Answer
See the detailed answer below.
Work Step by Step
$$\color{blue}{\bf [a]}$$
In a diffraction grating, the bright fringe position is given by
$$y_m=L\tan\theta_m\tag 1$$
And
and the author told us to use the small-angle approximation which means that $\tan\theta_m\approx\sin\theta_m$,
We also know, in a diffraction grating, that
$$d\sin\theta_m=m\lambda\tag{$m=0,1,2,3,...$}$$
Hence,
$$\sin\theta=\dfrac{m\lambda}{d}$$
Plugging into (1) since $\tan\theta_m\approx\sin\theta_m$,
$$y_m=\dfrac{m\lambda L}{d} \tag 2$$
We know that the fringe spacing is given by
$$\Delta y=y_{m+1}-y_m$$
Therefore,
$$\Delta y=\dfrac{(m+1)\lambda L}{d} -\dfrac{m\lambda L}{d} $$
$$\boxed{\Delta y= \dfrac{ \lambda L}{d} }$$
$$\color{blue}{\bf [b]}$$
Now we have a new diffraction grating that has a separation distance between its slit of $d'=\Delta y$ (the boxed formula above).
So to demonstrate that the film’s diffraction pattern is a reproduction of the original diffraction grating, we need to proof that $\Delta y'=\dfrac{\lambda L}{d'}$
By the same approach, from (2),
$$y_m'=\dfrac{m\lambda L}{d'}$$
So,
$$\Delta y'=y_{m+1}-y_m$$
Therefore,
$$\Delta y'=\dfrac{(m+1)\lambda L}{d'} -\dfrac{m\lambda L}{d'} $$
$$ \Delta y'= \dfrac{ \lambda L}{d'} \tag 3$$
$$\Delta y'= \dfrac{ \color{red}{\bf\not} \lambda \color{red}{\bf\not} L d}{\color{red}{\bf\not} \lambda \color{red}{\bf\not} L}=d$$
And hence, $$d=d'$$
Plugging into (3);
$$\boxed{ \Delta y'= \dfrac{ \lambda L}{d} }$$
which is as same as the original pattern.
