4. Visible light is incident perpendicularly on a diffraction grating of 200 rulings/mm. What are the (a) longest, (b) second longest, and (c) third longest wavelengths that can be associated with an intensity maximum at Ѳ = 30°?

To solve this problem, we need to use the formula for diffraction grating, which is:

dsin(θ) = mλ

where:

• d is the distance between the slits in the grating,
• θ is the angle of diffraction,
• m is the order of the diffraction (m = 1 for the longest wavelength, m = 2 for the second longest, etc.),
• λ is the wavelength of the light.

First, we need to convert the grating density from rulings per millimeter to distance between slits in meters. We have 200 rulings/mm, so d = 1/200 mm = 5*10^-6 m.

Then, we can solve the formula for λ to find the wavelengths for each order of diffraction:

(a) For the longest wavelength (m = 1), we have:

λ = dsin(θ) / m λ = 510^-6 m * sin(30°) / 1 λ = 2.5*10^-6 m = 2500 nm

(b) For the second longest wavelength (m = 2), we have:

λ = dsin(θ) / m λ = 510^-6 m * sin(30°) / 2 λ = 1.25*10^-6 m = 1250 nm

(c) For the third longest wavelength (m = 3), we have:

λ = dsin(θ) / m λ = 510^-6 m * sin(30°) / 3 λ = 0.83*10^-6 m = 833.33 nm

So, the longest, second longest, and third longest wavelengths that can be associated with an intensity maximum at Ѳ = 30° are 2500 nm, 1250 nm, and 833.33 nm, respectively.