Respuesta :
Answer:
[tex]0.54^{\circ}[/tex]
3.99322032 mm
Explanation:
n = Lines per mm = 125
Seperation between slits is given by
[tex]d=\dfrac{1}{n}\\\Rightarrow d=\dfrac{1}{125}\\\Rightarrow d=0.008\ mm[/tex]
[tex]\lambda_1[/tex] = 498 nm
[tex]\lambda_2[/tex] = 569 nm
We have the expression
[tex]dsin\theta_1=m\lambda_1[/tex]
For first maximum m = 1
[tex]\theta_1=sin^{-1}\dfrac{m\lambda_1}{d}\\\Rightarrow \theta_1=sin^{-1}\dfrac{1\times 498\times 10^{-9}}{0.008\times 10^{-3}}\\\Rightarrow \theta_1=3.57^{\circ}[/tex]
[tex]\theta_2=sin^{-1}\dfrac{m\lambda_2}{d}\\\Rightarrow \theta_2=sin^{-1}\dfrac{1\times 569\times 10^{-9}}{0.008\times 10^{-3}}\\\Rightarrow \theta_2=4.08^{\circ}[/tex]
Angular separation is given by
[tex]\Delta \theta=\theta_2-\theta_1\\\Rightarrow \Delta \theta=4.08-3.57\\\Rightarrow \Delta \theta=0.54^{\circ}[/tex]
Angular separation is [tex]0.54^{\circ}[/tex]
Now
[tex]\lambda_1[/tex] = 589 nm
[tex]\lambda_2[/tex] = 589.59 nm
[tex]\Delta \lambda=\lambda_2-\lambda_1\\\Rightarrow \Delta \lambda=589.59-589\\\Rightarrow \Delta \lambda=0.59]\ nm[/tex]
We have the relation
[tex]\dfrac{\lambda}{\Delta \lambda}=mN\\\Rightarrow N=\dfrac{\lambda}{m\Delta \lambda}\\\Rightarrow N=\dfrac{589}{2\times 0.59}\\\Rightarrow N=499.15254[/tex]
Width is given by
[tex]w=\dfrac{N}{n}\\\Rightarrow w=\dfrac{499.15254}{125}\\\Rightarrow w=3.99322032\ mm[/tex]
The width is 3.99322032 mm
The angular separation \theta of the first maxima of these spectral lines generated by this diffraction grating is 0.54°
The width which this grating needs to be to allow you to resolve the two lines 589.00 and 589.59 nanometers is 3.99322032 mm
Calculations and Parameters:
n = Lines per mm
= 125
The Separation between slits is given by:
d= 1/n
d= 1/125
= 0.008mm.
Where
line 1 = 498nm
line 2 = 569nm
The first maximum m= 1 will be:
θ1= 3.57°
θ2= 4.08°
The angular separation would be:
θ2- θ1= 0.54°.
Now, to find the width is:
w= N/n
= 499.15254/125
= 3.99322032 mm.
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