Question

To illustrate one of the ideas of holography in a simple way, consider a diffraction grating with slit spacing $d$. The small-angle approximation is usually not valid for diffraction gratings, because $d$is only slightly larger than $\lambda$, but assume that the $\lambda /d$ratio of this grating is small enough to make the small-angle approximation valid.

a. Use the small-angle approximation to find an expression for the fringe spacing on a screen at distance $L$behind the grating.

b. Rather than a screen, suppose you place a piece of film at distance $L$ behind the grating. The bright fringes will expose the film, but the dark spaces in between will leave the film unexposed. After being developed, the film will be a series of alternating light and dark stripes. What if you were to now “play” the film by using it as a diffraction grating? In other words, what happens if you shine the same laser through the film and look at the film’s diffraction pattern on a screen at the same distance $L$? Demonstrate that the film’s diffraction pattern is a reproduction of the original diffraction grating

Short answer

(a) The expression for the fringe spacing on a screen is$Y=\frac{\lambda L}{d}$

(b) The diffraction is the scattering pattern created by the film., i.e.$Y_2=d$

Step-by-step solution

Find the expression for the fringe spacing (part a)

We can deduce the following from the grating equation:

$d\sin \left({\theta }_1\right)=\lambda$

According to the small-angle estimate,

$\sin \left({\theta }_1\right)\approx {\theta }_1$

If a simple figure is required, this angle is just one that is such that

$\sin \left({\theta }_1\right)=\frac{Y}{L}$

where $Y$is the first fringe's vertical position. As a result, for$Y$, we can get

$d\frac{Y}{L}=\lambda \Rightarrow Y=\frac{\lambda L}{d}$

Find the diffraction for the fringe (part b)

(a) Next consider the diffraction caused by the shining light on the film: the spacing between the grating "slits" will be $Y$, the range to the screen will be $L$, and the first fringe created by the film will have a distance of $Y_2$from the centre. We can conclude the foregoing from the grating equation:

$Y\sin \left({\theta }_1\right)=\lambda$

We use the critical angle approximation once again to express the angle's sine as

$\sin \left({\theta }_1\right)=\frac{Y_2}{L}$

As a result of the diffraction equation, we get the following:

$Y_2=\frac{\lambda L}{Y}=\frac{\lambda L}{\frac{\lambda L}{d}}=d$

It indicates that the grating will become the image formed.