Question

Using a diffraction grating with a spacing of 0.00040 cm, a line appears 28.5 cm from the central line on the screen. The screen is 2.00 m from the grating. What is the wavelength of the light?

Group of answer choices

4.8×10−7 m

3.6×10−7 m

6.6×10−7 m

5.7×10−7 m

Answer 1

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**.