A light ray in medium (Refractive index $= 5 / 3$ ) making an angle of $30^{\circ}$ with normal,enters another medium. The angle in other medium is ${sin}^{- 1} (5 / 6)$ . How many minimum degrees the incident angle must be increased such that the ray is completely reflected.

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Solution

From Snell's law
$n_{1} sin i = n_{2} sin r$
$\frac{5}{3} sin 30^{\circ} = n_{2} [\frac{5}{6}]$
$\frac{5}{3} [\frac{1}{2}] = n_{2} [\frac{5}{6}]$
$n_{2} = 1$
For total internal reflection :-
$sin θ > sin θ_{C}$
$sin θ > \frac{μ_{2}}{μ_{1}}$
$sin θ > \frac{1}{[\frac{5}{3}]}$
$sin θ > \frac{3}{5}$
$θ > 37^{\circ}$
So $[37^{\circ} - 30^{\circ} = 7^{\circ}]$ incident angle must be increased.

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A light ray in medium (Refractive index =5/3) making an angle of 30∘ with normal,enters another medium. The angle in other medium is sin–1(5/6). How many minimum degrees the incident angle must be increased such that the ray is completely reflected.

From Snell's law n1sini=n2sinr 53sin30∘=n2[56] 53[12]=n2[56] n2=1 For total internal reflection :- sinθ>sinθC sinθ>μ2μ1 sinθ>1[53] sinθ>35 θ>37∘ So [37∘−30∘=7∘] incident angle must be increased.

A light ray in medium (Refractive index $= 5 / 3$ ) making an angle of $30^{\circ}$ with normal,enters another medium. The angle in other medium is ${sin}^{- 1} (5 / 6)$ . How many minimum degrees the incident angle must be increased such that the ray is completely reflected.