An experiment is set-up with a monochromatic light source, a double slit, and a

Question

An experiment is set-up with a monochromatic light source, a double slit, and a screen, similar to what is shown on page 7-2 of your lab manual. The experimenter measures and sketches the intensity pattern shown below. The scale along the bottom of the figure is in centimeters. What is the distance 2x_3 between the third maxima in the intensity pattern? the central peak is the "zeroth maximum", the two peaks closest to this on either side are the "first maxima", the next two peaks are the "second maxima", etc. (Units required, and you must be within 1 mm to be correct.) 2x_3 = The slit-to-screen distance is D = 1.800 m, and the slit separation is 0.173 mm. Use the formula for double-slit diffraction maxima and your value of 2x_3 to find the light source's wavelength A. (Units required.) If the light source's wavelength were set to 501 nm, but everything else left the same, what would the distance between the second maxima 2x_2 become? (Units required.) When A = 501 nm then 2x_2 =

Explanation / Answer

part a:

from the graph 2*x3=distance between the two third maxima=2*1.9=3.8 cm

part b:

D=1.8 m

slit separation=d=0.173 mm=173*10^(-6) m

for 3rd order maxima, distance from central maxima=3*lambda*D/d

==>1.9*0.01=3*lambda*1.8/(173*10^(-6))

==>lambda=1.9*0.01*173*10^(-6)/(3*1.8)=6.087*10^(-7) m

part c:

new wavelength=lambda=501 nm=501*10^(-9) m

distance of second maxima from central maxima=2*lambda*D/d

=2*501*10^(-9)*1.8/(173*10^(-6))=0.010425 m

then 2*x3=2*0.010425=0.020850 m

=2.085 cm

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