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(B) According to Snell's law,for any layer $n$,$\mu_0 \sin i = \mu(n) \sin r_n$. For total internal reflection to occur at the upper surface of layer

Vedclass · Accepted Answer

$5$

Vedclass · Answer

(B) According to Snell's law,for any layer $n$,$\mu_0 \sin i = \mu(n) \sin r_n$. For total internal reflection to occur at the upper surface of layer $n$,the angle of refraction $r_n$ must be $90^\circ$,so $\sin r_n = 1$. Thus,$\mu_0 \sin i = \mu(n)$. Given $i > 30^\circ$,we have $\sin i > \sin 30^\circ = 0.5$. Therefore,$\mu(n) = \mu_0 \sin i > 0.5 \mu_0$. Substituting the expression for $\mu(n)$: $\mu_0 - \frac{\mu_0}{4n - 18} > 0.5 \mu_0$ $1 - \frac{1}{4n - 18} > 0.5$ $0.5 > \frac{1}{4n - 18}$ $4n - 18 > 2$ $4n > 20$ $n > 5$. Since the question asks for the layer where $TIR$ takes place,and the refractive index must decrease for $TIR$,we check the condition $\mu(n) $\mu_0 - \frac{\mu_0}{4n - 18}  0$. This implies $4n - 18 > 0$,or $n > 4.5$. For $n=5$,$\mu(5) = \mu_0 - \frac{\mu_0}{20-18} = \mu_0 - 0.5 \mu_0 = 0.5 \mu_0$. Since $\sin i > 0.5$,the condition $\mu_0 \sin i = \mu(n)$ is satisfied at $n=5$.

Similar Questions

In an optical fibre, the core and cladding are made with materials of refractive indices $1.5$ and $1.414$ respectively. To observe total internal reflection, what will be the range of the incident angle with the axis of the optical fibre?

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