Respuesta :
Answer:
Part a)
[tex]f = 4.76 \times 10^{14} Hz[/tex]
Part b)
[tex]d = 3.48 \times 10^{-4} m[/tex]
Part c)
[tex]\theta = 0.311 degree[/tex]
Explanation:
Part a)
As we know that the speed of light is given as
[tex]c = 3 \times 10^8 m/s[/tex]
[tex]\lambda = 630 nm[/tex]
now the frequency of the light is given as
[tex]f = \frac{c}{\lambda}[/tex]
so we have
[tex]f = \frac{3 \times 10^8}{630 \times 10^{-9}}[/tex]
[tex]f = 4.76 \times 10^{14} Hz[/tex]
Part b)
Position of Nth maximum intensity on the screen is given as
[tex]y_n = \frac{n\lambda L}{d}[/tex]
so here we know for 3rd order maximum intensity
[tex]y_3 = 0.76 cm[/tex]
n = 3
L = 1.4 m
[tex]0.76 \times 10^{-2} = \frac{3(630 \times 10^{-9})(1.4)}{d}[/tex]
[tex]d = 3.48 \times 10^{-4} m[/tex]
Part c)
angle of third order maximum is given as
[tex]d sin\theta = 3 \lambda[/tex]
[tex]3.48 \times 10^{-4} sin\theta = 3(630 \times 10^{-9})[/tex]
[tex]\theta = 0.311 degree[/tex]
