ssiding

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Aug 20, 2008
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You have two adajacent transparent media. The speed of ligght through medium 1 is v1, and the speed of light through medium 2 is v2. If v1 < v2, then total internal reflection will occur at the interface between these media if a beam of light is

A. incident in medium 1 and strikes the interface at an angle of incidence greater than arcsin (v1/v2)

B. incident in medium 1 and strikes the interface at an angle of incidence greater than arcsin (v2/v1)

C. incident in medium 2 and strikes the interface at an angle of incidence greater than arcsin (v1/v2)

D. incident in medium 2 and strikes the interface at an angle of incidence greater than arcsin (v2/v1)
 

soby10

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Apr 27, 2008
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Since n=c/v and v2>v1 so n1>n2. Total internal reflection occurs when light passes through medium of larger refractive index to a smaller refractive index and the refractive ray bends away from the normal, angle of refraction increases and so does angle of incidence. When angle of incidence is at a critical angle the angle of refraction is 90. So when angle of incidence is larger than critical angle total internal reflection occurs so there is no refracted light, all light is in medium. The def. of critical angle is sin crit. angle= n2/n1 (n1>n2). Therefore, light is travelling from bigger index n1 to n2 the smaller index. Since v2 is inversely related to n2 and v1 is inversely related to n1 ,(c/ v2)/(c/v1) ---> (c/v2) *(v1/c) = v1/v2 so angle is at arcsin v1/v2 answer is A.
 

isaacmn

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May 5, 2007
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total internal reflection occurs when you have a light incident on a medium of high refractive index to low. The light will be trapped and internal reflection. Also when theta incidence is greater than theta critical.

agree with the last post. if v1<v2 n1>n2 ..which elimiated C and D
 
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