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Bukhovtsev-et-al-Problems-in-Elementary-Physics

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434<br />

ANSWERS AND SOLUTIONS<br />

cent slits Is<br />

6=BD-AC=dsln tp-dsln a<br />

These waves add up and Intensify each other when<br />

d (s<strong>in</strong> cp-s<strong>in</strong> 8)=kl<br />

where k= I, 2, 3, ... for the maxima ly<strong>in</strong>g at the right of the centr<strong>al</strong> one<br />

(Jl=O) and k=~I, -2, -3, ••• for those ly<strong>in</strong>g at its left.<br />

The maximum order of the spectrum will be observed when q>=-9()O.<br />

Thus d (-1- ~ ) =kA. Hence, k=-6. The spectrum of the sixth order<br />

may be observed. The m<strong>in</strong>us sign shows that the spectrum lies to the left<br />

of the centr<strong>al</strong> one.<br />

802. As follows from the formula d (s<strong>in</strong> q>-s<strong>in</strong> 8)=kA (see the solution<br />

to Problem 801), the period of the grat<strong>in</strong>g will be m<strong>in</strong>imum with tangenti<strong>al</strong><br />

Incidence of the rays: 9=90°. In this case d d ~ . Therefore, the period of<br />

the grat<strong>in</strong>g should satisfy the <strong>in</strong>equ<strong>al</strong>ity d~ ~ •<br />

803. In the gener<strong>al</strong> case, as shown <strong>in</strong> the solution of Problem 801, the<br />

sought condition will be<br />

d (s<strong>in</strong> cp-s<strong>in</strong> 9)=kA<br />

It may be rewritten as<br />

2dcos CPt 9 s<strong>in</strong> cP 2 9=kA<br />

q>+9 T 9

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