In a double slit pattern ($\lambda$ = 6000 Å), the first order and tenth order maxima fall at 12.50 mm and 14.75 mm from a particular reference point. If $\lambda$ is changed to 5500 Å, find the position of zero order and tenth order fringes, other arrangements remaining the same.

Distance between 10 fringes is 

              $9\omega =$$\left(14.75-12.50\right)$ mm 

                    $=2.25$ mm 

Fringe width, $\omega =0.25 mm$

When the wavelength is changed from 6000 Å to 5500 Å, the new fringe width will become,

      ${\omega }^{\text{'}}=$$\left(\frac{5500}{6000}\right)\omega$

            $=\left(\frac{5500}{6000}\right)\left(0.25\right)$

      As Fringe width  $\alpha \lambda$

         ${\omega }^{\text{'}}=0.23$mm

The position of central (or zero order) maxima will remain unchanged. Earlier it was at a position,

     $Y_0=Y_1-\omega$     $=\left(12.50-0.25\right)=12.25 mm$

        The new position of tenth order maxima will be 

        $Y_{10}=Y_0+10$$$\begin{gathered} \omega \text{'}=\left(12.25\right)+\left(10\right)\left(0.23\right) \\ =14.55 mm \end{gathered}$$