Answer
a). Location of minima = $y_{n}=(n+\frac{1}{2})\frac{vL}{d}$, n=1,2,3, .....
Distance between adjacent minima is $\frac{vL}{d}$
b). Path length difference = $\frac{5}{2}\frac{v}{d}$
Work Step by Step
a). To derive a relation which gives location of minima in Young's double slit experiment.
v=wavelength
$\Delta L=\frac{mv}{2}$, m=1,3,5, ....
$dsin\theta=\Delta L=\frac{mv}{2}$, m=1,3,5 ....
Again, $dsin\theta=nv$, n=0,1,2,3 .....
Comparing above two , $m=(2n+1)$
or, $dsin\theta=(n+\frac{1}{2})v$, n=0,1,2,3,.... Condition for minima
Taking the approximation for small $\theta$, $tan\theta=sin\theta=\frac{y}{L}$
So, $d\frac{y_{n}}{L}=(n+\frac{1}{2})v$, n=1,2,3, ....
or, $y_{n}=(n+\frac{1}{2})\frac{vL}{d}$, n=1,2,3, .....
This gives location of minima.
The distance between adjacent minima = $y_{n+1}-y_{n}=(n+1+\frac{1}{2})\frac{vL}{d}-(n+\frac{1}{2})\frac{vL}{d}=\frac{vL}{d}$
b). For third order minimum we need to find path length difference between that location and the two slits.
From a, $y_{n}=(n+\frac{1}{2})\frac{vL}{d}$
or, $\frac{y_{n}}{L}=(n+\frac{1}{2})\frac{v}{d}$
For third minimum, n=2, $\frac{y_{2}}{L}=(2+\frac{1}{2})\frac{v}{d}=\frac{5}{2}\frac{v}{d}$
