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Chapter 5: Q16E (page 1212)

Monochromatic light of wavelength 592 nm from a distant source passes through a slit that is 0.0290 mm wide. In the resulting diffraction pattern, the intensity at the center of the central maximum= 0° is 4.00 * 10-5 W/m^2 . What is the intensity at a point on the screen that corresponds to= 1.20°?

Short Answer

Expert verified

The intensity at a point on the screen that corresponds to θ= 1.20° is I=2.54×10-8W/m2.

Step by step solution

01

Intensity of light

We are given

λ=592nm=592×10-9ma=0.0290mm=0.0290×10-3mI0=4.0×10-5W/m2θ=1.20°=π150rad

We know that the intensity is given by

I=I0sinβ2β2β=2πasinθλ

Substitute the intensity formula above

I=I0sinπasinθλπasinθλ2I=4.0×10-5×sinπ×0.029×10-3sinπ150529×10-9π×0.029×10-3sinπ150529×10-92

02

Conclusion

After converting angles into rad mode.

I=2.54×10-8W/m2

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