Find an expression for the positions y 1 of the first-order fringes of a diffrac

Question

Find an expression for the positions y1 of the first-order fringes of a diffraction grating if the line spacing is large enough for the small-angle approximation tansin to be valid. Your expression should be in terms of d, L, and .

Express your answer in terms of some or all of the variables d, L, .

y1=

Use your expression from part A to find an expression for the separation y on the screen of two fringes that differ in wavelength by .

Express your answer in terms of some or all of the variables d, L, .

y =

Rather than a viewing screen, modern spectrometers use detectorssimilar to the one in your digital camerathat are divided into pixels. Consider a spectrometer with a 333 line/mm grating and a detector with 100 pixels/mm located 12 cm behind the grating. The resolution of a spectrometer is the smallest wavelength separation min that can be measured reliably. What is the resolution of this spectrometer for wavelengths near 550 nm, in the center of the visible spectrum? You can assume that the fringe due to one specific wavelength is narrow enough to illuminate only one column of pixels.

Express your answer with the appropriate units.

min =

Explanation / Answer

Condition for maximum is:

d sin(theta) = m lambda

for m = 1

d sin(theta) = lambda

for small angles,

tan(theta) = sin(theta) = theta = y/L

d y/L = lambda

y = lambda L/d

b)delta y = delta (lambda) L/d

c)d = 333 lines/mm = 0.33 lines/m ; L = 12 cm = 0.12 m

y = 550 x 10^-9 x 0.12/0.33 = 200 x 10^-9 m

Hence, y = 200 x 10^-9 m

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