Super User is a question and answer site for computer enthusiasts and power users. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

I scanned a film and now I have a brown jpeg. Is there any software to convert the image to a real photo? Or how can I scan a film and get the photos?

share|improve this question
What's a 'real photo'? Do you mean it's come out in negative? Or in one-tone sepia? – Mark Pim Jan 15 '10 at 9:32
i scanned the film with the original software (EPSON) and it came out in one-tone sepia – Remus Rigo Jan 15 '10 at 9:45
up vote 3 down vote accepted

Normally, this is done during the scan and not afterwards. Nearly every scanner software has the option to select to scan negatives.

Alternatively, you could try to invert the negative and then use color-correction on it (a photo negative is not an exact inversion of the image).

share|improve this answer
i inverted the colors, used auto-level (color-correction). Now the colors are almost ok, but the image is 423x327 pixels and it looks dusty and verry bad if i resize it – Remus Rigo Jan 15 '10 at 9:47
you can't get more out than you put in upfront. Low res to Highres conversion while adding detail on works in TV :-) Use a highres image to convert and keep that resolution. Just do the lowres for checking your settings to preserve time/interactiviy – JumpingPA Jan 15 '10 at 9:55
@Jumpy: Wait, you mean we can't extract the license plate from a car a half mile away on the MPEG-distorted 640x480 video? ;) --- @Remus: Jumpy is right...scan it in a higher resolution. – Bobby Jan 15 '10 at 10:05
In CSI they can! They have better software then you do ;-) – Ivo Flipse Jan 15 '10 at 10:46

If you want good quality images from negative films, you need a film scanner (e.g. Nikon Coolscan).

For the software part, Adobe Photoshop is able to invert negative and remove orange dominant color.

share|improve this answer

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .