In Young's double slit experiment , one of the slit is wider than other, so that amplitude of the light from one slit is double of that from other slit. If ${{\rm I}_m}$ be the maximum intensity, the resultant intensity ${\rm I}$ when they interfere at phase difference $\phi $ is given by :

Question

${{{I_m}} \over 9}\left( {4 + 5\cos \,\phi } \right)$

${{{I_m}} \over 3}\left( {1 + 2{{\cos }^2}\,{\phi \over 2}} \right)$

${{{I_m}} \over 3}\left( {1 + 4{{\cos }^2}\,{\phi \over 2}} \right)$

${{{I_m}} \over 9}\left( {1 + 8{{\cos }^2}\,{\phi \over 2}} \right)$

Solution

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