Consider a wedge that is 30 cm long and 8 cm thick. Wedge's attenuation is 0.3 /cm. Derive the equation and calculate the average transmission through this wedge.

Elements Of Electromagnetics
7th Edition
ISBN:9780190698614
Author:Sadiku, Matthew N. O.
Publisher:Sadiku, Matthew N. O.
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Consider a wedge that is 30 cm long and 8 cm thick. Wedge's attenuation is 0.3 /cm. Derive the equation and calculate the average transmission through this wedge.

**Problem Statement:**

Consider a wedge that is 30 cm long and 8 cm thick. The wedge's attenuation is 0.3 per cm. Derive the equation and calculate the average transmission through this wedge.

**Diagram Explanation:**

The diagram shows a right-angled wedge with the length (base) of 30 cm and the height (thickness) of 8 cm. The vertical arrows represent the incoming rays that will pass through the wedge.

**Calculation:**

1. The wedge causes attenuation to any ray passing through it, described by the attenuation coefficient, which is given as 0.3 per cm.

2. To compute the transmission, consider a differential element at a distance "x" along the length of the wedge. The thickness "t" at this point is given by the relation:

   \[
   t(x) = \frac{8}{30}x = \frac{4}{15}x
   \]

3. The intensity "I(x)" of a ray passing through a thickness "t(x)" is governed by the exponential attenuation law:

   \[
   I(x) = I_0 e^{-0.3 \cdot t(x)}
   \]

   Where \(I_0\) is the initial intensity of the ray.

4. Substitute \(t(x)\) into the equation:

   \[
   I(x) = I_0 e^{-0.3 \cdot \left(\frac{4}{15}x\right)}
   \]

5. Average transmission can be computed by integrating \(I(x)\) over the range 0 to 30 cm and dividing by the total length:

   \[
   \text{Average Transmission} = \frac{1}{30} \int_{0}^{30} e^{-0.08x} \, dx
   \]

6. Solving this integral:

   \[
   \int e^{-0.08x} \, dx = -\frac{1}{0.08} e^{-0.08x} = -12.5 e^{-0.08x}
   \]

7. Evaluate from 0 to 30:

   \[
   \left[-12.5 e^{-0.08x}\right]_{0}^{30} = -12.5 \left(e^{-2.4} - 1\right)
   \]

8. Calculate:

   \[
   =
Transcribed Image Text:**Problem Statement:** Consider a wedge that is 30 cm long and 8 cm thick. The wedge's attenuation is 0.3 per cm. Derive the equation and calculate the average transmission through this wedge. **Diagram Explanation:** The diagram shows a right-angled wedge with the length (base) of 30 cm and the height (thickness) of 8 cm. The vertical arrows represent the incoming rays that will pass through the wedge. **Calculation:** 1. The wedge causes attenuation to any ray passing through it, described by the attenuation coefficient, which is given as 0.3 per cm. 2. To compute the transmission, consider a differential element at a distance "x" along the length of the wedge. The thickness "t" at this point is given by the relation: \[ t(x) = \frac{8}{30}x = \frac{4}{15}x \] 3. The intensity "I(x)" of a ray passing through a thickness "t(x)" is governed by the exponential attenuation law: \[ I(x) = I_0 e^{-0.3 \cdot t(x)} \] Where \(I_0\) is the initial intensity of the ray. 4. Substitute \(t(x)\) into the equation: \[ I(x) = I_0 e^{-0.3 \cdot \left(\frac{4}{15}x\right)} \] 5. Average transmission can be computed by integrating \(I(x)\) over the range 0 to 30 cm and dividing by the total length: \[ \text{Average Transmission} = \frac{1}{30} \int_{0}^{30} e^{-0.08x} \, dx \] 6. Solving this integral: \[ \int e^{-0.08x} \, dx = -\frac{1}{0.08} e^{-0.08x} = -12.5 e^{-0.08x} \] 7. Evaluate from 0 to 30: \[ \left[-12.5 e^{-0.08x}\right]_{0}^{30} = -12.5 \left(e^{-2.4} - 1\right) \] 8. Calculate: \[ =
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