Monochromatic light with wavelength 690 nm is incident on a slit with width 0.0329 mm. the distance from the slit to a screen is 3.2 m. consider a point on the screen 1.3 cm from the central maximum. calculate (a) θ for that point, (b) α, and (c) the ratio of the intensity at that point to the intensity at the central maximum.

Step-by-step explanation:

We have given,

distance of the slit from the screen(d) = 3.2m

distance from the central maximum(m) = 1.3cm = 0.013m

Now,

(a). The expression for the angle of the inclination of the wave is ( )

Thus the angle of inclination of the wave is

(b). The expression for the angle

after solving the equation we get,

Thus the angle = 0.61 radian

(c). the expression for the ratio of the intensity is

Thus the ratio of the intensity is 0.0305

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idc

step-by-step explanation:

idk