Suppose that you have a reflection diffraction grating with n= 125 lines per millimeter. Light from a sodium lamp passes through the grating and is diffracted onto a distant screen. PART A
Two visible lines in the sodium spectrum have wavelengths 498\rm nm and 569 \rm nm. What is the angular separation \Delta \theta of the first maxima of these spectral linesgenerated by this diffraction grating?
PART B
How wide does this grating need to be to allow you to resolvethe two lines 589.00 and 589.59 nanometers, which are a well knownpair of lines for sodium, in the second order (m=2)?

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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