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# A beam of unpolarized light is incident on an air-glass interface derive, using a suitable ray diagram that light reflected from the interface is completely polarised, when $\mathrm{tan}{i}_{p}=?$

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Solution

## Step 1: Given dataWhen unpolarized light is incident on a transparent refracting medium, at a particular angle of incidence the reflected light and refracted light are perpendicular to each other that angle of incident is known as a polarizing angle.The reflected light is completely plane polarised.Let ${i}_{p}$ be the polarising angle.Let $r$ be the angle of refraction.Let the refractive index of air be ${\mu }_{1}=1$.Let the refractive index of glass be ${\mu }_{2}=\mu$.Therefore, the reflected ray and refracted ray are perpendicular to each other.Step 2: Formula UsedAccording to Brewster’s law,${i}_{p}+r=90°$$r=90°–{i}_{p}$According to Snell’s law${\mu }_{1}\mathrm{sin}i={\mu }_{2}\mathrm{sin}r$$\therefore \mu =\frac{\mathrm{sin}i}{\mathrm{sin}r}$Step 3: Brewster's AngleUpon substituting the values we get,$\mu =\frac{\mathrm{sin}{i}_{p}}{\mathrm{sin}\left(90°-{i}_{p}\right)}$$⇒\mu =\frac{\mathrm{sin}{i}_{p}}{\mathrm{cos}{i}_{p}}$$⇒\mu =\mathrm{tan}{i}_{p}$This relation is known as the Brewster angle.

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