Discover how IDNLearn.com can help you find the answers you need quickly and easily. Our community provides timely and precise responses to help you understand and solve any issue you face.
Sagot :
To determine the color of the light whose second-order bright band forms an angle of [tex]\(13.5^\circ\)[/tex] in a diffraction grating with 175 lines per mm, you can follow these steps:
1. Calculate the distance between lines on the grating:
[tex]\[ d_{\text{mm}} = \frac{1}{175} \text{ mm} \][/tex]
This yields:
[tex]\[ d_{\text{mm}} = 0.005714285714285714 \text{ mm} \][/tex]
2. Convert the distance from millimeters to meters:
[tex]\[ d_{\text{m}} = d_{\text{mm}} \times 10^{-3} \][/tex]
This yields:
[tex]\[ d_{\text{m}} = 0.005714285714285714 \times 10^{-3} \text{ m} \][/tex]
[tex]\[ d_{\text{m}} = 5.7142857142857145 \times 10^{-6} \text{ m} \][/tex]
3. Convert the angle from degrees to radians:
[tex]\[ \theta_{\text{rad}} = \theta_{\text{deg}} \times \frac{\pi}{180} \][/tex]
For [tex]\(\theta_{\text{deg}} = 13.5^\circ\)[/tex]:
[tex]\[ \theta_{\text{rad}} = 13.5 \times \frac{\pi}{180} \][/tex]
[tex]\[ \theta_{\text{rad}} = 0.23561944901923448 \text{ radians} \][/tex]
4. Use the diffraction grating formula to calculate the wavelength:
The general formula for diffraction grating is:
[tex]\[ d \sin(\theta) = n \lambda \][/tex]
For the second-order ([tex]\(n = 2\)[/tex]):
[tex]\[ \lambda = \frac{d \sin(\theta)}{n} \][/tex]
Substituting the given values:
[tex]\[ \lambda = \frac{5.7142857142857145 \times 10^{-6} \times \sin(0.23561944901923448)}{2} \][/tex]
[tex]\[ \lambda = 6.669867538740154 \times 10^{-7} \text{ m} \][/tex]
5. Convert the wavelength from meters to nanometers:
[tex]\[ \lambda_{\text{nm}} = \lambda \times 10^9 \][/tex]
[tex]\[ \lambda_{\text{nm}} = 6.669867538740154 \times 10^{-7} \times 10^9 \][/tex]
[tex]\[ \lambda_{\text{nm}} = 666.9867538740154 \text{ nm} \][/tex]
6. Determine the color based on the wavelength:
Using the given wavelength-color relationship:
- Red: 650 nm
- Yellow: 570 nm
- Green: 510 nm
- Violet: 400 nm
The calculated wavelength of 666.9867538740154 nm falls closest to the "Red" range.
Therefore, the most likely color of the light is red.
1. Calculate the distance between lines on the grating:
[tex]\[ d_{\text{mm}} = \frac{1}{175} \text{ mm} \][/tex]
This yields:
[tex]\[ d_{\text{mm}} = 0.005714285714285714 \text{ mm} \][/tex]
2. Convert the distance from millimeters to meters:
[tex]\[ d_{\text{m}} = d_{\text{mm}} \times 10^{-3} \][/tex]
This yields:
[tex]\[ d_{\text{m}} = 0.005714285714285714 \times 10^{-3} \text{ m} \][/tex]
[tex]\[ d_{\text{m}} = 5.7142857142857145 \times 10^{-6} \text{ m} \][/tex]
3. Convert the angle from degrees to radians:
[tex]\[ \theta_{\text{rad}} = \theta_{\text{deg}} \times \frac{\pi}{180} \][/tex]
For [tex]\(\theta_{\text{deg}} = 13.5^\circ\)[/tex]:
[tex]\[ \theta_{\text{rad}} = 13.5 \times \frac{\pi}{180} \][/tex]
[tex]\[ \theta_{\text{rad}} = 0.23561944901923448 \text{ radians} \][/tex]
4. Use the diffraction grating formula to calculate the wavelength:
The general formula for diffraction grating is:
[tex]\[ d \sin(\theta) = n \lambda \][/tex]
For the second-order ([tex]\(n = 2\)[/tex]):
[tex]\[ \lambda = \frac{d \sin(\theta)}{n} \][/tex]
Substituting the given values:
[tex]\[ \lambda = \frac{5.7142857142857145 \times 10^{-6} \times \sin(0.23561944901923448)}{2} \][/tex]
[tex]\[ \lambda = 6.669867538740154 \times 10^{-7} \text{ m} \][/tex]
5. Convert the wavelength from meters to nanometers:
[tex]\[ \lambda_{\text{nm}} = \lambda \times 10^9 \][/tex]
[tex]\[ \lambda_{\text{nm}} = 6.669867538740154 \times 10^{-7} \times 10^9 \][/tex]
[tex]\[ \lambda_{\text{nm}} = 666.9867538740154 \text{ nm} \][/tex]
6. Determine the color based on the wavelength:
Using the given wavelength-color relationship:
- Red: 650 nm
- Yellow: 570 nm
- Green: 510 nm
- Violet: 400 nm
The calculated wavelength of 666.9867538740154 nm falls closest to the "Red" range.
Therefore, the most likely color of the light is red.
Your presence in our community is highly appreciated. Keep sharing your insights and solutions. Together, we can build a rich and valuable knowledge resource for everyone. For dependable answers, trust IDNLearn.com. Thank you for visiting, and we look forward to assisting you again.
