Question 173: Waves Optics

Question

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 :

Accepted Answer

Let $${a_1} = a,\,{I_1} = a_1^2 = {a^2}$$
$${a_2} = 2a,\,{I_2} = a_2^2 = 4{a^2}$$
Therefore $${{m I}_2} = 4{{m I}_1}$$
$${I_r} = {I_1} + {I_2} + 2\sqrt {{I_1}{I_2}\cos \phi } $$
$${I_r} = {I_1} + 4{I_1} + 2\sqrt {4I_1^2} \,\cos \phi $$
$$ \Rightarrow {I_r} = 5{I_1} + 4{I_1}\cos \phi \,\,\,\,\,\,\,\,\,\,\,\,\,\,...\left( 1 ight)$$
Now, $${I_{\max }} = {\left( {{a_1} + {a_2}} ight)^2} = {\left( {a + 2a} ight)^2} = 9{a^2}$$
$${I_{\max }} = 9{I_1} \Rightarrow {I_1} = {{{{\mathop{m I}
olimits} _{max}}} \over 9}$$
Substituting in equation $$\left( 1 ight)$$
$${I_r} = {{5{I_{\max }}} \over 9} + {{4{I_{\max }}} \over 9}\cos \phi $$
$${I_r} = {{{I_{\max }}} \over 9}\left[ {5 + 4\cos \phi } ight]$$
$${I_r} = {{{I_{\max }}} \over 9}\left[ {5 + 8{{\cos }^2}{\phi  \over 2} - 4} ight]$$
$${I_r} = {{{I_{\max }}} \over 9}\left[ {1 + 8{{\cos }^2}{\phi  \over 2}} ight]$$

Answer

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

Answer

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

Answer

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

Answer

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

Waves Optics

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 :

Choose the correct answer: