He-Ne light is a gas laser in which He is excited, and those ions then collide with Neon to release energy. Of the most common spectrums of this laser, 633 nm is the most common wavelength.
d sinTheta=m lambda
d = 633*E-9/sin40.7
so, the question: d sinGamma=m lambda
633E-9/sin40.7 * sinGamma=418E-9
solve for The angle Gamma.
b. 980khz*wavelenth= speed light
solve fodr wavelength, then the size of the antenna is half that.
1) If the first order maximum of He-Ne light exits a diffraction grating at an angle of 40.7o, at what angle will the first order maximum of violet light with a wavelength of 418 nm exit?
What is He-Ne light?????
2) The most efficient antennae have a size of half the wavelength of the radiation they are emitting. How long should an antenna be to broadcast at 980 kHz?
2 answers
HeNe laser operates at a wavelength of 632.8 nm in the red part of the visible spectrum
d•sinφ = k•λ,
for He-Ne:
d•sin40.7º =1•632.8•10^-9….(1)
For violet light:
d•sinφ=1•418•10^-9 ……(2)
Divide(1) by (2)
d•sin40.7º/ d•sin φ =632.8•10^-9/418•10^-9,
sinφ = sin40.7º •418/632.8 = 0.433.
φ =arcsin 0.433 =25.68º.
λ = c/f = 3•10^8/980000 =305.9 m
L = λ/2 =153 m.
d•sinφ = k•λ,
for He-Ne:
d•sin40.7º =1•632.8•10^-9….(1)
For violet light:
d•sinφ=1•418•10^-9 ……(2)
Divide(1) by (2)
d•sin40.7º/ d•sin φ =632.8•10^-9/418•10^-9,
sinφ = sin40.7º •418/632.8 = 0.433.
φ =arcsin 0.433 =25.68º.
λ = c/f = 3•10^8/980000 =305.9 m
L = λ/2 =153 m.
5 answers