In a double-slit experiment, the fourth-order maximum for a wavelength of 450 nm occurs at an angle of $\mathit{\theta }\mathbf{=}\mathbf{90}\mathbf{°}$. (a) What range of wavelengths in the visible range (400 nm to 700 nm) are not present in the third-order maxima? To eliminate all visible light in the fourth-order maximum, (b) should the slit separation be increased or decreased and (c) what least change is needed?

The wavelength of light is ${\lambda }_4=450\text{nm}$

The order of fringe for fourth order is $m_4=4$

The angle for the fourth order fringe is $\theta =90°$

The order of fringe for third order is $m_3=3$

The phase difference of fringe pattern varies with the path difference. For minimum phase difference path difference should be minimum and vice versa.

The separation between slits is given as:

$d\sin \theta =m_4{\lambda }_4$

Substitute all the values in the above equation.

$\begin{array}{rcl}d\left(\sin 90°\right)& =& \left(4\right)\left(450\text{nm}\right)\\ d& =& 1800\text{nm}\end{array}$

The lowest wavelength absent in visible range is given as:

$d\sin \theta =m_3{\lambda }_3$

Substitute all the values in equation.

$\begin{array}{rcl}\left(1800\text{nm}\right)\left(\sin 90°\right)& =& \left(3\right){\lambda }_3\\ {\lambda }_3& =& 600\text{nm}\end{array}$

The wavelength from $600\text{nm to 700 nm}$ will be absent in the visible range.

Therefore, the wavelength from $600\text{nm to 700 nm}$will be absent in the visible range.

The slit separation varies inversely with wavelength of light. The wavelength of light for third order fringe increases so slit separation for third order wavelength of light decreases.

Therefore, the slit separation for third order wavelength decreases.

The separation for lowest wavelength of visible range is given as:

$d_l\sin \theta =m_4{\lambda }_l$

Here, ${\lambda }_l$ is the lowest wavelength of visible range and its value is $400\text{nm}$.

Substitute all the values in equation.

$\begin{array}{rcl}{d}_l\left(\sin 90°\right)& =& \left(4\right)\left(400\text{nm}\right)\\ d_l& =& 1600\text{nm}\end{array}$

The least change needed is given as:

$D=d-d_l$

Substitute all the values in equation.

$\begin{array}{rcl}D& =& 1800\text{nm}-1600\text{nm}\\ D& =& 200\text{nm}\end{array}$

Therefore, the least change needed in separation is $200\text{nm}$.