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In the figure shown for an angle of incidence i at the top of the surface, what is the minimum refractive index for total internal reflection at the vertical surface. |
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Answer» SOLUTION :The RAY will total INTERNALLY reflect at the vertical surface if `theta gt theta_(c)` Now, `r=(90^(@)-theta)` and Snell.s law is `sin i=mu sin r` `(sini)/(mu)=sin(90^(@)-theta) implies cos theta=(sin i)/(mu)` or `sin theta=sqrt(1-cos^(2)theta)=sqrt(1-(sin^(2)i)/(mu^(2)))`  If `theta gt theta_(C)` , then `sin theta gt sin theta_(C)` ( As `sin theta` is an increasing function for `0 lt theta lt 90^(@))` `sqrt(1-(sin^(2)i)/(mu^(2)))gt(1)/(mu)` `1-(sin^(2)i)/(mu^(2))gt(1)/(mu^(2))` `mu^(2)-sin^(2)i gt 1 "or" (mu^(2)-1)gtsin^(2)i` If total internal reflection has to be larger for all value, the above inequality must be SATISFIED for all `(sin^(2)i)_("MAX")=1` `impliesmu^(2)-1gt1 "or" mu gtsqrt(2)` This total internal reflection phenomenon is used in fibre optics to bend light in a curved path. 
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