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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掳 is 4.00 * 10-5 W/m^2 . What is the intensity at a point on the screen that corresponds to= 1.20掳?

Short Answer

Expert verified

The intensity at a point on the screen that corresponds to 胃= 1.20掳 is I=2.54脳10-8W/m2.

Step by step solution

01

Intensity of light

We are given

位=592nm=592脳10-9ma=0.0290mm=0.0290脳10-3mI0=4.0脳10-5W/m2胃=1.20掳=蟺150rad

We know that the intensity is given by

I=I0sin尾2尾2尾=2蟺asin胃位

Substitute the intensity formula above

I=I0sin蟺asin胃位蟺asin胃位2I=4.0脳10-5脳sin蟺脳0.029脳10-3sin蟺150529脳10-9蟺脳0.029脳10-3sin蟺150529脳10-92

02

Conclusion

After converting angles into rad mode.

I=2.54脳10-8W/m2

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