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Subtracting the values ​​indicating the density of the filters: Just as the values ​​indicating the density of a group of filters can be added to each other, these values ​​can also be subtracted. Assuming that the first color image (which was prepared for the initial test) had received a correction of (20-40), then we saw for the purposes of color correction to subtract the values ​​indicating the density of previously used filters by (2010), it is possible to subtract this value according to the following : - From = 10 - 20 is subtracted from it.

10 - 20

20 - 10 = . <-: 20 - 30 = = 1. — 0. Thus, the necessary filters become. Sometimes it may happen that the value to be subtracted is greater than the value of the original previously used filters, such as the latter being (10 - 50), while the value of the previously used filters was (20 - 30) Then the following follows: The original value of the previously used filters: The value of the filters to be subtracted. Since the number 50 cannot be subtracted from the number 20, therefore a new value is added to the value of the first previously used set of filters equal to the difference between the number to be subtracted and the number subtracted from it. Thus, the original value becomes the previously used filters: to subtract from it the new value to be subtracted, which is = 50 - 10, so the value of the filters suitable for the new reform becomes = from it, i.e. the difference between 30 - 20 - 50 7.7.0. =

That is, color correction in the following image requires the use of a smaller filter

Its density is .and, and another scarlet has a density of 30

e Over Correction

We mentioned earlier in more than one place on pages 308, 310, 311, that it is necessary to remove the colored smear from the positive images, to place in the path of the rays emitted by the light source one or more filters of the same color as the smear, and on condition that the density of this smear is Filters are equal to the intensity of the swab. So what would happen, for example, if the removal of the crimson tinge from the positive images required the use of a filter (-30-), but the photographer had misjudged the intensity of (-80-)? The anointing, the scarcity of the intensity, then the bitter situation

In order to know the impact of this error (which is known as over correction), let us follow the consequences of it, which we show as follows:

If the density of the scarlet filter increases to the appropriate limit, there is no doubt that it will absorb a larger part of the green rays, which would greatly reduce the extent of the sensitivity of the sensitive layer to green rays (in sensitive paper), which is the layer that bears scarlet color problems, and accordingly we find The picture has lacked the scarlet color, and (on the other hand) the influence of both yellow and cyan colors prevails in it, so they are dominant over it. And since the sum of yellow + cyan is green, so we find the image covered with a green tinge of 1.

We conclude from the foregoing that if the correction of the color of the positive image increases in error, then it is certain that a new shade will arise from colors that are complementary to the color of the original shade that we wanted to correct first.
And from

From this rule we deduce the following facts:

1- If the correction of the yellow tinge increases, a new blue tinge will be created.

scarlet

green.

Cyan

red.

Zarqa

d yellow. D D D

green

e scarlet.

Red is Cyan. > -9 D D

What do we do if the correction increases an error?

Scarcity of his amount) - - 50 smears, and put bitter to demonstrate this, we will use the following example: - Supposing that it appears to us from examining the colored positive repellent image that it has a yellow smear covering it, then there is no doubt that this case requires removing this smear with a yellow filter that we place in a x-ray path Light source while typing or enlarging. Supposing that the removal of this yellow smear necessitated that we use a scarcity of density (20), but the photographer made a mistake in his estimate of the intensity (which led to the emergence of a new blue smear) i.e. of a color complementary to the original smear, and its density at that time is equal to the difference between the correct candidate, which is ( - - 20) and the filter used is wrong and it is ( - (0. -

Here we are thinking of removing this blue tinge from the following positive images, and the correct way that comes to mind at first sight is that the two filters have a blue color together. These two filters are dsian and scarlet, but for their intensity, it must be equal to the difference between the filter we used, which is (-20) and the filter we used wrongly, which is (50). That is, the intensity of each of the two filters that make up the blue color must be (30). We are looking for the correct one. - He should
Thus, we see that the required set of candidates are:

Added to it (i.e. a cyan filter with a density of 30), (i.e. a crimson filter with a density of 30) also, which is the value of the new filters that must be added to the previous first filter - 30 30 = its use is wrong and it is) -

.(0.

Thus, the value of the set of valid filters can be calculated to remove the blue tinge that covered the image as a result of increasing the correction as follows:

(It is the yellow filter in the amplifier after increasing the error correction). Added to it 30 by 30 (they are the blue and scarlet candidates, which are sufficient to remove the blue tinge). -

5030 30 =

20

This is any set of filters that ensure the new patch. This is after subtracting the number 30 from the group of filters, as the mistake is that we use three filters at once (for the reasons previously explained on p. 312). from

Thus, we conclude from the foregoing that the results will be equal if we directly decrease the intensity of the yellow filter (-50) that we used wrongly the first time, so we make it (-20) directly, or if we add to it an amount of filters that form the blue color complementary to its color (which are the filters Cyan and scarlet). The calculation of the values ​​of all these filters will be equal in the end.