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In double slit experiment illuminated by coherent 600 nm light, the slit separation is 0.2 mm and slit to screen distance is 2 m. At distance y=6mm, what is the average intensity on the screen of the maximum? With Explaining please. |
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Answer» I have not write, but this answer is correctExplanation:ANSWERSeparation is given byy= dnλD whered=3mm=3×10 −3 D=2mλ 1 =480nm=480×10 −9 mλ 2 =600nm=600×10 −9 MN 1 =N 2 =5Y 2 −y 1 =?So,y 1 = dnλ 1 D y 1 = 3×10 −3 5×480×10 −9 ×2 y 1 =1.6×10 −3 mAlso, y 2 = dnλ 2 D y 2 = 3×10 −3 5×600×10 −9 ×2 y 2 =2×10 −3 mAs y 2 >y 1 y 2 −y 1 =2×10 −3 −1.6×10 −3 =4×10 − 4mTherefore the separation on the screen between the fifth ORDER bright fringes of the two INTERFERENCE patterns is 4×10 −4 m |
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