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
Use the equation.
When a ray enters a medium with a greater refraction index, it refracts and gets closer to the normal line, N
okay.. and that means n= 1.00 like it says in the question..
okay i get that.
the tricky part, is the angle 35 degrees with respect to the normal or with respect to the water surface
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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?
water has a refractive index of 1.333, and air has a refractive index of about 1
ohh okay.
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okay. im following a bit.
we can use the equation $$ \Large { n_1 \sin\theta_1 = n_2 \sin \theta_2 } $$
you can use the equation $$ \Large { n_1 \sin\theta_1 = n_2 \sin \theta_2 \\ \therefore \\ 1.33 \sin (35^o) = 1 \sin \theta_2 } $$
1.33(sin theta 1) = 1.00 (sin theta 2) right?
make sure you are in degree mode if you are using a calculator
okay i got 0.762.
good
so that's the answer? 0.763?
$$ \Large { n_1 \sin\theta_1 = n_2 \sin \theta_2 \\ \therefore \\ 1.33 \sin (35^o) = 1 \sin \theta_2 \\ \iff \\ \\0.76285 = \sin \theta_2 } $$
we have to use inverse sine of 0.762
I got 49.716.
$$ \Large { n_1 \sin\theta_1 = n_2 \sin \theta_2 \\ \therefore \\ 1.33 \sin (35^o) = 1 \sin \theta_2 \\ \iff \\ \\0.76285 = \sin \theta_2 \\ \sin^{-1}(0.76285) = \theta_2 } $$
yes thats correct
yay!! thank you!! :)
you dont mind that i deleted some of the other posts, which were not relevant
Not at all. :)
Thanks for your help. :)
is it correct?
Yes it was :)
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
okay :)
but angle of incidence is always with respect to the normal :)
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