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Unformatted text preview: 1 a for long wavelengths, / 1 B hc k T = So / 1 / B hc k T B e hc k T  = 2 2 / 5 5 4 2 2 1 2 1 B B B hc k T k T ck T hc hc I e hc = = = 1 2 3 4 5 6 5 10 15 20 The red one is RayleighJeans function, the Blue one is Planck distribution. The X axis is the wavelength. 2 (a) 5 1 x x y e = ( 29 4 5 2 5 1 1 x x x x x e y e e = Let y = , so we get the maximum point( we know from the question one that, it should be maximum point, not the minimum, so we dont need to derive the second order derivative) ( 29 max max 5 5 x x e = There are many ways of using the Newton method, here I give one example. With ( 29 ( ) 5 5 x f x x e = , and initial guess (1) 5 x = , we have x( 1 )= 5 x( 2 )= 4.9663102650045726645 x( 3 )= 4.965115686301458572 x( 4 )= 4.965114231746430276 x( 5 )= 4.965114231744276304 x( 6 )= 4.965114231744276304 x( 7 )= 4.96511423174427630 x( 8 )= 4.96511423174427630 x( 9 )= 4.96511423174427630 x( 10 )= 4.9651142317442763 max 4.965 x...
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This note was uploaded on 03/31/2011 for the course PHY 126 taught by Professor Hobunchan during the Spring '11 term at HKUST.
 Spring '11
 HoBunCHAN
 Physics

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