Which of these is the theoretical maximum number of bright fringes visible in the diffraction pattern when green (I = 500 nm) light shines through double slits separated by 2.1 * 10^-6m?

Consider a single-slit diffraction pattern for =589 nm, projected on a screen that is 1.00 m from a slit of width 0.25 mm. How far from the center of the pattern are the centers of the first and second dark fringes?

A beam of monochromatic green light is diffracted by a slit of width 0.550 mm. The diffraction pattern forms on a wall 2.06 m beyond the slit. The distance between the positions of zero intensity on both sides of the central bright fringe is 4.10 mm. Calculate the wavelength of the light.

In Figure P27.7 (not to scale), let L = 1.20 m and d = 0.120 mm and assume the slit system is illuminated with monochromatic 500-nm light. Calculate the phase difference between the two wave fronts arriving at P when (a) = 0.500 and (b) y = 5.00 mm. (c) What is the value of for which the phase difference is 0.333 rad? (d) What is the value of for which the path difference is /4?

Eight slits equally separated by 0.149 mm is uniformly illuminated by a monochromatic light at =523 nm. What is the intensity of a secondary maxima compared to that of the principal maxima?

Red light (wavelength 632.8 nm in air) from a Helium-Neon laser is incident on a single slit of width 0.05 mm. The entire apparatus is immersed in water of refractive index 1.333. Determine the angular width of the central peak.

What is the angular width of the central fringe of the interference pattern of (a) 20 slits separated by d=2.0103 mm? (b) 50 slits with the same separation? Assume that =600 nm.

A single slit of width 2100 nm is illuminated normally by a wave of wavelength 632.8 nm. Find the phase difference between waves from the top and one third from the bottom of the slit to a point on a screen at a horizontal distance of 2.0 m and vertical distance of 10.0 cm from the center.

Show that the distribution of intensity in a double-slit pattern is given by Equation 36.9. Begin by assuming that the total magnitude of the electric field at point P on the screen in Figure 36.4 is the superposition of two waves, with electric field magnitudes E1=E0sintE2=E0sin(t+) The phase angle in in E2 is due to the extra path length traveled by the lower beam in Figure 36.4. Recall from Equation 33.27 that the intensity of light is proportional to the square of the amplitude of the electric field. In addition, the apparent intensity of the pattern is the time-averaged intensity of the electromagnetic wave. You will need to evaluate the integral of the square of the sine function over one period. Refer to Figure 32.5 for an easy way to perform this evaluation. You will also need the trigonometric identity sinA+sinB=2sin(A+B2)cos(AB2)

For 600-nm wavelength light and a slit separation of 0.12 mm, what are the angular positions of the first and third maxima in the double slit interference pattern?