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Mathematics 7 Online
OpenStudy (anonymous):

WILL MEDAL FOR CORRECT ANSWER: A light wave travels through water (n = 1.33) at an angle of 35. What angle does it have when it passes from the water into air (n = 1.00)? A. 25.5 B. 25.5 C. 0.763 D. 49.7

OpenStudy (anonymous):

Use the equation.

OpenStudy (perl):

When a ray enters a medium with a greater refraction index, it refracts and gets closer to the normal line, N

OpenStudy (anonymous):

okay.. and that means n= 1.00 like it says in the question..

OpenStudy (anonymous):

okay i get that.

OpenStudy (perl):

the tricky part, is the angle 35 degrees with respect to the normal or with respect to the water surface

OpenStudy (perl):

|dw:1427491197186:dw|

OpenStudy (anonymous):

okay im confused with n again. is it going to be 1.00 or 1.33? or does it goes from 1.33 into the 35 degree angle to 1.00?

OpenStudy (perl):

water has a refractive index of 1.333, and air has a refractive index of about 1

OpenStudy (anonymous):

ohh okay.

OpenStudy (perl):

|dw:1427491433006:dw|

OpenStudy (anonymous):

okay. im following a bit.

OpenStudy (perl):

we can use the equation $$ \Large { n_1 \sin\theta_1 = n_2 \sin \theta_2 } $$

OpenStudy (perl):

you can use the equation $$ \Large { n_1 \sin\theta_1 = n_2 \sin \theta_2 \\ \therefore \\ 1.33 \sin (35^o) = 1 \sin \theta_2 } $$

OpenStudy (anonymous):

1.33(sin theta 1) = 1.00 (sin theta 2) right?

OpenStudy (perl):

make sure you are in degree mode if you are using a calculator

OpenStudy (anonymous):

okay i got 0.762.

OpenStudy (perl):

good

OpenStudy (anonymous):

so that's the answer? 0.763?

OpenStudy (perl):

$$ \Large { n_1 \sin\theta_1 = n_2 \sin \theta_2 \\ \therefore \\ 1.33 \sin (35^o) = 1 \sin \theta_2 \\ \iff \\ \\0.76285 = \sin \theta_2 } $$

OpenStudy (perl):

we have to use inverse sine of 0.762

OpenStudy (anonymous):

I got 49.716.

OpenStudy (perl):

$$ \Large { n_1 \sin\theta_1 = n_2 \sin \theta_2 \\ \therefore \\ 1.33 \sin (35^o) = 1 \sin \theta_2 \\ \iff \\ \\0.76285 = \sin \theta_2 \\ \sin^{-1}(0.76285) = \theta_2 } $$

OpenStudy (perl):

yes thats correct

OpenStudy (anonymous):

yay!! thank you!! :)

OpenStudy (perl):

you dont mind that i deleted some of the other posts, which were not relevant

OpenStudy (anonymous):

Not at all. :)

OpenStudy (anonymous):

Thanks for your help. :)

OpenStudy (perl):

is it correct?

OpenStudy (anonymous):

Yes it was :)

OpenStudy (perl):

great, so they used the angle with the normal in both cases. but sometimes the angle will be with respect to the water or surface. so watch out for that

OpenStudy (anonymous):

okay :)

OpenStudy (perl):

but angle of incidence is always with respect to the normal :)

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