![]() If we have an incoming wave travelling in the $xy$ plane, incident upon an interface plane defined by $y=0$, with amplitude $E_i$ at incidence angle $\theta_i$ that produces a reflected wave at angle $\theta_r$ and a transmitted wave at $\theta_t$ to the normal, then you can write down the following equation (where I assume the electric field is $z$-polarised with its direction also parallel to the interface plane). perpendicular to the normal to the surface) must be the same immediately either side of the interface. The alternative "textbook" electromagnetism answer is to use the boundary condition for the electric field either side of an interface - this is that the component of electric field parallel to the interface (i.e. Any optics textbook should have the calculation, or a quick Google found an example here. This is basically the Huygen's construction, and if you do the sums for a surface you can show that the overall scattering is only non-zero when the angle of reflection is equal to the angle of incidence. Add lots more to make a 2D surface, then add more layers of silver atoms below, and you're building up a system where the overall light scattering is the sum of individual scattering from huge numbers of individual silver atoms. Now add lots of atoms in a row, and you get something like a diffraction grating. (I'm oversimplifying because two atoms would be too closely spaced to act as Young's slits, but bear with me.) Now the light isn't simply isotropically scattered, but instead it's scattered into preferred directions. Each atom will scatter isotropically, so in effect we have two closely spaced emitters of light and the system behaves like a Young's slits setup. it will scatter the light equally in all directions.īut suppose we have two silver atoms side by side. The starting point it that a single silver atom is far smaller than the wavelength of light, so any scattering from it will be isotropic i.e. Without realising it you have stumbled across the Huygens-Fresnel principle. The coordinates of A, B, C are given asįind reflected position of triangle i.e., to the x-axis.That's a good question. The last step is the rotation of y=x back to its original position that is counterclockwise at 45°.Įxample: A triangle ABC is given. ![]() After it reflection is done concerning x-axis. Reflection about line y=x: The object may be reflected about line y = x with the help of following transformation matrixįirst of all, the object is rotated at 45°. This is also called as half revolution about the origin.Ĥ. In this value of x and y both will be reversed. In the matrix of this transformation is given below Reflection about an axis perpendicular to xy plane and passing through origin: The following figure shows the reflection about the y-axisģ. The object will lie another side of the y-axis. Here the values of x will be reversed, whereas the value of y will remain the same. Reflection about y-axis: The object can be reflected about y-axis with the help of following transformation matrix The object will lie another side of the x-axis.Ģ. Following figures shows the reflection of the object axis. ![]() In this transformation value of x will remain same whereas the value of y will become negative. Reflection about x-axis: The object can be reflected about x-axis with the help of the following matrix Reflection about an axis perpendicular to xy plane and passing through the originġ.The mirror image can be either about x-axis or y-axis. It is a transformation which produces a mirror image of an object.
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