Physics doubt - URGENT!!!

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Oakenari
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#1
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#1
Question: A diffraction grating is used to measure the wavelength of monochromatic light, as shown in the diagram. (not present). The spacing of the slits in the grating is 1.00x10^-6m. The angle between the first order diffraction maxima is 70.0 degrees. What is the wavelength of the light?
A 287nm
B 470nm
C 574nm
D 940nm
Answer: C
I'm unsure about how they got this answer. Can anyone help me please? I have my exam tomorrow.
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Eimmanuel
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(Original post by Oakenari)
Question: A diffraction grating is used to measure the wavelength of monochromatic light, as shown in the diagram. (not present). The spacing of the slits in the grating is 1.00x10^-6m. The angle between the first order diffraction maxima is 70.0 degrees. What is the wavelength of the light?
A 287nm
B 470nm
C 574nm
D 940nm
Answer: C
I'm unsure about how they got this answer. Can anyone help me please? I have my exam tomorrow.
You just need to apply the diffraction grating formula:
d sin θ = n λ
where d = 1.00x10^-6m,
θ = 35 degrees = 70/2
n = 1.

https://www.cyberphysics.co.uk/topic...difraction.htm
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Oakenari
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#3
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#3
Thank you. That makes much more sense to me. My mistake was with the angle.
(Original post by Eimmanuel)
You just need to apply the diffraction grating formula:
d sin θ = n λ
where d = 1.00x10^-6m,
θ = 35 degrees = 70/2
n = 1.

https://www.cyberphysics.co.uk/topic...difraction.htm
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scotfield12
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#4
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#4
A parallel beam of red light of wavelength 630 nm is incident normally (900) on a diffraction grating of 450 lines per millimetre.Calculate the number of principal bright fringes observed.help please
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Callicious
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#5
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#5
(Original post by Skully -_-)
d sinθ = n λ ; d=1.00 x 10–6 m and θ=70°/2 = 35° (angle between the first order maxima is 70° so the angle between n=0 and n=1 is 35°) and n=1 –> 1.00 x 10-6 x sin 35 = 1 λ –> λ= 5.74 x 10-7 m (3 s.f.) = 574 nm.Option C is the correct answer.
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