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
B
I02
C
I02
D
I04

Solution

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

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