### Discussion

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##### Andy C.

University of Michigan - Ann Arbor

LB
##### Jared E.

University of Winnipeg

### Video Transcript

So we're asked to figure out the maximum distance D that will be able to see a fish that is a distance below the surface, which I call H of 2.25 meters, considering the fact that the index every fraction of Aaron survey is one in the index of refraction of water and sub ws 1.33 So I drop this right triangle over here and you can see that the critical angle fate A. C uh, if we find that we can use tricking a metric identities to find deed given that we know h so they to see the critical angle is defined as the inverse sine of the ratio of the index of refraction of the medium, the lightest traveling into which is air divided by the medium in which the light of is traveling, out of which is water. Such sense of w so calculating this out, we find that this is equal to 48.75 degrees now that we know that we can use our triggered a metric identity which says that the tangent of the critical angle is equal to the distance d divided by the height H or the depth of the fish h. So therefore, the distance which we want to calculate D is going to be equal to h times the tangent of the critical angle we just found they wanna see so carrying out This calculation we find at that distance is equal to 2.56 meters. This could be boxed in. Who wants to read? Draw that meters. This can be boxed in as our solution to our question.

University of Kansas
##### Andy C.

University of Michigan - Ann Arbor

LB
##### Jared E.

University of Winnipeg