Question

Gratings A grating has exactly 8000 slits uniformly spaced over \(2.54 \mathrm{cm}\) and is illuminated by light from a mercury vapor discharge lamp. What is the expected angle for the third-order maximum of the green line $(\lambda=546 \mathrm{nm}) ?$

Short answer

The expected angle for the third-order maximum of the green line (λ = 546 nm) is approximately 31.4°.

Step-by-step solution

Identify the grating equation and the given variables

In this exercise, we will use the grating equation:$$m\lambda = d\sin\theta$$Where \(m\) is the order of the maximum, \(\lambda\) is the wavelength of light, \(d\) is the slit separation, and \(\theta\) is the angle we need to find. We are given: - The green line wavelength: \(\lambda = 546\mathrm{nm} = 5.46 \times 10^{-7}\mathrm{m}\) (converted to meters) - The total distance of 8000 slits: \(2.54\mathrm{cm} = 2.54 \times 10^{-2}\mathrm{m}\) (converted to meters) - The order of the maximum we are looking for: \(m = 3\)

Calculate the slit separation, d

To find the slit separation, we can simply divide the total distance by the number of slits:$$d = \frac{2.54 \times 10^{-2}\mathrm{m}}{8000}$$Calculating this value gives us:$$d = 3.175 \times 10^{-6}\mathrm{m}$$

Solve the grating equation for the angle, θ

Now that we have the slit separation, we can plug in the values into the grating equation and solve for the angle:$$3(5.46 \times 10^{-7}\mathrm{m}) = (3.175 \times 10^{-6}\mathrm{m})\sin\theta$$Divide by \(3.175 \times 10^{-6}\mathrm{m}\) on both sides to isolate \(\sin\theta\):$$\sin\theta = \frac{3(5.46 \times 10^{-7}\mathrm{m})}{3.175 \times 10^{-6}\mathrm{m}}$$Calculate the value of \(\sin\theta\):$$\sin\theta \approx 0.5164$$

Find the angle θ

Finally, to find the angle, we take the inverse sine of the calculated value:$$\theta = \sin^{-1}(0.5164)$$Calculating this value gives us:$$\theta \approx 31.4°$$ The expected angle for the third-order maximum of the green line \((\lambda=546\mathrm{nm})\) is approximately \(31.4°\).