Question #354d2

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Question #354d2

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Answer 1 · 2018-01-14T05:36:15

41.81 degrees

Explanation:

enter image source here Critical angle of light is that one,if at that angle light falls at an interface,while going from a solid to lighter media,it will not enter the media next to the interface,rather go along the interface,

So,using Snell's law we can say,
$u 1 sin θ$ $u1 sin theta$ = $u 2 sin α$ $u2 sin alpha$
Given, $u 2$ $u2$ = $(\frac{3}{2})$ $(3/2)$
And $sin θ$ $sin theta$ = 1
So, $sin α$ $sin alpha$ = $\frac{2}{3}$ $2/3$ or $α$ $alpha$ = 41.81 degrees

Answer 2 · 2018-01-14T05:38:21

the critical angle is $i_{c} = 42.067$ $i_c=42.067$

Explanation:

critical angle $i_{c}$ $i_c$ is given as
$sin (i_{c}) = \frac{1}{n_{21}}$ $sin(i_c)=1/n_21$ where $n_{21} = \frac{n_{2}}{n_{1}}$ $n_21=n_2/n_1$
$n$ $n$ is the refractive index of that particular media
$h e r e n_{2} = \frac{3}{2} and n_{1} = 1$ $here n_2=3/2 and n_1=1$ substituting in the formula we get
$sin (i_{c}) = (\frac{1}{\frac{3}{2}}) = (\frac{2}{3}) \Rightarrow i_{c} = {sin}^{- 1} (\frac{2}{3}) = {sin}^{- 1} (0.67)$ $sin(i_c)=(1/(3/2))=(2/3)rArri_c=sin^-1(2/3)=sin^-1(0.67)$
$\Rightarrow i_{c} = 42.067$ $rArri_c=42.067$

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Question #354d2

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