Respuesta :
The equation we use is mλ=dsinθ for intensity maximas. We are given at the first maximum (m=1), it occurs at 17.8 degrees. Thus we can solve for d by substituting known values into our equation.
(1) (632.8*10^-9m)=dsin(17.8) => d = 2.07*10^-6m
Next we want to find the angle at the second maximum (m=2) so we need to solve for θ.
(2) (632.8*10^-9m) = (2.07*10^-6m)sinθ
θ=37.69 degrees
Hopes this helps!
P.S. I hope this is right. If not sorry in advance.
(1) (632.8*10^-9m)=dsin(17.8) => d = 2.07*10^-6m
Next we want to find the angle at the second maximum (m=2) so we need to solve for θ.
(2) (632.8*10^-9m) = (2.07*10^-6m)sinθ
θ=37.69 degrees
Hopes this helps!
P.S. I hope this is right. If not sorry in advance.
The process of diffraction is to composed different light having different wavelengths into the light of the same wavelength. The angle at which the second bright light spot is [tex]37.69^o[/tex]
What is diffraction?
The process of diffraction is to composed different light having different wavelengths into the same wavelength. light components are diffracted at pitches that are defined by the individual wavelengths.
The relation between mass wavelength and the angle at which the bright spots occurs will be,
[tex]\rm {m\lambda= d sin\theta}[/tex]
For ( m=1 ) , [tex]\theta = 17.8 ^o[/tex]
[tex]\rm {1\times 632.8 \times 10^{-9}=d sin17.8^o}[/tex]
[tex]d = 2.07 \times 10^{-6} m[/tex]
For ( m =2 ) , [tex]d = 2.07 \times 10^{-6} m[/tex]
[tex]2\times 632.8= 2.07\times 10^-6 sin \theta[/tex]
[tex]\rm {sin \theta = 0.611388 }[/tex]
[tex]\theta = 37.69^o[/tex]
Therefore the angle at which the second bright light spot is [tex]37.69^o[/tex]
To learn more about diffraction refer to the link ;
https://brainly.com/question/1812927
