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The cross section of a glass prism has t...

The cross section of a glass prism has the form of an isosceles triangle. One of th erefracting faces is silvered. A ray of light falling normally on the other refracting face, being reflected twice, emerges through the base of the prism perpendicular to it. Find the angles of the prism.

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The incident ray BC at normal incidence is reflected at silvered face, along DE and at E it again sufferes reflection along EF. Since the ray emerges normally from the base, therefore the ray EF must fall normally on the base and emerges along EG.
We find `i=A`.
Also, `beta=d`
Since `EN_(2)||CD,beta=2i` (alternate angles)
`:. alpha=2A (beta=alpha,i=A)` (i)
Also, `2alpha+A =180^(@)`
(`:'` Sum of angles of a triangle = `180^(@)`) (ii)
Solving Eqs. (i) and (ii), we get A =`36^(@) , alpha=72^(@)` .
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