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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 $位$, but assume that the $位/d$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 鈥減lay鈥� 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鈥檚 diffraction pattern on a screen at the same distance $L$? Demonstrate that the film鈥檚 diffraction pattern is a reproduction of the original diffraction grating

Step by step solution

01

Find the expression for the fringe spacing (part a)

We can deduce the following from the grating equation:

$d\sin 鈦�\left({胃}_1\right)=位$

According to the small-angle estimate,

$\sin 鈦�\left({胃}_1\right)鈮�{胃}_1$

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

$\sin 鈦�\left({胃}_1\right)=\frac{Y}{L}$

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

$d\frac{Y}{L}=位鈬�Y=\frac{位L}{d}$

02

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:

$Y\sin 鈦�\left({胃}_1\right)=位$

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

$\sin 鈦�\left({胃}_1\right)=\frac{Y_2}{L}$

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

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

It indicates that the grating will become the image formed.

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