What Is the Longer Wavelength in a Diffraction Grating Experiment?

[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.

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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?