JEE MAIN - Physics (2012 - No. 7)
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 :
$${{{I_m}} \over 9}\left( {4 + 5\cos \,\phi } \right)$$
$${{{I_m}} \over 3}\left( {1 + 2{{\cos }^2}\,{\phi \over 2}} \right)$$
$${{{I_m}} \over 3}\left( {1 + 4{{\cos }^2}\,{\phi \over 2}} \right)$$
$${{{I_m}} \over 9}\left( {1 + 8{{\cos }^2}\,{\phi \over 2}} \right)$$
Explanation
Let $${a_1} = a,\,{I_1} = a_1^2 = {a^2}$$
$${a_2} = 2a,\,{I_2} = a_2^2 = 4{a^2}$$
Therefore $${{\rm I}_2} = 4{{\rm I}_1}$$
$${I_r} = {I_1} + {I_2} + 2\sqrt {{I_1}{I_2}\cos \phi } $$
$${I_r} = {I_1} + 4{I_1} + 2\sqrt {4I_1^2} \,\cos \phi $$
$$ \Rightarrow {I_r} = 5{I_1} + 4{I_1}\cos \phi \,\,\,\,\,\,\,\,\,\,\,\,\,\,...\left( 1 \right)$$
Now, $${I_{\max }} = {\left( {{a_1} + {a_2}} \right)^2} = {\left( {a + 2a} \right)^2} = 9{a^2}$$
$${I_{\max }} = 9{I_1} \Rightarrow {I_1} = {{{{\mathop{\rm I}\nolimits} _{max}}} \over 9}$$
Substituting in equation $$\left( 1 \right)$$
$${I_r} = {{5{I_{\max }}} \over 9} + {{4{I_{\max }}} \over 9}\cos \phi $$
$${I_r} = {{{I_{\max }}} \over 9}\left[ {5 + 4\cos \phi } \right]$$
$${I_r} = {{{I_{\max }}} \over 9}\left[ {5 + 8{{\cos }^2}{\phi \over 2} - 4} \right]$$
$${I_r} = {{{I_{\max }}} \over 9}\left[ {1 + 8{{\cos }^2}{\phi \over 2}} \right]$$
$${a_2} = 2a,\,{I_2} = a_2^2 = 4{a^2}$$
Therefore $${{\rm I}_2} = 4{{\rm I}_1}$$
$${I_r} = {I_1} + {I_2} + 2\sqrt {{I_1}{I_2}\cos \phi } $$
$${I_r} = {I_1} + 4{I_1} + 2\sqrt {4I_1^2} \,\cos \phi $$
$$ \Rightarrow {I_r} = 5{I_1} + 4{I_1}\cos \phi \,\,\,\,\,\,\,\,\,\,\,\,\,\,...\left( 1 \right)$$
Now, $${I_{\max }} = {\left( {{a_1} + {a_2}} \right)^2} = {\left( {a + 2a} \right)^2} = 9{a^2}$$
$${I_{\max }} = 9{I_1} \Rightarrow {I_1} = {{{{\mathop{\rm I}\nolimits} _{max}}} \over 9}$$
Substituting in equation $$\left( 1 \right)$$
$${I_r} = {{5{I_{\max }}} \over 9} + {{4{I_{\max }}} \over 9}\cos \phi $$
$${I_r} = {{{I_{\max }}} \over 9}\left[ {5 + 4\cos \phi } \right]$$
$${I_r} = {{{I_{\max }}} \over 9}\left[ {5 + 8{{\cos }^2}{\phi \over 2} - 4} \right]$$
$${I_r} = {{{I_{\max }}} \over 9}\left[ {1 + 8{{\cos }^2}{\phi \over 2}} \right]$$
Comments (0)
