In interference, I_(max)/I_(min) = alpha , find (a) A_(max)/A_(min) (b) A_1/A_2 (c)I_1/I_2 .
Answers
Answer:
MATHS
The maximum value of (cosα
1
).(cosα
2
)...(cosα
n
), under the restrictions 0≤α
1
,α
2
,...α
n
≤
2
π
and (cotα
1
).(cotα
2
)...(cotα
n
)=1 is :
December 27, 2019avatar
Nilraj Hirani
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ANSWER
The given relation implies that
$$\displaystyle \prod _{ i=1 }^{ n }{ \cos{ \alpha }_{ i }= } \prod _{ i=1 }^{ n }{ \sin{ \alpha }_{ i } }$$
$$\therefore\quad \displaystyle\prod _{ i }^{ n }{ { \cos }^{ 2 }\alpha _{ i }=\prod _{ i=1 }^{ n }{ \cos{ \alpha }_{ i }{ \sin\alpha }_{ i } } = } \prod _{ i=1 }^{ n }{ \left( \dfrac { \sin{ \alpha }_{ i } }{ 2 } \right) }$$
Where $$0\le { \alpha }_{ i }'<\dfrac { \pi }{ 2 }$$ or $$0\le { 2\alpha }_{ i }\le \pi$$.
$$\therefore\quad \displaystyle\prod _{ i=1 }^{ n } \cos ^{ 2 }{ { \alpha }_{ i } } =\dfrac { 1 }{ { 2 }^{ n } } $$ as max value of $$\sin { \theta } =1$$
Or $$\displaystyle\prod _{ i=1 }^{ n } \cos { { \alpha }_{ i } } =\dfrac { 1 }{ { 2 }^\cfrac{ n}{2 } } $$
Explanation: