PHYSICS ******************10 POINTS*********
1.A 4.9-m-wide swimming pool is filled to the top. The bottom of the pool
becomes completely shaded in the afternoon when the sun is 21DEGREE above
the horizon.How deep is the pool? 2.The glass core of an
optical fiber has an index of refraction 1.61. The index of refraction of
the cladding is 1.48. What is the maximum angle a light ray can make
with the wall of the core if it is to remain inside the fiber? essentially a small box with a hole in the front (the pupil) and ''film'' at the back (the retina). The distance from the pupil to the retina is approximately 24mm}.Suppose you look at a 180-cm-tall friend who is standing 7.4 {rm m} in front of you. Assuming your eye functions like a pinhole camera, what will be the height, in mm}, of your friend’s image on your retina? 4.A narrow beam of light with wavelengths from 450 DEGREE to 700nm incident perpendicular to one face of a 30.00D EGRE prism made of crown glass, for which the index of refraction ranges from n =1.533 to ;N=1.517 for those wavelengths. thanks!!!!!! 10 points like usual!
Lookie, somebody goes to SFU. lol.
I have the for the first two questions:
1. Use snell is law: n1sin(theta1)=n2sin(theta2)
the refraction index (n1) is 1 and refraction index (n2) is 1.33.
so.(1)(sin69)=1.33sin(theta2) now use tan(theta2)=4.3/d and solve for d. 2. (theta)= inverse sin (n1/n2) then 90-(theta)
as usual, finding a way not to get an education.
yeah small
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