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These rainbows exist and our eyes do see them - they are just extremely thin, plus our brain filters out this information, but the edges of bright objects on dark background in the mirror do have a thin rainbow along them. Our eye is using this rainbow to infer the image comes from a mirror. I once saw an exhibit at a modern art museum - a seemingly black-...


2

Keep in mind that $R,T$ as defined in your text are the coefficients of the electric field, not of the intensity. The reason for the relation $R+1=T$ is simply that the electric field should be continuous across the boundary. Outside the material, you have an incident wave (coefficient of $1$) and the reflected wave (coefficient of $R$). Inside the material ...


1

You can see this from the Lensmaker's equation for thin lenses $$\frac{1}{f} = \frac{n'-n}{n} \left( \frac{1}{R_1} - \frac{1}{R_2} \right)$$ $R_1,R_2$ are the curvature radii of the surfaces of the lens. $n^\prime$ is the refractive index of the interior of the lens, whereas $n$ is the refractive index of the medium outside the lens. So for $n\approx 1$ (...


1

Feynman is trying to say that the principle of least time implies that the indices $n_{ij}$ are ratios of speeds, probably by a reasoning similar to the following: If the paths of the light in the two materials are $s_1$ and $s_2$ and the speeds of light in the materials are $v_1$ and $v_2$, then the principle of least time says that the total travel time $\...


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