A single slit diffraction pattern is formed with white light. For what wavelength of light the $4^{\text {th }}$ secondary maximum in diffraction pattern coincides with the $3^{\text {rd }}$ secondary maximum in the pattern of light of wavelength ' $\lambda$ ' ?

To determine the condition for secondary maxima coincidence in single-slit diffraction, use the general formula $a \sin \theta = (n + \frac{1}{2}) \lambda$, where $a$ is the slit width, $\theta$ is the diffraction angle, $n$ is the order (starting at $n=1$ for the first secondary maximum), and $\lambda$ is the wavelength.

For the $4^{\text{th}}$ secondary maximum of wavelength $\lambda'$ at angle $\theta_1$:
$a \sin \theta_1 = (4 + \frac{1}{2}) \lambda' = \frac{9}{2} \lambda'$

For the $3^{\text{rd}}$ secondary maximum of wavelength $\lambda$ at angle $\theta_2$:
$a \sin \theta_2 = (3 + \frac{1}{2}) \lambda = \frac{7}{2} \lambda$

Since the angular positions coincide, $\sin \theta_1 = \sin \theta_2$, equating the two expressions:
$\frac{9}{2} \lambda' = \frac{7}{2} \lambda$

Multiply both sides by 2:
$9 \lambda' = 7 \lambda$

Solve for $\lambda'$:
$\lambda' = \frac{7}{9} \lambda$

The wavelength for which the $4^{\text{th}}$ secondary maximum coincides with the $3^{\text{rd}}$ secondary maximum of wavelength $\lambda$ is $\frac{7 \lambda}{9}$.

The option is B.