I was reading about how CD-ROMs work and I came to this:

Although it might seem simplest to use a pit to record a 0 and a land to record a 1, it is more reliable to use a pit/land or land/pit transition for a 1 and its absence as a 0,so this scheme is used.

Now, I'd like to know why?

  • $\begingroup$ I just want to ask something....is not really an answer but a question....I am very confused about this...because actually the beam is prpependicular to the surface...and the light when reflects has a 180 degree phase change....so actually every incoming Ray with every reflected Ray are out of phase so we will always have destructive interference....can someone help please....I really want to understand how this works $\endgroup$
    – Gio lou
    Commented Jan 11, 2016 at 14:31

3 Answers 3


To elaborate on ratchet freak's answer; CD-ROMs work by sensing the intensity of the light reflected from the CD as it is spinning. More light being reflected stands for a 1 and less light being reflected stands for a 0 (or vice versa). One way to encode the information would be to have highly reflective surfaces for the 1's and dark patches for the 0's. Printing dark patches at the size of the laser spot size is actually used in writable CDs, but the technique suffers from degradation over time.

Instead commercial CDs rely on the property of interference to create dark patches. When the laser beam is in the transition region from land to pit, part of the laser beam reflects from the land and part from the pit. The depth of the pit is roughly 1/4 of a wavelength such that the portion of the beam which reflects from the pit has picked up an extra $\lambda/2$ of phase. This phase shift causes the beam falling on the photodetector to appear dark due to destructive interference.

The system used to detect this induced intensity modulation is a very simple one. It essentially consists of shining a laser at the disk and putting a photodiode on the reflected beam. Actually detecting the depth of the pits would require an interferometer which is significantly more complicated. This is because the interferometer would need a reference mirror whose position was stable to significantly better than the depth of the pits, ~250 nm. This is a difficult task to achieve inside of a disc drive which has a bunch of moving components (such as a disc spinning at 500 RPM) and may even be mounted in a vehicle which is moving and shaking around.

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Because of optical interference.

When the laser is fully in a pit or fully on land then it gets reflected cleanly and will be picked up by the sensor.

However, on the transition, half the laser is in the pit and half is in the land and the reflections will interfere with each other and the sensor won't pick anything up. The pit depth and laser color (wavelength) is picked so that this happens.

This means that the sensor cannot detect whether it is seeing a land or pit. It can only detect when it changes.


The explanation is incorrect. The laser is never more than half "in the pit". The reflection is land or 1/2 pit plus 1/2 non pit.

  • $\begingroup$ Are you saying that the laser illuminates a spot larger than the area of a pit? $\endgroup$
    – hazzey
    Commented Nov 8, 2015 at 7:39
  • 1
    $\begingroup$ A reference would improve this answer. $\endgroup$
    – Fred
    Commented Nov 8, 2015 at 11:19

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