Chapter 27, Problem 45 For a wavelength of 420 nm, a diffraction grating produce

Question

Chapter 27, Problem 45 For a wavelength of 420 nm, a diffraction grating produces a bright fringe at an angle of 22°. For an unknown wavelength, the same grating produces a bright fringe at an angle of 37°. In both cases the bright fringes are of the same order m. What is the unknown wavelength? Bright fringe m (unknown wavelength) Bright fringe m (known wavelength) entral bright fringe (m = 0) Bright fringe m (known wavelength) Bright fringe m (unknown wavelength) Diffraction grating Screen Number Units the tolerance is +/-296

Explanation / Answer

In diffraction grating, the constructive interference (brightness) will occur if the difference in their two path lengths is an integral multiple of their wavelength (lambda)

d * sin (theta) = n (lambda)

Theta is the angle of emergence d is the distance between slits (d = 1 / N and N, called the grating constant) and n is the "order number of maxima", (n = + 1, 2, 3, ...)

let grating produce same order (n same) maxima for wavelengths [ lamda-1, and lamda-2] respectively at angles [ theta-1 and theta-2]. First wavelength is known (420nm) other unknown and angles are [ 22 deg and 37 deg]

we can write

d * sin (theta-1) = n (lamda-1) -----(1)

d * sin (theta-2) = n (lamda-2) ----(2)

divide (1) and (2)

(lamda-2) /(lamda-1) = sin (theta-2) /sin (theta-1)

(lamda-2) /(lamda-1) = sin (37) /sin (22)

(lamda-2) = 420 nm * [sin (37) /sin (22)]

(lamda-2) = 420 nm * [0.6018 /0.3746]

(lamda-2) = 420 nm * [1.6065]

(lamda-2) = 674.74 nm

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