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Explain the Young's double slit experimental setup and obtain the equation for path difference. |
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Answer» Solution :(i) An opaque screen with two small openings called double slit `S_1` and `S_2` is kept equidistance from a SOURCE S. (ii) The width of each slit is about 0.03 mm and they are separated by a distance of about 0.3 mm. (iii) As `S_1` and `S_2` are equidistant from S, the light Superposition PRINCIPLE waves from S reach `S_1` and `S_2` in-phase. So, `S_1` and `S_2` act as COHERENT sources which are the requirement of obtaining interference pattern. Experimental setup: (i) WAVEFRONTS from `S_1` and `S_2` spread out and overlapping takes place to the right side of double slit. (ii) When a screen is placed at a distance of about 1 meter from the slits, alternate bright and dark fringes which are equally spaced appear on the screen. (iii) These are called interference fringes or bands. 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 as shown in Figure.  (iv) These two waves constructively interfere and bright fringe is observed at O. This is called central bright fringe.(v) The fringes disappear and there is uniform illumination on the screen when one of the slits is covered. This shows clearly that the bands 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 wavelength `lamda`. A screen is placed parallel to the double slit at a distance D from it. (ii) The mid-point of `S_1` and `S_2` is C and the i mid-point of the screen O is equidistant from `S_1` and `S_2` P is any point at a distance y from O.(iii) The waves from `S_1` and `S_2` meet at P either n-phase or out-of-phase depending uponthe path difference between the two waves. (iv) The path difference `delta` between the light waves from `S_1` and `S_2` to the point P is, ` delta= S_2P - S_1 P` (v) A perpendicular is dropped from the point `S_1` to the line `S_2P ` at M to find the path difference more precisely. ` delta = S_2 P - MP= S_2M` The angular position of the point P from C is `theta. angle OCP = theta` From the geometry, the angles `angleOCP` and `angleS_2S_1M` are equal. ` angle OCP = angle S_2 S_1 M= theta` In right angle triangle `Delta S_1 S_2 M` , the path difference, ` S_2 M = d sin theta` ` delta = d sin theta` If the angle `theta` is small, `sin theta ~~ tan theta ~~ theta` .From the right angle triangle `DeltaOCP, tan theta = y/D` The path difference,` delta= (dy)/(D)` Based on the condition on the path difference, the point P may have a bright or dark fringe. |
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