In a Young's double slit experiment, 10 fringes are observed in a...

The length of the segment on the screen remains constant. The total width of $n$ fringes is given by $y = n\beta = n\left(\dfrac{\lambda D}{d}\right)$.

Since $y$, $D$, and $d$ are constant, $n \propto \dfrac{1}{\lambda}$, which gives $n_1\lambda_1 = n_2\lambda_2$.

The initial number of fringes is $n_1 = 10$ and the initial wavelength is $\lambda_1 = 600\text{ nm}$.

The number of fringes increases by $5$, so the new number of fringes is $n_2 = 10 + 5 = 15$.

Substituting these values into the relation:
$10 \times 600 = 15 \times \lambda_2$
$\lambda_2 = \dfrac{6000}{15} = 400\text{ nm}$.

Answer: $400\text{ nm}$