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

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390<br />

ANSWERS AND SOLUTIONS<br />

Then. accord<strong>in</strong>g to the law of refraction,<br />

s<strong>in</strong> a nt<br />

s<strong>in</strong> ~ = no<br />

s<strong>in</strong> p n'<br />

--=s<strong>in</strong><br />

y nl<br />

s<strong>in</strong> 'V n"<br />

s<strong>in</strong> ~=n'<br />

Fig. 503<br />

s<strong>in</strong> q> n 2<br />

s<strong>in</strong>, = n(n)<br />

s<strong>in</strong> ~ na<br />

s<strong>in</strong>x=n2<br />

Upon multiply<strong>in</strong>g these equations we" g<strong>et</strong><br />

s<strong>in</strong> a ns<br />

SiiiX= no<br />

Hence, the angle at which the beam leaves the plate<br />

x=arcs<strong>in</strong> (:: ~<strong>in</strong> a)<br />

depends only on the angle of <strong>in</strong>cidence of the beam on the plate and on the<br />

refraction <strong>in</strong>dices of the media on both sides of the plate. In particular, if<br />

ng=no. then x=a..

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