Light of wavelength 600.0 nm is incident on a single slit of width 0.100 mm. The distance between the slit and the screen is 2.40 m. What is the width of the central maximum?

Respuesta :

Answer:

It is 0.0288

Explanation:

The equation of single slit interference phenomenon for light is:

[tex]a\sin\theta=m\lambda [/tex] (1)

With a the width of the slit, λ the wavelength of the light, m the order of the minimum and θ the angular position of the minimum, because the distance (x) between screen and slit is so bigger respect the distance (y) of the first minima respect the center of the screen we can use the approximation:

[tex]\sin\theta\approx\tan\theta=\frac{y}{x} [/tex] (2)

by (2) on (1):

[tex]a\frac{y}{x}=m\lambda [/tex]

solving for y:

[tex]y=\frac{m\lambda x}{a} [/tex]

Because a maximum is approximately between two minima, the first maximum is between the minima of order m=1 and m=-1

[tex]y=\frac{1(600\times10^{-9})(2.4)}{0.100\times10^{-3}}=0.0144m  [/tex]

because central maximum is symmetric respect the center of the screen we only have to find the distance to the minimum of order m=1 and multiply by two:

[tex]width_{maximum}=2y=2(0.0144)=0.0288 m [/tex]