Question #4e709

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Question #4e709

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Answer 1 · 2016-06-02T11:29:50

9 maxima measured from the normal

Explanation:

The equation for determining the orders of maxima seen is:

$n λ = d sin θ$ $nlamda=dsintheta$

Where

$n$ $n$ = order of maxima (0,1,2,3..)
$d$ $d$ = separation of slits in grating= $\frac{1}{2 \times 10^{5}} = 5 \times 10^{- 6} m$ $1/(2times10^5)=5times10^-6m$
$θ$ $theta$ =the angle at which the interference patterns are seen (measured normal to the grating)

$λ$ $lamda$ =wavelength = $625 \times 10^{- 9} m$ $625times10^-9m$

The maximum angle( $θ$ $theta$ ), that is possible is 90 degrees, and ${sin 90}^{0} = 1$ $sin90^0=1$

The largest possible order of maxima is therefore $n λ = d$ $nlamda=d$

So in this example:

$n \cdot (625 \times 10^{- 9}) = (5 \times 10^{- 6})$ $n*(625times10^-9)=(5times10^-6)$

$n = 8$ $n=8$

Including the zero order this would give 9 maxima measured from the normal (in practice, 8 observed either side of the zero order central maxima)

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Question #4e709

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