Physics - Wave Optics Question with Solution | TestHub

Let the position of the $n^{\text{th}}$ order maximum from the reference point be $x_n$.

$x_n=x_0+n\beta$

For ${\lambda }_1=6000\mathring{A}$:

$x_1=x_0+{\beta }_1=14.50\text{ mm}$

$x_{10}=x_0+10{\beta }_1=16.75\text{ mm}$

Subtracting the first equation from the second:

$9{\beta }_1=16.75-14.50=2.25\text{ mm}$

${\beta }_1=0.25\text{ mm}$

The position of the zero order maximum (central maximum) is:

$x_0=x_1-{\beta }_1=14.50-0.25=14.25\text{ mm}$

When the wavelength is changed to ${\lambda }_2=5500\mathring{A}$, the new fringe width ${\beta }_2$ is:

${\beta }_2={\beta }_1\times \frac{{\lambda }_2}{{\lambda }_1}=0.25\times \frac{5500}{6000}=\frac{11}{48}\text{ mm}\approx 0.23\text{ mm}$

The position of the zero order maximum remains unchanged at $14.25\text{ mm}$ since the central maximum position is independent of wavelength.

The position of the tenth order maximum for the new wavelength is:

$x_{10}^{\prime }=x_0+10{\beta }_2\approx 14.25+10(0.23)=14.25+2.30=16.55\text{ mm}$

Answer: $14.25\mathrm{mm},16.55\mathrm{mm}$