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Vaselesa [24]
3 years ago
6

ver shines light up to the surface of a flat glass-bottomed boat at an angle of 30 relative to the normal. If the index of refr

action of water and glass are 1.33 and 1.5, respectively, at what angle (in degrees) does the light leave the glass (relative to its n

Physics
1 answer:
Free_Kalibri [48]3 years ago
7 0

Answer:

\beta = 41.68°

Explanation:

according to snell's law

\frac{n_w}{n_g} = \frac{sin\alpha}{sin30 }

refractive index of water n_w is 1.33

refractive index of glass  n_g  is 1.5

sin\alpha = \frac{n_w}{n_g}* sin30

sin\alpha = 0.443

now applying snell's law between air and glass, so we have

\frac{n_g}{n_a} = \frac{sin\alpha}{sin\beta}

sin\beta = \frac{n_g}{n_a} sin\alpha

\beta = sin^{-1} [\frac{n_g}{n_a}*sin\alpha]

we know that sin\alpha = 0.443

\beta = 41.68°

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