question_answer

Two sources of intensity \[{{I}_{1}}\] and \[{{I}_{2}}\] undergo interference in Young's double slit experiment. Show that \[\frac{{{I}_{\max }}}{{{I}_{\min }}}={{\left( \frac{{{a}_{1}}+{{a}_{2}}}{{{a}_{1}}-{{a}_{2}}} \right)}^{2}}\], where \[{{a}_{1}}\] and \[{{a}_{2}}\] are the amplitudes of disturbance for two sources \[{{S}_{1}}\], and \[{{S}_{2}}\].

Answer:

                When two light waves of amplitudes \[{{a}_{1}}\], and \[{{a}_{2}}\] and having phase difference \[\phi \] interfere, the resultant amplitude is \[a=\sqrt{{{a}_{1}}^{2}+{{a}_{2}}^{2}+2{{a}_{1}}{{a}_{2}}\cos \phi }\] As intensity \[\propto \] \[{{\left( \text{amplitude} \right)}^{\text{2}}}\] \[\therefore \]  \[I=k{{a}^{2}}=k({{a}_{1}}^{2}+{{a}_{2}}^{2}+2{{a}_{1}}{{a}_{2}}\cos \phi )\] When \[\phi =0,\cos \phi =1\]and the intensity is maximum, \[{{I}_{\max }}=k({{a}_{1}}^{2}+{{a}_{2}}^{2}+2{{a}_{1}}{{a}_{2}}\times 1)=k{{({{a}_{1}}+{{a}_{2}})}^{2}}\] When \[\phi =\pi ,\cos \phi =-1,\]the intensity is minimum, \[{{I}_{\min }}=k({{a}_{1}}^{2}+{{a}_{2}}^{2}-2{{a}_{1}}{{a}_{2}})=k{{({{a}_{1}}-{{a}_{2}})}^{2}}\] \[\therefore \]  \[\frac{{{I}_{\max }}}{{{I}_{\min }}}=\frac{k{{({{a}_{1}}+{{a}_{2}})}^{2}}}{k{{({{a}_{1}}-{{a}_{2}})}^{2}}}={{\left( \frac{{{a}_{1}}+{{a}_{2}}}{{{a}_{1}}-{{a}_{2}}} \right)}^{2}}.\]