The correct option is D [ 3+2√( 2 )] ^ 2 S _3 = 4 I _o S _4 = 2 I _o I _max = 4I _o + 2I _o + 2√(4I _o2I _o) = (6 + 4√(2))I _o I _min ...

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Vector Component

If z=λ D4d

Question

If $z = \frac{λ D}{4 d}$

{[3 - 2 \sqrt{2}]}^{2}

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{[3 + \sqrt{2}]}^{2}

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{[3 - \sqrt{2}]}^{2}

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{[3 + 2 \sqrt{2}]}^{2}

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Solution

The correct option is D ${[3 + 2 \sqrt{2}]}^{2}$
$S_{3} = 4 I_{o}$
$S_{4} = 2 I_{o}$

$I_{m a x} = 4 I_{o} + 2 I_{o} + 2 \sqrt{4 I_{o} 2 I_{o}} = (6 + 4 \sqrt{2}) I_{o}$
$I_{m i n} = 4 I_{o} + 2 I_{o} - 2 \sqrt{4 I_{o} 2 I_{o}} = (6 - 4 \sqrt{2}) I_{o}$

Hence $\frac{I_{m a x}}{I_{m i n}} = [3 + 2 \sqrt{2}]^{2}$

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If z=λD4d

The correct option is D [3+2√2]2S3=4IoS4=2IoImax=4Io+2Io+2√4Io2Io=(6+4√2)IoImin=4Io+2Io−2√4Io2Io=(6−4√2)IoHence ImaxImin=[3+2√2]2

If $z = \frac{λ D}{4 d}$