A beam of light, originally travelling through air (with index refraction n=1),

Question

A beam of light, originally travelling through air (with index refraction n=1), hits the surface of a glass block (n=1.5) at point A in the figure below and is partially reflected and partially refracted.

b. The refracted light continues as shown until it hits (at point B) the interface between the glass block and a block of ice stuck to the side of it. If the light undergoes total internal reflection at B, find the maximum allowed value of the index of refraction of the ice block.

Explanation / Answer

From the Snell's law at point A

n1sintheta1 =n2 sintheta2

1 sintheta =1.5*sin(30) ===> theta =sin-1(1.5sin(30)) =48.59degrees

We know that in reflection, angle of incidenace is equal to angle of reflection then theta =alpha= 48.59degrees.

The refractive index of ice is n2 =?

The glass block of refrctive index is n1 =1.5

For a light to get total internally reflect it must pass from denser medium to rarer then angle of incidence must be greater than the critical angle.

From the corresponding angles the angle of incidence atpoint becomes 60 degrees then for a light ray to reflect or it must go with the interface then refraction becomes 90degrees then from the snells law at point B

n1sintheta1 =n2 sintheta2

1.5sin(60) =n2sin90 ( sin90 =1)

Then n2 =1.5sin60 =1.299

The maximum allowed value of the index of refraction of the ice block is n2 =1.299

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