Question

A laser beam of wavelength 740 nm shines through a diffraction grating that has 750 lines/mm and observed on a screen 1.4 m behind the grating.

Answer 1

Here is the full question

A laser beam of wavelength 740 nm shines through a diffraction grating that has 750 lines/mm and observed on a screen 1.4 m behind the grating.

a) How many bright fringes can be observed on a screen?

b) What is the distance between m = 0 and m = 1 bright fringes?

Answer:

a) 3

b) 0.94 m

Explanation:

The wavelength $( \lambda)$ of the laser beam =  740 nm = 7.40 × 10 ⁻⁹ m

The width of the slit (d) = $\frac{1}{number of lines per nm (N)}$

= $\frac{1}{750*10^3lines/m}$

= $1.33*10^{-6}m$

Bright fringe is denoted by $m^{th}$ and the expression for calculating it is given as;

dsin 90° = $m \lambda$

$d = m \lambda$

$m =\frac{d}{\lambda}$

$m=\frac{1.33*10*10^{-6}}{740*10^{-9}}$

m = 1.80

m ≅ 2

Hence, the maximum number of the bright fringes (N) = 2(1) +1 = 3

b)

To determine the condition of maxima (m) grating is:

dsinθ = mλ

Since m= 0; the bright fringe formation is at the center and there is no point calculating that; since all value from the result will point back to zero

For m=1; then d = 1.33 × 10⁻⁶ ; λ = 740 × 10 ⁻⁹ m

dsinθ = mλ

substituting our values ; we have:

(1.33 × 10⁻⁶)sin θ = (1)(740 × 10 ⁻⁹ m)

sin θ = $\frac{(1)(740*10^{-9}m)}{(1.33*10^{-6})}$

sin  θ = 0.5564

θ = sin⁻¹ (0.5564)

θ = 33.81°

For distance (d) at which maxima (m) =1; we have :

tan θ = $\frac{d}{1.4m}$

tan (33.81°) = $\frac{d}{1.4m}$

d = tan (33.81°) (1.4m)

d = 0.9376 m

d = 0.94 m

∴ the distance  between m = 0 and m = 1 bright fringes will be:

S =  0.94 m - 0 m

S =  0.94 m