Diffraction Grating wavelength

[nomorenomore]

Homework Statement

A beam of light comprises two wavelengths is passed through a transmission diffraction grating. When viewed at an angle of 21.1° to the incident direction, the second order maximum for one wavelength is overlapped with the third order maximum for the other wavelength. The shorter wavelength is 424nm.

i) Calculate the longer wavelength. (3M)
ii) Determine the number of lines per metre in the diffraction grating. (2M)
iii.) Determine the angle(s) (other than 21.1°) at which overlapping occur. (2M)

Homework Equations

For constructive interference, d*sinθ = mλ, m = 0, ±1, ±2...
For destructive interference, d*sinθ = (m + 1/2)λ, m = 0, ±1, ±2...

The Attempt at a Solution

I think my main problem here is not sure what "When viewed at an angle of 21.1° to the incident direction" means. How should I hand this, please?

 

[ehild]

Homework Statement

A beam of light comprises two wavelengths is passed through a transmission diffraction grating. When viewed at an angle of 21.1° to the incident direction, the second order maximum for one wavelength is overlapped with the third order maximum for the other wavelength. The shorter wavelength is 424nm.

i) Calculate the longer wavelength. (3M)
ii) Determine the number of lines per metre in the diffraction grating. (2M)
iii.) Determine the angle(s) (other than 21.1°) at which overlapping occur. (2M)

Homework Equations

For constructive interference, d*sinθ = mλ, m = 0, ±1, ±2...
For destructive interference, d*sinθ = (m + 1/2)λ, m = 0, ±1, ±2...

The Attempt at a Solution

I think my main problem here is not sure what "When viewed at an angle of 21.1° to the incident direction" means. How should I hand this, please?

21.1° is the angle the diffracted rays in the question deviate from the initial ray (θ in your equations) .

 

[orsanyuksek]

1. The diffraction formula for a maximum order is

m . λ = d . (sin α + sin β)

where

m : Diffraction order

λ: Wavelength

α: Angle of incidence

β: Exit angle

d: The grating constant

i) The grating constant, the angle of incidence and the exit angle for both wavelengths are the same. That means the right side of the equation is the same.

m_short . λ_short = m_longer . λ_longer

To justify the equation the third order must belong to the shorter wavelength.

3 . 424 nm = 2 . λ_longer

λ_longer = 636 nm

ii) mλ = d (sin α + sin β)

3 . 424 nm = d (sin 0° + sin 21,1°)

d = 283 lines / mm

iii.) If m = 1 (m=2) is given, the exit angle for the first (second) order is calculated etc.

-----------------------

Örsan Yüksek

 

[ehild]

i) The grating constant, the angle of incidence and the exit angle for both wavelengths are the same. That means the right side of the equation is the same.3 . 424 nm = 2 . λ_longer

λ_longer = 636 nm

d = 283 lines / mm

Correct so far.

iii.) If m = 1 (m=2) is given, the exit angle for the first (second) order is calculated etc.

I do not understand what you mean.
Find those orders p and q so as pλ1=qλ2. You know that λ2/λ1=3/2, therefore q/p =2/3. p and q are integers, so p must be divisible by 3. What angle do you get with the same grating and 484 nm wavelength if p=6? Is there a higher order?

 

[HalfLostAndSearching]

I would first start by clarifying the given information and assumptions. From the statement, we can assume that the beam of light is passing through a transmission diffraction grating, meaning that the grating is transparent and allows light to pass through it. The statement also mentions two different wavelengths, one shorter (424nm) and one longer (unknown). The incident direction is the direction in which the beam of light is approaching the diffraction grating. The angle of 21.1° is the angle at which the observer is viewing the diffraction pattern.

Now, to answer the questions:

i) To calculate the longer wavelength, we can use the equation for constructive interference, where d is the spacing between the lines of the diffraction grating, θ is the angle at which the observer is viewing the diffraction pattern, and m is the order of the maximum. We know that the second order maximum for the shorter wavelength (424nm) overlaps with the third order maximum for the longer wavelength. So, we can set up the following equation:

d*sin(21.1°) = 2*424nm = 848nm

d*sin(21.1°) = 3*λ

Solving for λ, we get:

λ = 848nm/3 = 282.67nm

Therefore, the longer wavelength is 282.67nm.

ii) To determine the number of lines per meter in the diffraction grating, we can use the equation d = 1/N, where N is the number of lines per meter. We already know the spacing between the lines (d) and the angle at which the observer is viewing the diffraction pattern (21.1°). So, we can rearrange the equation to solve for N:

N = 1/d = 1/(d*sin(21.1°))

Substituting in the value for d from part (i), we get:

N = 1/282.67nm*sin(21.1°) = 0.0031lines/nm

Therefore, the number of lines per meter in the diffraction grating is 0.0031lines/nm.

iii) To determine the angle(s) at which overlapping occurs, we can use the equation for constructive interference, where d is the spacing between the lines of the diffraction grating, θ is the angle at which the observer is viewing the diffraction pattern