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A ray of light of intensity I is incident on a parallel glass slab at a point A as shown.It undergoes partial reflection and refraction. At each reflection 25% of incident energy is reflected. The rays AB and A' B'undergo interference. The ratio I_(max)/I_(min) is : |
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Answer» ` 4 : 1` LIGHT TRANSMITTED along `AC = (3I)/(4)` then REFLECTED along `CA. = (1)/(4) xx (3I)/(4)= (3I)/(16)` Reflected at A. = `1/4 xx (3I)/(16) = (3I)/(64)` `therefore` Intensity of light transmitted along A.B. is `I_(2) = (3I)/(16) -(3I)/(64) =(9I)/(64)` `therefore (I_(1))/(I_(2)) = (a^(2))/(b^(2)) = (I)/(4) xx (64)/(9I) = (16)/(9)` `therefore (a^(2))/(b^(2)) = (16)/(9)`, so (a)/(b) = 4/3 `(I_(max))/(I_(min)) = ((a + b)^(2))/((a- b)^(2)) = ((4 + 3)/(4 -3))^(2) = (49)/(1)` |
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