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Question

In a Young's double slit experiment, the intensity at the central maximum is $${ I }_{ 0 }$$. The intensity at a distance $$\beta /4$$ from the central maximum is ($$ \beta $$ is fringe width )


A
I0
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B
I02
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C
I02
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D
I04
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Solution

The correct option is B $$\dfrac { I_{ 0 } }{ 2 } $$
Central max intensity $$=I_0$$
$$\begin{array}{l} \beta =\dfrac { { \lambda D } }{ d }  \\ Y=\dfrac { { \lambda D } }{ { 4d } }  \\ \Delta x=\dfrac { { Yd } }{ D } \Rightarrow \Delta x=\dfrac { { \lambda D\, \, d } }{ { 4d\, \, D } } =\dfrac { \lambda  }{ 4 }  \\ \emptyset =\dfrac { { 2\pi  } }{ \lambda  } \times \dfrac { \lambda  }{ 4 } =\dfrac { \pi  }{ 2 }  \\ I={ I_{ 1 } }+{ I_{ 2 } }+2\sqrt { { I_{ 1 } }{ I_{ 2 } }\cos  \emptyset  }  \\ \emptyset =0,I={ I_{ 0 } },{ I_{ 1 } }={ I_{ 2 } }={ I_{ s } } \\ { I_{ 0 } }=4{ I_{ s } }\to \left( i \right)  \\ now\, \, los\, \, \emptyset =\dfrac { \pi  }{ 2 }  \\ \Rightarrow I=2{ I_{ s } }\to \left( { ii } \right)  \\ \therefore from\left( i \right) +\left( { ii } \right) , \\ I=\dfrac { { { I_{ 0 } } } }{ 2 }  \end{array}$$

Physics
NCERT
Standard XII

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