A beam of light passes from quartz to zircon at an angle of 19°. What is the angle of refraction? What is the critical angle when passing from zircon to quartz?

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Answer 1 · 2016-06-03T08:49:07

$r = {14.88}^{o}$ $r=14.88 ^o$

$θ_{C} = 51.98^{o}$ $theta_C=51.98" "^o$

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$assuming n_{q} = 1.544, n_z=1.960, sin 19 = 0.326$ $"assuming "n_q=1.544 " , " " n_z=1.960 " ,"sin 19=0.326$

$using Snell law:$ $"using Snell law:"$

$n_{q} \cdot sin i = n_{z} \cdot sin r$ $n_q*sin i=n_z*sin r$

$1.544 \cdot sin 19 = 1.960 \cdot sin r$ $1.544*sin 19=1.960* sin r$

$1.544 \cdot 0.326 = 1.960 \cdot sin r$ $1.544*0.326=1.960*sin r$

$0.503344 = 1.960 \cdot sin r$ $0.503344=1.960*sin r$

$sin r = \frac{0.503344}{1.960}$ $sin r=(0.503344)/(1.960)$

$r = {14.88}^{o}$ $r=14.88 ^o$

$critical angle for light passing from zircon to quartz:$ $"critical angle for light passing from zircon to quartz:"$

$θ_{C} = critical angle$ $theta_C=" critical angle"$

${sin θ}_{C} \cdot n_{z} = sin 90 \cdot n_{q}$ $sin theta_C*n_z=sin 90*n_q$

$s o; sin 90 = 1$ $so; sin 90=1$

${sin θ}_{C} \cdot n_{z} = 1 \cdot n_{q}$ $sin theta_C*n_z=1*n_q$

${sin θ}_{C} = \frac{n_{q}}{n_{z}} = \frac{1.544}{1.960}$ $sin theta_C=n_q/n_z=(1.544)/(1.960)$

$θ_{C} = 51.98^{o}$ $theta_C=51.98" "^o$