To find the wavelength of the light using the diffraction grating equation, we can use the following formula:
\[ d \sin(\theta) = m \lambda \]
where:
- \(d\) is the spacing of the grating,
- \(\theta\) is the angle of the diffraction maximum,
- \(m\) is the order of the maximum (1 for the first-order maximum, etc.),
- \(\lambda\) is the wavelength of the light.
Given Data:
- Grating spacing, \(d = 0.00040 , \text{cm} = 0.00040 \times 10^{-2} , \text{m} = 4.0 \times 10^{-6} , \text{m}\)
- Distance from the grating to the screen, \(L = 2.00 , \text{m}\)
- Distance of the line from the central line on the screen, \(y = 28.5 , \text{cm} = 0.285 , \text{m}\)
Step 1: Calculate \(\theta\)
We can use the tangent function to find \(\theta\):
\[ \tan(\theta) = \frac{y}{L} \]
Using the values:
\[ \tan(\theta) = \frac{0.285}{2.00} \]
Calculating \(\tan(\theta)\):
\[ \tan(\theta) = 0.1425 \]
Now, calculate \(\theta\):
\[ \theta = \tan^{-1}(0.1425) \approx 8.2^\circ \]
Step 2: Calculate \(\sin(\theta)\)
Using the angle \(\theta\) to find \(\sin(\theta)\):
\[ \sin(8.2^\circ) \approx 0.1423 \]
Step 3: Use the diffraction equation
Assuming we are looking for the first-order maximum (\(m=1\)):
\[ d \sin(\theta) = m \lambda \] \[ \lambda = \frac{d \sin(\theta)}{m} \]
Substituting the values:
\[ \lambda = \frac{(4.0 \times 10^{-6} , \text{m}) \cdot (0.1423)}{1} \]
Calculating \(\lambda\):
\[ \lambda \approx 5.692 \times 10^{-7} , \text{m} \]
Conclusion
After rounding, we find:
\[ \lambda \approx 5.7 \times 10^{-7} , \text{m} \]
Thus, the closest answer to the wavelength of light is:
5.7×10−7 m.
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