Final answer:
To show that second and third-order spectra overlap for white light on a diffraction grating, use the diffraction grating formula. By comparing the maximum wavelengths for each order, one can conclude that overlap will occur since the sine function has a maximum value of 1, ensuring the second-order maximum will always precede the third-order minimum.
Step-by-step explanation:
To show that the second and third-order spectra overlap for white light falling on a diffraction grating, we can use the formula for diffraction grating which relates the wavelength of the light (λ), the diffraction order (m), the grating constant (d), and the diffraction angle (θ):
mλ = d sin(θ)
For the overlap between the second and third-order spectra to occur, the longest wavelength of the second order (λ = 700 nm for m=2) must be at or beyond the shortest wavelength of the third order (λ = 400 nm for m=3).
Setting the equations equal to each other for overlap:
2(700 nm) = d sin(θ_2)
3(400 nm) = d sin(θ_3)
Simplifying, we get:
1400 nm = d sin(θ_2)
1200 nm = d sin(θ_3)
Since sin(θ_3) must be less than or equal to 1, θ_3 will always be less than θ_2, thus confirming that there is overlap regardless of the value of d.