(Hindi) Young double slit experiment

Lesson 6 of 17 • 5 upvotes • 11:38mins

Rajesh Kumar

Young double slit experiments. Thomas young in 1801 reported his experiment on the interference of light. young double hole experiment one of them two waves interfered to give a pattern of bright and dark areas . the variation of intensity on the screen shows the interference taking place between the light waves reaching the screen from the two pin pinholes. if light of single wavelength used then pattern of bright and dark areas are sharply defined . in young double slit experiment S1 and S2 these are two long parallel slits , source is parallel beam of monochromatic light . we have to calculate electric field at point P on the screen due to two waves. and finally we calculate the resultant of two electric field at point P. then we calculate phase difference between these two electromagnetic waves arriving at B. phase difference phase (¢) is equal to 2pi by Lambda in to path difference (∆x). the resultant field at point P is calculated by in the form of resultant vector for first vector is electric field amplitude E1, electric field amplitude E2 and resultant electric field isE०, for bright fringe constructive interference for the phase difference is equal to 2 Pi ,in other form path difference Delta x is equal to n Lambda ,for dark fringe of destructive interference phase difference is equal to( 2 n + 1 )Pie or Delta x is equal to (N + half )Lambda. at the centre of the screen bright fringe ,for n is equal to zero .how we can calculate intensity variation if amplitude of both electric wave or electric field E 1 and E 2 , equal to E0' . then resultant amplitude is square of E0' (1 + cos Lambda ). intensity is directly proportional to the square of the amplitude . so intensity I is equal to 2 I in to ( oneplus cause delta) or equal to 4I cos square Delta by2. the resultant intensity at centre Delta equal to zero so be on maximum intensity at centre for bright fringe . and for dark fringe phase difference (2n+ pi). intensity for dark fringe is equal to zero . so on the screen intensity of the fringe, vary from 0 to 4I'. fringe width separation between two consecutive bright or dark fringe is known as fringe width separation between the slits small d and between slit and a screen is capital D if capital D very very greater than a small D then path difference Delta x is equal to small d multiply sin theta that theta is the angle between slit to the centre of the screen line and centre of the slit to point P . that is POB is equal to theta. which is also equal to S1 S2 is equal to theta then path difference Delta is equal to dsin theta , theta is very small. so we can also write in approximation deltax is equal to tan theta the whole Delta x is equal to a small D into by divided by capital D for bright fringe Delta x is equal to nLambda ,is equal to small d sintheta . Capital D there to in capital D Lambda divided by small D that is at both side of the centre in upward and downward direction. both side on Plus and minus side of centre of the screen so bar. if d between the slit increases ringe width decreases , if slit between distance between slit is very very greater than D then fringewidth is small, very very less than one that is very very less than 1 then maxima and minima so in general difficult to observe.