Question

Someone please help me on the second part to this question... thanks in advance!

A physicist illuminates a 0.56 mm-wide slit with light characterized by i = 540 nm, and this results in a diffraction pattern forming upon a screen located 124 cm from the slit assembly. Compute the width of the first and second maxima (or bright fringes) on one side of the central peak. (Enter your answer in mm.) W1 = 1.20 ✓ mm (1st maxima) W2 = 2.40 Other than the central maximum, the width of any low-order bright fringe will be equal to the distance between adjacent dark fringes. Use the formula for the position of a dark fringe in a single-slit diffraction pattern. Apply it to adjacent fringes on the same side of the central peak (that is, [Am] = 1, with both m values positive or both negative). Take care with units, mm (2nd maxima) Need Help? Read It

Answer 1

Use the equation for diffraction of light and find the distance between the two adjacent minima to find the width of the maxima as shown below

Width of and maxima is distance between 2nd and 3rd minimum. So, dsino = na {d=slit width, n = wavelength for and minima, n=2 - sing2 = 22 for 3rd minima, n=3 1 sings = 32 So, y = Dtano {D = 124 cm - y = distance of minimum So, width = Y3-Y2 from center. W2 =D (tanda-tanda) Since Oz and Og are very small tano, ~ Sines and tan02= sino So, W₂ = Y, Y₂ = D(sinos - sinoz) =0( 37 - 2A) = Da = (1.24m)(540nm) = 1.2 mm 0:56mm