Helium atoms emit light at several wavelengths. light from a helium lamp illuminates a diffraction grating and is observed on a screen 50.00 cm behind the grating. the emission at wavelength 501.5 nm creates a first-order bright fringe 21.90cm from the central maximum. what is the wavelength of the bright fringe that is 31.60cm from the central maximum?

wavelength = 667.94 nm

Explanation:

Given data

screen distance L = 50.00 cm = 0.50 m

wavelength = 501.5 nm

bright fringe d = 21.90 cm = 0.2190 m

bright fringe  D = 31.60 cm = 0.3160 m

to find out

wavelength

solution

we will apply here triangle law

tan (θ) = d / L

(θ) = tan^-1 ( 0.219 / 0.50)

(θ) = 23.65 degree

so

d sin (θ) = 1 ( wavelength )

d = 501.5 / sin (23.65)

d = 1250.1 nm

and

for another angle

tan (θ) = d / L

(θ) = tan^-1 ( 0.316 / 0.50)

(θ) = 32.3 degree

so

d sin (θ) = 1 ( wavelength )

wavelength =1250 sin (32.3)

wavelength = 667.94 nm

3. dark matter

empty space could also be a simple vacuum