A laser emitting light with a wavelength of 560 nm is directed at a single slit, producing an interference pattern on a screen that is 3.0 m away. The central maximum is 5.0 cm wide. Determine the width of the slit and the distance between adjacent maxima.
What would the effect on this pattern be, if
the width of the slit was smaller?
the screen was moved further away?
a larger wavelength of light was used?

How would this interference pattern differ if the light was shone through a double slit?
diffraction grating?

a) a = 6.72 10⁻⁵ m,  b)  the slit (a) is smaller, which represents a wider pattern

Explanation:

In is a diffraction experiment since we have a single slit, it is explained by the equation

          a sin θ = m λ

where a is the width of the slit

The diffraction pattern is characterized by a very intense central maximum, with a value of 5.0 cm, therefore the distance from the center to the first zero is y = 5.0 / 2 cm = 2.5 10⁻² m

let's use trigonometry to enter the angle

         tan θ = y / L

         tan θ = sin θ / cos θ = sin θ

         sin θ = y / L

substituting into the equation

          a y / L = m λ

the first maximum occurs for m = 1

          a = λ L / y

let's calculate

          a = 560 10⁻⁹ 3.0 / 2.5 10⁻²

          a = 6.72 10⁻⁵ m

b) if the width a of the slit (a) is smaller

        sin θ = m λ / a

therefore the sinus increases, which implies a greater angle, which represents a wider pattern

c) if the distance to the screen (L) goes away

           y = m λL / a

If L increases the width of the pattern they also increase of course the intensity must be less

d) If the wavelength increases

    In this case the width of the pattern also increases

e) What happens if the light passes through two slits in this case we have two diffraction patterns one centered in each slit and the resulting pattern is the sum of these patterns, this sum gives the double slit interference that characterizes a series of slits of equal height

muscle endurance

explanation:

i can do 50 in a row ( but i get so tired after that)

double the distance = divide strength by 4