Answer Happy

In this problem we are going to use phasors to derive the formula for the intensity I of a double slit pattern on a distant screen as a function of 0. We shall start be considering the electric field of the light arriving at the screen. At any point these are out of phase with each other by some angle . We can write the two electric fields as: E₁ = Ensin(wt) E2 = Eosin(wt + 6) In this problem, ignore any effects from the diffraction pattern of the light around the slits. (i.e. the slits are so narrow that each one radiates equally over the angles we consider.) Part 1) Using phasors or otherwise, derive an equation relating E, the electric field amplitude due to the two waves to E, and . E = Part 2) Now use the relationship between intensity and electric fields: //= E3 to obtain the an expression for the total intensity on the screen, I, in terms of the intensity of each light In and p. I = Part 3) Write an expression for the phase difference between the electric fields at a point on the screen for light with a wavelength X, passing through slits with a slit length d at an angle as shown on the diagram. (The screen is distant so we may approximate the two rays as parallel.) d‡t 0 Figure 1. Two slits are separated by a distance d. The angle to the horizontal from the centre of the two slits is 0. $