InterviewSolution
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InterviewSolution
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Describe Young's slit experiment to produce interference pattern due to a monochromatic source of light. Deduce the expression for the fringe width. |
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Answer» Solution :A SIMPLE schematic diagram of Young.s double-slit experiment to produce interference pattern due to a monochromatic source of light is shown in figure. Let `S_(1) and S_(2)` be two narrow slits, a small distance .d. apart, illuminated by a monochromatic light source S of wavelength `lamda`, kept equidistant from the slits `S_(1) and S_(2)`. Consider a point P on the screen MN, situated at a normal distance .D. from the slits, at a distance .x. from the symmetrical central point O of screen. Path difference between light waves reaching point P from two slit sources is `S_(2) P-S_(1)P`, where `(S_(2)P)^(2)=D^(2)+(x+(d)/(2))^(2) and (S_(1)P)^(2)=D^(2)+(x-(d)/(2))^(2)` `therefore (S_(2)P)^(2)-(S_(1)P)^(2)=[D^(2)+(x+(d)/(2))^(2)]-[D^(2)+(x-(d)/(2))^(2)]=2xd` `implies(S_(2)P-S_(1)P)(S_(2)P+S_(1)P)=2xd`  WIDTH="80%"> or `(S_(2)P-S_(1)P)=(2xd)/((S_(2)P+S_(1)P))` If x and d are very very small as compared to D, then `(S_(2)P+S_(1)P)` may be considered as 2D. hence, `[S_(2)P-S_(1)P]=(2xd)/(2D)=(xd)/(D)` For constructive interference, path difference MUST be an integer multiple of `lamda` i.e., `(xd)/(D)=nlamda`, where `n=0,1,2,3,`. etc. `impliesx=(nDlamda)/(d)` i.e., positions of various maxima (bright bands) will be given by : For destructive interference, path difference must be an odd multiple of `(lamda)/(2)` i.e., `(xd)/(D)=(2n-1)(lamda)/(2)`, where n=1,2,3, . . etc. `impliesx=((2n-1)Dlamda)/(2d)` i.e., positions of various minima (dark bands) will be given by: `x_(1).=(Dlamda)/(2d),x_(2).=(3Dlamda)/(2d),x_(3).=(5Dlamda)/(2d)` . . . Fringe width of the fringe pattern is defined as the distance between two SUCCESSIVE maxima or successive minima. therefore, considering two successive maxima, we have Fringe width `beta=x_(n+1)-x_(n)=((n+1)Dlamda)/(d)-(nDlamda)/(d)=(Dlamda)/(d)` |
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