Respuesta :
Answer:
The total number of maxima produced is [tex]m_T = 7[/tex] maxima
Explanation:
From the question we are told that
The number of lines per cm is [tex]n = 5000 \ lines/cm[/tex]
The wavelength of the light is [tex]\lambda = 633 nm = 633 *10^{-9} \ m[/tex]
Now the distance between the lines is mathematically evaluated as
[tex]d = \frac{1}{n}[/tex]
substituting values
[tex]d = \frac{1}{5000}[/tex]
[tex]d = \frac{1 *10^{-2}}{5000}[/tex] N/B - this statement convert it from cm to m
[tex]d = 2 *10^{ -6} \ m[/tex]
Generally the condition for diffraction i mathematically represented as
[tex]dsin(\theta ) = m \lambda[/tex]
at maximum [tex]\theta = 90 ^o[/tex]
[tex]d sin (90) = \lambda m[/tex]
here m is the number of maxima
Thus making m the subject we have
[tex]m = \frac{d sin (90)}{ \lambda }[/tex]
So [tex]m = \frac{2*10^{-6} sin (90)}{ 633 *10^{-9}}[/tex]
[tex]m = 3.2[/tex]
=> m =3
Now the total number of maxima would include the bright fringe(3) and dark fringe (3) plus the central maxima (1)
Thus
[tex]m_T = 3 + 3 +1[/tex]
[tex]m_T = 7[/tex] maxima