# Please solve and also explain the reason during the same... Let two waves from S1 and S2 interfere at P.   The two waves are given by where δ is the phase difference between two waves The resultant is given by (student should workout the required trignometry/algebra here )   Above equation shows that superposition of two sinusoidal waves with same frequency but with phase difference gives superposition of two sinusoidal waves of same frequency but different amplitudes. Where E is the resultant amplitude and φ is the resultant phase   By squaring and adding the above two equations we get since Intensity I proportional to E2, we can write If I1 = I2 = I0 and δ = 0, we get maximum Imax = 4I0 If I1 = I2 = I0 and δ = 2π, we get minimum Imin = 0 If I1 = I2 = I0 and δ  ranging from 0 to 2π, we get intensity between Imax   and Imin which is given by I = (1+cosδ)2I0 . When this intensity is averaged out for δ = 0 to 2π, we get average value 2I0   Hence the option (A) in the question is correct answer Answered by  | 4th Dec, 2017, 12:18: PM Related Videos This video explains Young's double slit experiment for interference. This video establishes few mathematical expressions pertaining to interfere... This video provides expert tips and contains practice problems on interfere... \$(window).on('load', function(){ if(showAds){ googletag.cmd.push(function() { googletag.defineSlot('/1039154/TopperLearning_WEB/TL_WEB_728x90_BTF', [[970, 90],[728, 90],[468, 60]], 'div-gpt-ad-1539848643692-article-detail-btf').addService(googletag.pubads()).setCollapseEmptyDiv(true); //googletag.pubads().enableSingleRequest(); // googletag.enableServices(); }); } }); \$(window).on('load', function(){ if(showAds){ googletag.cmd.push(function() { googletag.display('div-gpt-ad-1539848643692-article-detail-btf'); }); } }); All Questions Ask Doubt Question 15 kindly answer kindly provide answer pl In young double slit experiment, slit with ratio is 3:2 then find out the ratio of the maximum and minimum intensity of interference of light. How is phase difference related to path difference? In Young's double slit experiment, the distance between the slit is kept smaller than the wavelength of the light. What is the effect? Why is Interference pattern not detected when the two coherent sources are far apart? The phase difference between two light waves emerging from the slits of Young’s experiment is radians. Will the central fringe be bright or dark? What will be the effect on the interference fringes obtained in Young's double slit experiment if the monochromatic light is replaced by white light source? In a Young's double slit experiment, the interval between the slits is 0.200 mm.For the light of wavelength 6000 A0, interference fringes are formed on a screen at a distance of 0.800 m. a) what is the distance of second dark fringe from the central fringe? b) What is the distance of second bright fringe from the central fringe? Queries asked on Sunday & after 7pm from Monday to Saturday will be answered after 12pm the next working day.

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