Answer
a) $1.25\times 10^{-4} \;\rm m$
b) $3.89\times 10^{-7}\;\rm nm$
Work Step by Step
a)
In the first case, we focus on the bright bands and we know that the path length difference for this case is given by multiple numbers of wavelengths.
We know that the position of the bright fringe on the screen is given by
$$d\sin\theta=m\lambda$$
And we assume that the angle $\theta$ is too small which means $\sin\theta\approx \tan\theta$
And $\tan\theta$, from the figure below, is given by $\tan\theta=\dfrac{\Delta x}{L}$.
Thus, for $\Delta x$, we need to solve for $\Delta m$ since we are given the separation distance between two bright fringes.
Hence,
$$d\dfrac{\Delta x}{L}=\overbrace{\Delta m}^{=1}\;\;\lambda$$
Working for $d$;
$$d=\dfrac{L \lambda}{\Delta x}$$
Plugging the known;
$$d=\dfrac{5\times 5\times 10^{-7}}{2\times 10^{-2}}=\color{red}{\bf 1.25\times 10^{-4}}\;\rm m$$
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b) First of all, we need to find the position on the screen of the fourth-order minimum for $\lambda=5\times 10^{-7}\;\rm nm$.
The position of the minima in double-slit is given by
$$d\sin\theta=\left(m+\frac{1}{2}\right)\lambda$$
Using the same approach above of $\sin\theta$;
$$ \dfrac{x d}{L}=\left(m+\frac{1}{2}\right)\lambda $$
And for the first case above, when $m=3$ for the fourth order minimum;
$$ \dfrac{x_1d}{L}=\left(3+\frac{1}{2}\right)\lambda_1 $$
$$ \dfrac{x_1d}{L}=3.5\lambda_1\tag 1 $$
For the second unknown wavelength $\lambda_2$
$$ \dfrac{x_2d}{L}=\left(m+\frac{1}{2}\right)\lambda_2 $$
And when $m=4$ for the fifth order minimum;
$$ \dfrac{x_2d}{L}=\left(4+\frac{1}{2}\right)\lambda_2 $$
$$ \dfrac{x_2d}{L}=4.5\lambda_2\tag 2 $$
Noting that these two orders are at the same location on the screen which means that $x_1=x_2$, and hence the left sides of the two equations (1) and (2) are equal.
Therefore, the right sides are equal as well.
$$3.5\lambda_1=4.5\lambda_2$$
Solving for $\lambda_2$;
$$\lambda_2=\dfrac{3.5\lambda_1}{4.5}=\dfrac{3.5\times 5\times 10^{-7} }{4.5}$$
$$\lambda_2=\color{red}{\bf 3.89\times 10^{-7}}\;\rm nm$$
