Question

A diffraction grating has 3 200 rulings/cm. On a screen 2.50 m from the grating, it is found that for a particular order m, the maxima corresponding to two closely spaced wavelengths of sodium (589.0 nm and 589.6 nm) are separated by 1.21 mm. Determine the value of m.
m = ____

m is not 3,4 or 5

PLEASE ONLY ANSWER IF YOU KNOW ITS CORRECT

Answer 1

Diffraction grating, d = 3200 rulings/cm = [(0.01) / (3200)] m

d = 3.12 x 10^-6 m

Under the assumption that sin $\theta$ $\approx$ tan $\theta$ = (y / L)

We know that, $\lambda$ ₁ correspond to the wavelength that is higher on the screen, while $\lambda$ ₂ is lower on the screen.

For a bright fringes, we have

d sin $\theta$ ₁ = m $\lambda$ ₁

d (y₁ / L) = m $\lambda$ ₁                                                  { eq.1 }

And

d sin $\theta$ ₂ = m $\lambda$ ₂

d (y₂ / L) = m $\lambda$ ₂

d [y₁ - (1.21 x 10^-3 m)] / L = m $\lambda$ ₂                                           

d (y₁ / L) - (1.21 x 10^-3 m) (d / L) = m $\lambda$ ₂                                              { eq.2 }

inserting the value of eq.1 in eq.2 & we get -

m $\lambda$ ₁ - (1.21 x 10^-3 m) (d / L) = m $\lambda$ ₂   

m ( $\lambda$ ₁ - $\lambda$ ₂) = (1.21 x 10^-3 m) (d / L)

m [(589.6 - 589) x 10^-9 m] = [(1.21 x 10^-3 m) (3.12 x 10^-6 m)] / (2.5 m)

m = (1.51008 x 10^-9 m) / (0.6 x 10^-9 m)

m = 2.5