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A point sources `S` emitting light of wavelength `600nm` is placed at a very small height `h` above the flat reflecting surface `AB`(see figure).The intensity of the reflected light is`36%` of the intensity.interference firnges are observed on a screen placed parallel to the reflecting surface a very large distance `D` from it. (A)What is the shape of the interference fringes on the screen? (B)Calculate the ratio of the minimum to the maximum to the maximum intensities in the interference fringes fromed near the point `P` (shown in the figure) (c) if the intenstities at point `P` corresponds to a maximum,calculate the minimum distance through which the reflecting surface `AB` should be shifted so that the intensity at `P` again becomes maximum. |
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Answer» Correct Answer - (i). Circular (ii). `(1)/(16)` (iii). 300 nm. (i). S is a point source, fringes formed will be circular (ii). Ratio of minimum and maximum intensities intensity of light direct from source `I_(1)=I_(0)`(say) Intensity after reflection `I_(2)=0.36I_(0)` `therefore(I_(min))/(I_(max))=((sqrt(I_(1))sqrt(I_(2)))/(sqrt(1))+sqrt(I_(2)))^(2)=((0.4)/(1.6))^(2)=(1)/(16)` (iii). Shift of AB for same intensity if intensity at P correspionds to maximum it means that constructive interference occurs at P. `therefore` Path difference between direct waves from S and reflected waves, from reflector AB, is `nlamda` Let AB is shifted by x (towards P or away from P) `therefore` Additional path difference introduced =2x For minimum value of `x_(1),n=1` `therefore` Path difference `=1xxlamda=600nm` `2x=600nmimpliesx=300nm` |
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