what is the approximate wavelength of a light whose first - order bright band forms a diffraction angle of…

what is the approximate wavelength of a light whose first - order bright band forms a diffraction angle of 45.0° when it passes through a diffraction grating that has 500.0 lines per mm?\n236 nm\n353 nm\n943 nm\n1414 nm

what is the approximate wavelength of a light whose first - order bright band forms a diffraction angle of 45.0° when it passes through a diffraction grating that has 500.0 lines per mm?\n236 nm\n353 nm\n943 nm\n1414 nm

Answer

Explanation:

Step1: Calculate the grating spacing

The number of lines per unit length $N = 500.0$ lines/mm. The grating - spacing $d$ is the reciprocal of the number of lines per unit length. So $d=\frac{1}{N}=\frac{1}{500.0\ lines/mm}=2\times10^{-3}\ mm = 2\times10^{-6}\ m$.

Step2: Use the grating equation

The grating equation is $d\sin\theta = m\lambda$, where $m$ is the order of the bright - band, $\theta$ is the diffraction angle, $\lambda$ is the wavelength of light. Here, $m = 1$ (first - order bright band) and $\theta=45.0^{\circ}$, and $d = 2\times10^{-6}\ m$. We can solve for $\lambda$: $\lambda=\frac{d\sin\theta}{m}$. Substitute the values: $\lambda=\frac{2\times10^{-6}\ m\times\sin45.0^{\circ}}{1}$. Since $\sin45.0^{\circ}=\frac{\sqrt{2}}{2}\approx0.707$, then $\lambda=\frac{2\times10^{-6}\ m\times0.707}{1}=1.414\times10^{-6}\ m = 1414\ nm$.

Answer:

1414 nm