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In Youngs double slit experiment, to increase the fringe width |
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Answer» Solution :Experimental setup (i) Wavelenght from `S_(1) and S_(2)` spread out and overlapping takes place to the right side of double slit. When a screen is placed at a distance of about I meter from the slits, alternate bright and dark fringes which are equally spaced appear on the screen. These are called interference fringes or bands. (ii) Using an eyepiece the fringes can be seen directly. At the center point O on the screen, WAVES from `S_(1) and S-(2)` travel equal distances and arrive in phase. These two wavesconstructively interfere and bright FRINGE is observed at O. This is called central bright fringe. (iii) The fringes DISAPPEAR and there is unifrom illumination on the screen when one of the slits is covered. This shows clearly tha the band are due to interference. Equation for path difference (i) Let d be the distance between the double slits`S_(1) and S_(2)` which act as coherent sources of wavelenght `lambda`. A screen is placed parallel to the double slit at a distance D from it. The mid - point of `S_(1) and S_(2)` is C and the midpoint of the screen O is equidstant from `S_(1) and S_(2)`. P is any point at a distance y from O. (ii) The waves from `S_(1) and S_(2)` meet at P either inphase or out of - phase depending upon the pathdifference between the two waves. (iii) The path differcne `delta` between the light waves from `S_(1) and S_(2)` to the point P is, `delta = S_(2)P - S_(1)P` A perpendicular is drpped from the point `S_(1)` to the line `S_(2)`P at Mto find the difference more precisey. `delta = S_(2) P - MP = S_(2)M` The angular position of the pointfromC is `theta, angle OCP = theta`. From the geometry, the angles `angleOCP and angle S_(2)S_(1)M` are equal. `angleOCP = angle S_(2)S_(1)M = theta` In right angle triangle `DeltaS_(1)S_(2)M`, t he path difference, `S_(2)M = d sin theta` `delta = d sin theta` If the angle `theta` is small, `sin theta = sin theta = theta`, From the right angle triangle `Delta OCP, TAN theta = (y)/(D)` The path difference, `delta = (dy)/(D)`  
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