Q16E - University Physics with Modern Physics

Question

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°?

Step-by-Step Solution

Answer

The intensity at a point on the screen that corresponds to  $\theta$= 1.20° is $I=2.54\times {10}^{-8}W/m^2.$

Step 1: Intensity of light

We are given

$$\begin{gathered} \lambda =592nm=592\times {10}^{-9} m \\ a=0.0290mm=0.0290\times {10}^{-3}m \\ {I}_0=4.0\times {10}^{-5} W/m^2 \\ \theta =1.20°=\frac{\pi }{150} rad \end{gathered}$$

We know that the intensity is given by

                                                

$$\begin{gathered} I=I_0\left(\frac{\sin {\displaystyle \frac{\beta }{2}}}{\frac{\beta }{2}}\right) \\ \beta =\frac{2\pi a\sin \theta }{\lambda } \end{gathered}$$

 

Substitute the intensity formula above

$$\begin{gathered} I=I_0{\left(\frac{\sin \left({\displaystyle \frac{\pi a\sin \theta }{\lambda }}\right)}{\frac{\pi a\sin \theta }{\lambda }}\right)}^2 \\ I=4.0\times {10}^{-5}\times {\left(\frac{\sin \left({\displaystyle \frac{\pi \times 0.029\times {10}^{-3}\sin \left({\displaystyle \frac{\mathrm{\pi }}{150}}\right)}{529\times {10}^{-9}}}\right)}{\frac{\pi \times 0.029\times {10}^{-3}\sin \left({\displaystyle \frac{\mathrm{\pi }}{150}}\right)}{529\times {10}^{-9}}}\right)}^2 \end{gathered}$$

 

Step 2: Conclusion

After converting angles into rad mode.

$I=2.54\times {10}^{-8}W/m^2$