Light from Element X through a diffraction Grating

[kikko]

Homework Statement

Light emitted by Element X passes through a diffraction grating having 1200 lines/mm. The diffraction pattern is observed on a screen 75.0 cm behind the grating. Bright fringes are seen on the screen at distances of 56.2 cm, 65.9 cm, and 93.5 cm from the central maximum. No other fringes are seen.

What is the maximum value of m for each of these diffracted wavelengths?
This part will be visible after you complete previous part(s).
What are the wavelengths of light emitted by Element X?

Homework Equations

$$\theta$$_m = m($$\lambda/d$$) (angles of bright fringes)
y_m = (m$$\lambda$$L)/d
dsin$$\theta$$_m = m$$\lambda$$
y_m - Ltan$$\theta$$_m

The Attempt at a Solution

\theta

No clue for this one. I tried plugging various things into the second equation, but got lost. I also don't understand how it can have multiple $$\lambda$$.

 

[kikko]

bump?

 

[SammyS]

What does each variable represent?

θ_m, m, λ, d, y_m, L

 

[SammyS]

Homework Statement

Light emitted by Element X passes through a diffraction grating having 1200 lines/mm. The diffraction pattern is observed on a screen 75.0 cm behind the grating. Bright fringes are seen on the screen at distances of 56.2 cm, 65.9 cm, and 93.5 cm from the central maximum. No other fringes are seen.

What is the maximum value of m for each of these diffracted wavelengths?
This part will be visible after you complete previous part(s).
What are the wavelengths of light emitted by Element X?

Homework Equations

$$\theta$$_m = m($$\lambda/d$$) (angles of bright fringes)
y_m = (m$$\lambda$$L)/d
dsin$$\theta$$_m = m$$\lambda$$
y_m - Ltan$$\theta$$_m

The Attempt at a Solution

\theta

No clue for this one. I tried plugging various things into the second equation, but got lost. I also don't understand how it can have multiple $$\lambda$$.

What is the range of wavelengths for the visible spectrum?

Find where the shortest wavelength (for the visible) would a appear on the screen, for m=1, m=2, ... until you can completely answer the first question.