Question

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 refraction 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

Answer 1

`\beta = 41.68°`

Step-by-step explanation:

according to snell's law

`(n_w)/(n_g) = (sin\alpha)/(sin30 )`

refractive index of water n_w is 1.33

refractive index of glass n_g is 1.5

`sin\alpha = (n_w)/(n_g)* sin30`

`sin\alpha = 0.443`

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

`(n_g)/(n_a) = (sin\alpha)/(sin\beta)`

`sin\beta = (n_g)/(n_a) sin\alpha`

`\beta = sin^(-1) [(n_g)/(n_a)*sin\alpha]`

we know that
`sin\alpha = 0.443`

`\beta = 41.68°`