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High.
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Here in this given problem there are three parts.
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The first one, this is having a slab of a transparent material.
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This is the slab of the transparent medium whose height, its side length is 20 cm means its height is also 20 cm.
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Now a ray of light is incident on it.
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Here suppose this is a ray of light which is incident at an angle of 25 degree then it goes into this medium then towards the normal as the medium is denser as compared with the surrounding medium means air and then again goes back to the same medium and at the same angle of emergence which is given as 25 degrees.
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Here this angle is 25 degree.
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This angle of refraction this is missing.
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We have to find it and this displacement of the ray of light from the normal.
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This is given as 2 .5 centimeter.
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So we get this right angle triangle.
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So in this right angle triangle, so in this right angle triangle using and this is the first part of the problem which we are doing in this right angle triangle using tangent 10r that is equal to perpendicular means 2 .5 divided by base and that base is 20 centimeter so here it comes out to be equal to 0 .1 so we get the value of this angle of refraction are as 10 inverse of 0 .125 and it comes out to be equal to 7 .125 degree.
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The angle of refraction.
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So to find index of refraction of this medium, we use snell's law, which says the ratio of sign of angle of incidence...