Answer:The angle at which the first-order maximum appears can be calculated using the formula:
sin(θ) = mλ/d
where θ is the angle at which the maximum appears, m is the order of the maximum (in this case, m=1 for the first-order maximum), λ is the wavelength of the light, and d is the spacing between adjacent lines on the diffraction grating.
We can rearrange this formula to solve for d:
d = mλ/sin(θ)
For the first-order maximum:
d = (1)(λ)/(sin(12.07°))
Now, to find the angle at which the second maximum appears, we can use the same formula but with m=2:
sin(θ) = 2λ/d
Rearranging this formula to solve for θ:
θ = arcsin(2λ/d)
Substituting the values we have already calculated:
θ = arcsin(2λ/[(1)(λ)/(sin(12.07°))])
Simplifying:
θ = arcsin(2sin(12.07°))
θ ≈ 24.17°
Therefore, the second maximum will appear at an angle of approximately 24.17°.
Explanation:
calculations:
1. We are given the number of lines per centimeter on the diffraction grating, which is 3550 lines/cm.
2.We are also given that the light forms a first-order maximum at an angle of 12.07°.
3.We can use the formula sin(θ) = mλ/d to find the spacing between adjacent lines on the diffraction grating. Rearranging this formula to solve for d, we get d = mλ/sin(θ).
4.Substituting the values m=1, λ (which we assume to be the wavelength of the light), and θ=12.07°, we can solve for d. We get d = (1)(λ)/(sin(12.07°)).
5.Next, we want to find the angle at which the second maximum appears. We can use the same formula, but with m=2: sin(θ) = 2λ/d.
6.Rearranging this formula to solve for θ, we get θ = arcsin(2λ/d).
7.Substituting the values we calculated for d and assuming the same wavelength of light, we can solve for the angle θ. We get θ = arcsin(2sin(12.07°)).
8.Finally, we evaluate this expression and get the answer: the second maximum will appear at an angle of approximately 24.17°.