Answer
$$\theta=arcsin(0.4-sin\psi)$$
Work Step by Step
In question 81, we are provided with the expression which related the angular position of the bright diffraction fringe $\theta$ and incidence angle $\psi$
$$d(sin\psi+sin\theta)=m\lambda$$
where $d$ is the spacing between gratings, $m$ is the order of diffraction and $\lambda$ is the incident wavelength.
We are given the information that $m=1$, $d=1.50 \mu m$ and $\lambda=600 nm$
Hence
$$sin\psi+sin\theta=\frac{m\lambda}{d}=\frac{1*600*10^{-9}m}{1.50*10^{-6}m}$$
Thus $$sin\psi+sin\theta=0.4$$
Thus, $$sin\theta=0.4-sin\psi$$
We then isolate $\theta$
$$\theta=arcsin(0.4-sin\psi)$$
Hence our $x$ value is $\psi$ and $y$ value is $\theta$ which can now be plotted using any graphic calculator. The resulting graph is attached with domain $$0\leq\psi\leq\frac{\pi}{2}$$
