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QJEEM12Pha8a0: JEE Main 2012 Physics - Class 12 JEEM12Pha8a0 Solution: In Young's double slit experiment , one of the slit is wider…

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 Im{{\rm I}_m} be the maximum intensity, the resultant intensity I{\rm I} when they interfere at phase difference ϕ\phi is given by :

Multiple Choice Options

  • A Im9(4+5cos⁡ ϕ){{{I_m}} \over 9}\left( {4 + 5\cos \,\phi } \right)
  • B Im3(1+2cos⁡2 ϕ2){{{I_m}} \over 3}\left( {1 + 2{{\cos }^2}\,{\phi \over 2}} \right)
  • C Im3(1+4cos⁡2 ϕ2){{{I_m}} \over 3}\left( {1 + 4{{\cos }^2}\,{\phi \over 2}} \right)
  • D Im9(1+8cos⁡2 ϕ2){{{I_m}} \over 9}\left( {1 + 8{{\cos }^2}\,{\phi \over 2}} \right)

Step-by-Step Verified Solution

Let a1=a, I1=a12=a2{a_1} = a,\,{I_1} = a_1^2 = {a^2}

a2=2a, I2=a22=4a2{a_2} = 2a,\,{I_2} = a_2^2 = 4{a^2}

Therefore I2=4I1{{\rm I}_2} = 4{{\rm I}_1}

Ir=I1+I2+2I1I2cos⁡ϕ{I_r} = {I_1} + {I_2} + 2\sqrt {{I_1}{I_2}\cos \phi }

Ir=I1+4I1+24I12 cos⁡ϕ{I_r} = {I_1} + 4{I_1} + 2\sqrt {4I_1^2} \,\cos \phi

⇒Ir=5I1+4I1cos⁡ϕ              ...(1)\Rightarrow {I_r} = 5{I_1} + 4{I_1}\cos \phi \,\,\,\,\,\,\,\,\,\,\,\,\,\,...\left( 1 \right)

Now, Imax⁡=(a1+a2)2=(a+2a)2=9a2{I_{\max }} = {\left( {{a_1} + {a_2}} \right)^2} = {\left( {a + 2a} \right)^2} = 9{a^2}

Imax⁡=9I1⇒I1=Imax9{I_{\max }} = 9{I_1} \Rightarrow {I_1} = {{{{\mathop{\rm I}\nolimits} _{max}}} \over 9}

Substituting in equation (1)\left( 1 \right)

Ir=5Imax⁡9+4Imax⁡9cos⁡ϕ{I_r} = {{5{I_{\max }}} \over 9} + {{4{I_{\max }}} \over 9}\cos \phi

Ir=Imax⁡9[5+4cos⁡ϕ]{I_r} = {{{I_{\max }}} \over 9}\left[ {5 + 4\cos \phi } \right]

Ir=Imax⁡9[5+8cos⁡2ϕ2−4]{I_r} = {{{I_{\max }}} \over 9}\left[ {5 + 8{{\cos }^2}{\phi \over 2} - 4} \right]

Ir=Imax⁡9[1+8cos⁡2ϕ2]{I_r} = {{{I_{\max }}} \over 9}\left[ {1 + 8{{\cos }^2}{\phi \over 2}} \right]

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