Question

For a fiber Bragg grating-based sensor, the shift in the Bragg wavelength gives information about the value of the measurand. For a Bragg wavelength of \(1550\) nm and effective index of \(1.44\), the grating period is nm (rounded off to two decimal places).

Hint:

Use the Bragg condition \(\lambda_B = 2 n_{eff} \Lambda\) and solve for the grating period \(\Lambda\).

Answer 1

Correct Answer: 538

Step 1: Recall the Bragg condition for a fiber Bragg grating.
A fiber Bragg grating reflects light strongly at the Bragg wavelength \(\lambda_B\), which is fixed by the effective refractive index \(n_{eff}\) of the fiber core and the physical period \(\Lambda\) of the grating. The first-order Bragg condition is
\[ \lambda_B = 2\, n_{eff}\, \Lambda \]
This comes from requiring the light reflected from successive grating planes to add up in phase, so the round trip over one grating period must equal one wavelength as measured inside the fiber.

Step 2: Rearrange the formula for the grating period.
\[ \Lambda = \frac{\lambda_B}{2\, n_{eff}} \]

Step 3: Substitute the given values.
With \(\lambda_B = 1550\) nm and \(n_{eff} = 1.44\):
\[ \Lambda = \frac{1550}{2 \times 1.44} = \frac{1550}{2.88} \]
\[ \Lambda = 538.194\ldots \text{ nm} \]

Final Answer:
Rounded off to two decimal places, the grating period is \(538.19\) nm. \[ \boxed{538.19 \text{ nm}} \]