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help? A ray of light travels from lead to vacuum. If the ray of light in the vacuum makes an angle of 71.0 degrees to the normal, what is the angle of incidence in the lead?
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we have to use this equasion: \[n _{1}\sin \theta _{1}=n _{2}\sin \theta _{2}\]
this is a physics question related to optics please use physics forum for this it will be more beneficial
yeah. i just realised i was in math. oops.
n1=2.01 for lead, and n2=1 for vacuum, θ2=71, use snell's law to find answer n1sinθ1=n2sinθ2 sinθ1=(n2/n1)sinθ2=(1/2.01)(sin71)=.471 θ1=inverse sine (.471) =27 degrees
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