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Why is second image always bright ? This question is so often asked in mcat examination but I find no legitimate answer to it .

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  • $\begingroup$ Can you please provide the exact text of the question? The mirror's thickness has little to do with it; which surface is metallized (or dielectric-coated) has a lot to do with it. $\endgroup$ – Carl Witthoft Oct 13 '14 at 11:43
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I'm not sure exactly what you mean butYou get a virtual image from the main mirror surface, but you also see another virtual image a little nearer to you (by an amount equal to twice the mirror's thickness) than the main one. This Fresnel reflexion has intensity $\left(\frac{n-1}{n+1}\right)^2$ times the incident wave one, where $n$ is the glass's refractive index. The main reflexion's intensity is $\left(1-\left(\frac{n-1}{n+1}\right)^2\right)^2$ times the intensity of the incident light - almost as intense as the incident light.

Then you get higher order reflexions. The third order, where the light reflected from the main mirror surface is Fresnel reflected back to the main mirror, then bounced back, has an intensity $\left(1-\left(\frac{n-1}{n+1}\right)^2\right)^2\,\times\,\left(\frac{n-1}{n+1}\right)^2$ the first Fresnel reflexion, i.e. almost as bright as the first "ghost". The $N^{th}$ ghost's intensity is $\left(1-\left(\frac{n-1}{n+1}\right)^2\right)^{2N-2}\,\times\,\left(\frac{n-1}{n+1}\right)^2$.

All of these equations are derived from the Fresnel Equations deployed to analyse multiple reflexions in the layers.

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    $\begingroup$ Seriously? They are asking that in mcat??? $\endgroup$ – CuriousOne Oct 13 '14 at 8:27

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