Shown above is the intensity pattern observed on a screen 1.6 m from an arrangem

Question

Shown above is the intensity pattern observed on a screen 1.6 m from an arrangement of N-slits. The separation between adjacent slits is 0.01 mm, and the light incident on the slits has a wavelength of 585 nm. We observe a single secondary intensity peak between each of the maxima. These secondary peaks are one ninth the intensity of the maxima. An arrow is drawn on the diagram indicating another position where the intensity is also one ninth of the maxima but which does not correspond to a secondary peak. What is the smallest distance from the center of the screen where you can observe such a point?

Explanation / Answer

here,

distance to screen, D = 1.6 m
wavelength, w = 585 nm = 585*10^-9 m
slit seperation, d = 0.01 mm = (0.01*10^-3) m

From Diffraction Experiment,
Smallest distance from centre to screen , x
x = m*w*D/d (m is order of diffraction)
x = (1 * (585*10^-9) * 1.6)/(0.01*10^-3)
x = 0.094 m

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