consider a square of wire side length d= 0.422 as shown in the figure 4. (Chapt
ID: 2039478 • Letter: C
Question
consider a square of wire side length d= 0.422 as shown in the figure 4. (Chapt 33, LO 4) A double-slit interference pattern is light incident upon two narrow slits spaced 0.239ted by monochromatic the 3rd and 4th maxima on a screen 1.75 m away is 3.08 mm amapart. The distance between (a) (S pts) What is the wavelength of the light? (b) (3 pts) If the wavelength of the light was shorter, would the distance between the 3rd and 4th maxima on a screen increase or decrease? (circle one) 5. (Chapt 33, LO 4) 687 nm light illuminates a single slit. The width of the bright single-slit central diffraction maximum on a screen 1.25 m away is 1.89 mm. (a) (5 pts) What is the width of the slit? (b) (3 pts) If the wavelength of the light was longer, would the width of the bright central maximum be greater or less? (circle one) 5 of 6Explanation / Answer
in interference patterns
use the formula
Y = mLR/d
here given Ymn = (m-n) * LR/d
wavelength L = Ymn * d/((m-n)*R)
L = 3.08 *10^-3 * 0.239 *10^-9/((4-3)*1.75)
L = wavelength = 420.64 nm
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as Y is directly porportional to L
if L increases, Y also increases
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5. '
again y = mLR/d
d = 1 * 687*10^-9 * 1.25/(1.89 *10^-3)
d = 0.454 mm
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if Wavelength is longer, Widht of bright Central maxima be greater
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