Show that the irradiance, Im, of the m^th subsidiary peak,...

Transcript

00:01 Hello students here in this question from the given function that is f of xy is equals to 4xy cube plus 1 i plus 6x square y square plus 2 e raise to the power 2 y j gap now from this we'll get del f over del x is equal to 4x y cube plus 1 and del f over del y is equal to 6 x square y square plus 2 into a raise to the power 2 y now take del f over del x is equals to 4x y q plus 1 which implies del f is equal to 4xy q plus 1 into del x now take integral on both side therefore we'll get after integrating this function with respect to x f of x y is equals to 2x square y q plus x plus h of y where h of y is constant of integration now differentiate this function with respect to y therefore we'll get del f over del y is equal to 6x square y square plus h dash y now from above put the value of del f over del y that is 6x square y square plus 2 erase to the power 2 y is equal to 6x square y square plus h dash y now here 6x square y square cancels out with 6x square y square therefore we'll get at dash y is equal to 2 into erase to the power to 2 x x square square 2y now after integrating this function with respect to y will get h of y is equal to e raised to the power 2y plus c now after putting the value of h above will get the potential function will be f of xy is equal to 2x square y cube plus x plus e raise to the power 2y plus c now, as we have to calculate integral over c f dr, that is equals to del f over c, dr, which is again equals to potential function f of r of pi by 2 minus f of r8 0, where r8 pi by 2 is equals to 2 sine 5, pi by 2, i cap plus 2 into t that is pi by 2 divided by 5 sine 4 5 5 by 2 by 2 which implies r8 pi by 2 is calculated as 2 .1 now r8 0 is calculated as 2 into sine 5 of 0 i cap plus 2 into 0 divided by pi 5 5 5 5 5 4 5 into 0 j which implies r8 0 is therefore integral fdr is equals to f of 21 minus fh 0 which implies where f of xy is the potential function that we have find above that is 2x2 y cube plus x plus x plus e raise to the power to y plus c which implies integral f dot d is equals to 2 into 4 into 1 plus 2 plus e raised to the power 2 plus c minus 0 plus 0 plus 0 plus 0 plus 0 plus c which implies 10 plus e raise to the power 2 plus c minus c here c can see out with c therefore we'll get 9 plus e raise to the power 2 therefore value of the integral that is f dot t r over c is calculated as 9 plus e -d -st -the -power 2 that is equals to 16 .380...

Question

Show that the irradiance, $I_m$, of the $m^{th}$ subsidiary peak, can be well approximated by $I_m = I(0) \left[ \frac{1}{(m + \frac{1}{2})\pi} \right]^2$

Question

Show that the irradiance, $I_m$, of the $m^{th}$ subsidiary peak, can be well approximated by $I_m = I(0) \left[ \frac{1}{(m + \frac{1}{2})\pi} \right]^2$

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