Answer
640 nm.
Work Step by Step
Equation 24–2a gives the condition for constructive interference, to be applied to the second-order blue light.
$$d sin\theta=m\lambda_{blue}=2(480nm)$$
We use this with equation 24–2b because the unknown wavelength and unknown m’ result in an interference minimum.
$$d sin\theta=(m’+\frac{1}{2})\lambda_x$$
Note that the slit separation and angle are the same in both situations.
$$2(480nm)= (m’+\frac{1}{2})\lambda_x $$
Enumerate the possibilities and see which wavelength falls within the visible spectrum, 400-700 nm.
For m’=0
$$2(480nm)= (0+\frac{1}{2})\lambda_x $$
$$\lambda_x =1920nm$$
For m’=1
$$2(480nm)= (1+\frac{1}{2})\lambda_x $$
$$\lambda_x =640nm$$
For m’=2
$$2(480nm)= (2+\frac{1}{2})\lambda_x $$
$$\lambda_x =384nm$$
The solutions will get smaller and smaller after this. The only solution in the visible region of 400-700 nm is 640 nm.