A white light shines through a diffraction grating that is ruled at 900 lines/cm. A detector is 1.2 cm long and is to be used to measure light between 564.6 and 571.2 nm. If the detector is to measure light from the 7 th order spectrum, how far from the grating should the screen with the detector on it be placed

Explanation:

Given that,

Diffraction grating is ruled by 900 lines/cm

The detector length is

h= 1.2cm =1.2/100= 0.012m

Short wavelength is λs= 564.6nm

Long wavelength is λl = 571.2nm

Question: how far from the grating should the screen with the detector on it be placed. x=?

Since there are 900 lines per centimeter, each line is separated by 1/900 of a centimeter.

The distance between slits is d is given as

d = 1/900 cm

d = 0.00111cm

d = 0.0000111m = 1.11×10^-5 m

Let us call the two angles θs for short-wavelength (564.6nm) and θl for long-wavelength (571.2nm). Diffraction grating is given as

d sin(θs)= mλl

sin(θs)= mλl / d

Since we are given 7th order, then m=7

sin(θs)= 7×564.6×10^-9 / 1.11×10^-5

sin(θs)= 0.3557

θs = arcsin(0.3557)

θs = 20.84°

Applying the same principle for long wavelength

d sin(θl)= mλl

sin(θl)= mλl / d

Since we are given 7th order, then m=7

sin(θl)= 7×571.2×10^-9 / 1.11×10^-5

sin(θl)= 0.3599

θl = arcsin(0.3599)

θl = 21.1°

Using trigonometry,

Then, tanθ = opposite/adjacent

Adjacent is the distance between the screen and the detector

Opposite Is the length of detector, which serves like a height

Tanθl = h/x

x= h / Tan(θl)

x = 0.012 / Tan(21.1)

x = 0.0311m

x =3.11cm

Check attachment for better understanding the how I used the trigonometric function and the triangle I used.

i already answered it by 34/s