Question

You are given four slabs of lossless dielectric, all with the same intrinsic impedance, \(\eta\), known to be different from that of free space. The thickness of each slab is \(\lambda / 4\), where \(\lambda\) is the wavelength as measured in the slab material. The slabs are to be positioned parallel to one another, and the combination lies in the path of a uniform plane wave, normally incident. The slabs are to be arranged such that the airspaces between them are either zero, one-quarter wavelength, or one-half wavelength in thickness. Specify an arrangement of slabs and airspaces such that \((a)\) the wave is totally transmitted through the stack, and \((b)\) the stack presents the highest reflectivity to the incident wave. Several answers may exist.

Short answer

Answer: For total transmission, the arrangement is: 1. Dielectric slab 2. λ/4 airspace 3. Dielectric slab 4. λ/4 airspace 5. Dielectric slab 6. λ/4 airspace 7. Dielectric slab For highest reflectivity, the arrangement is: 1. Dielectric slab 2. λ/2 airspace 3. Dielectric slab 4. λ/2 airspace 5. Dielectric slab 6. λ/2 airspace 7. Dielectric slab

Step-by-step solution

Finding the Transmission Coefficient

To determine the transmission coefficient of an arrangement, we first find the transmission coefficients of the individual dielectric-air interfaces and air-dielectric interfaces. Since the boundary conditions don't change, the transmission coefficients are the same for each interface. Therefore, we can find the overall transmission coefficient as the product of the transmission coefficients of the individual interfaces. We know that for a normally incident wave, the transmission coefficient, \(T\), is given by $$ T = \frac{2\eta_{2}}{\eta_{1} + \eta_{2}}, $$ where \(\eta_{1}\) and \(\eta_{2}\) are the intrinsic impedances of the first and second media, respectively. For total transmission, we want the overall transmission coefficient to be equal to 1. A set of four slabs and airspaces should be connected to allow this to happen.

Arrangement for Total Transmission

To achieve total transmission through the stack, we'll arrange the four dielectric slabs with quarter-wavelength airspaces between them: 1. Dielectric slab 2. \(\lambda/4\) airspace 3. Dielectric slab 4. \(\lambda/4\) airspace 5. Dielectric slab 6. \(\lambda/4\) airspace 7. Dielectric slab This arrangement helps to cancel out the reflections that occur at each interface, such that the transmitted wave builds up constructively, and thus all the wave energy passes through the stack.

Arrangement for Highest Reflectivity

Now, we are interested in an arrangement that presents the highest reflectivity for the incident wave. In this case, we want the magnitude of the overall reflection coefficient to be maximum. For the highest reflectivity, we'll arrange the dielectric slabs with half-wavelength airspaces between them: 1. Dielectric slab 2. \(\lambda/2\) airspace 3. Dielectric slab 4. \(\lambda/2\) airspace 5. Dielectric slab 6. \(\lambda/2\) airspace 7. Dielectric slab This arrangement leads to reflections that add up constructively and result in the maximum reflectivity of the stack.