Aslit 0.240mm wide is illuminated by parallel light rays with a wavelength 600nm . the diffraction pattern is observed on a screen that is 3.50m from the slit. the intensity at the center of the central maximum (theta = 0) is 4.10�10? 6 w/m^2. what is the distance on the screen from the center of the central maximum to the first minimum? what is the intensity at a point on the screen midway between the center of the central maximum and the first minimum? expert answer

To solve this exercise, it is necessary to apply the concepts on the principle of superposition, specifically on constructive interference,

Constructive interference (light spot) is defined by

Where,

m = The integer m is called the interference order and is the number of wavelengths by which the two paths differ.

d = Distance

For smaller angles ,

From the trigonometric properties it is understood that so the is - in this case - the length measured vertically by reason of distance, that is

Re-arrange to find y,

Replacing our values

PART B) To calculate the intensity it is necessary to find the angle between the previously calculated height and distance in order to calculate the phase angle, in other words,

Therefore phase angle is

The intensity formula would then be given by,

hi,

what occurs during a solar eclipse?

there are small tides across earth.

the moon is between earth and the sun.

p.s. im not sure if the first answer is correct because during a solar ecliphse there are spring tides and i dont know if they are the same.