1. Condition for no bending at interface $BC$

If a ray is incident on the common interface $BC$ and passes through without refraction for any angle of incidence, then the refractive indices of the two prisms must be equal for that wavelength:

$n_1=n_2$

This is because Snell's law gives

$n_1 \sin i = n_2 \sin r$

For the ray to go undeviated at the interface for any $i$, we must have $r=i$, hence

$n_1=n_2$

2. Given dispersion relations

$n_1 = 1.2 + \frac{10.8\times 10^{-14}}{\lambda^2}$

$n_2 = 1.45 + \frac{1.8\times 10^{-14}}{\lambda^2}$

Set them equal:

$1.2 + \frac{10.8\times 10^{-14}}{\lambda^2} = 1.45 + \frac{1.8\times 10^{-14}}{\lambda^2}$

3. Solve for $\lambda$

Bring constants to one side and variable terms to the other:

$\frac{10.8\times 10^{-14} - 1.8\times 10^{-14}}{\lambda^2} = 1.45 - 1.2$

$\frac{9.0\times 10^{-14}}{\lambda^2} = 0.25$

So,

$\lambda^2 = \frac{9.0\times 10^{-14}}{0.25}$

$\lambda^2 = 36\times 10^{-14} = 3.6\times 10^{-13}$

Taking square root:

$\lambda = \sqrt{36\times 10^{-14}}$

$\lambda = 6\times 10^{-7}\text{ m}$

Now convert to nm:

$1\text{ nm} = 10^{-9}\text{ m}$

$\lambda = 6\times 10^{-7}\text{ m} = 600\times 10^{-9}\text{ m} = 600\text{ nm}$

4. Final answer

$\boxed{600}$

The derived answer matches the stored correct answer.