A diver shines a flashlight upward from beneath the water (n = 1.33) at a 32.7° angle to the vertical. At what angle does the light leave the water?

Respuesta :

Answer:

45.93°

Explanation:

The angle of incidence is given as 32.7°

The refractive index of the water that is [tex]n_1=1.33[/tex]

Refractive index of the air that is [tex]n_2=1[/tex] (because the refractive index of air is 1 )

We have to find the angle at which the light leave the water means angle of refraction

So according to snell's law [tex]n_1sini=n_2sinr[/tex]

[tex]1.33sin32.7=1\times sinr[/tex]

[tex]sinr=0.7185[/tex]

r =45.93°

So the light leave the water at an angle of 45.93°

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