Question

In a double slit pattern $(\lambda =6000A^{\circ })$, the first order and tenth order maxima fall at $13.00\text{mm}$ and $15.25\text{mm}$ from a particular reference point. If $\lambda$ is changed to $5500A^{\circ }$, find the position of tenth order fringe (in mm) (answer the nearest integer), other arrangement remaining the same.

Answer 1

Distance between 10 fringes is $9_{\omega }=(15.25-13.00)\text{mm}=2.25\text{mm}$
$\therefore$ Fringe width $=0.25\text{mm}$
when the wavelength is changed from $6000A^{\circ }$
$\Rightarrow 5500A^{\circ },$
the new fringe width will become -
${\omega }^{\prime }=\left(\frac{5500}{6000}\right)\omega =\left(\frac{5500}{6000}\right)0.25$
as fringe width $\propto \lambda$
$\therefore {\omega }^{\prime }=0.23\text{m}$
The position of central maxima will remain unchanged.
$y_0=y_1-\omega =(13.00-0.25)=12.75\text{mm}$
The new position of tenth order maxima will be -
$Y_{10}=Y_0+10{\omega }^{\prime }$
put the values -
$Y_{10}=12.75+10\left(0.23\right)=12.75+2.3=15.05\text{mm}$
$\approx 15\text{mm}$