For a given grating,it is observed that a third-order line (m = 3)with $\lambda$ = 465.3 nm overlaps a second-order line (has the same $\theta$ as a line with m = 2).The wavelength of the second-order line is
A)304 nm
B)698 nm
C)913 nm
D)152 nm
E)372 nm

Question

Quizplus · Accepted Answer

Let's use the grating equation:

mλ = d(sinθ + sinϕ)

Where m is the order, λ is the wavelength, d is the grating spacing, θ is the angle of incidence, and ϕ is the angle of diffraction.

For the third-order line:

3(465.3 nm) = d(sinθ + sinϕ1)

For the second-order line:

2λ = d(sinθ + sinϕ2)

We know that both lines have the same F1F1F1S1θF1F1F10 (angle of incidence), so we can equate the values:

sinθ + sinϕ1 = sinθ + sinϕ2

sinϕ1 = sinϕ2

Now we can substitute sinϕ1 for sinϕ2 in the second equation:

2λ = d(sinθ + sinϕ1)

2λ = 3(465.3 nm)

λ = 697.95 nm

The closest answer choice is B, 698 nm.

For a Given Grating,it Is Observed That a Third-Order Line λ\lambdaλ

For a Given Grating,it Is Observed That a Third-Order Line $\lambda$