we can use the following formula: \[d = \frac{1}{N}\] where, - N is the number of lines per unit length (in this case, centimeters) So, first we need to find the number of lines per meter (N in meters), and then calculate the distance between lines (d). Number of lines per meter: \(N = 5000 * 100\) (since there are 100 cm in 1 m) Now, we can find the distance between lines: \[d = \frac{1}{5000 \times 100} = \frac{1}{5 \times 10^5}\] meters

The diffraction grating equation is: \[d\sin{\theta} = m\lambda\] Using the given values, we have: - \(d = \frac{1}{5 \times 10^5}\) meters - \(\theta = 45^{\circ}\) - \(m = 2\) (since it's the second-order maximum) Now, we can solve for the wavelength (\(\lambda\)): \[\frac{1}{5 \times 10^5} \times \sin{45^{\circ}} = 2\lambda\] \[\lambda = \frac{1}{5 \times 10^5 \times 2} \times \sin{45^{\circ}}\] Finally, we can calculate the wavelength of light: \[\lambda = \frac{\sin{45^{\circ}}}{10^6} = 7.07 \times 10^{-7}\] meters Therefore, the wavelength of light that has its second-order maximum at \(45.0^{\circ}\) when falling on a diffraction grating with 5000 lines per centimeter is approximately \(7.07 \times 10^{-7}\) meters.