Why does having a shorter wavelength allow more data on optical discs? What properties of light allow this?

Question

I know that shorter wavelength = higher frequency. But when people just say oh, the shorter wavelength allows for the data pits/bumps to be packed in tighter. Why? If anything, I'd think amplitude would matter more. Or am I just visualizing a wave of light in the wrong way? Can someone clear this up for me?

Ok, some more followup. What I was initially thinking was that the shorter wavelengths meant higher energy so that given the smaller pits/bumps, the pickup could still read the data like using a big mirror to reflect a flashlight or a small mirror to reflect a laser beam.

What I was talking about freq, wavelength, and amplitude was...let's say it's a water wave. When you see the ripples in water, you see the amplitude and the freq/wavelength. The shorter the wavelength, the higher the freq. But what does that matter? Optical discs don't rely on frequency measurements. It's either, there is light reflected or there isn't light reflected. Does anyone get what I'm saying?

If I had a 530nm hole, could light with a longer wavelength fit through?

What I don't get is why wavelength matters when, to me at least, it seems that amplitude should be how "big" a wave of light is just like an ocean wave. When you talk of a big wave, you talk about amplitude, not freq/waveleng

Also, the freq or speed of the disc doesn't matter. I don't think eyeonthescreen has it right either. Optical disks don't rely on freq/wavelength measurements to carry data. Look at

http://computer.howstuffworks.com/cd5.htm

They have a nice little demo.

Another analogy of sorts. Take your left hand and make a circle with your thumb and fingers. That's the pit/bump. Now with your left hand, do a wave. You can either alter the wavelength/freq of the wave or the amplitude. Now if the amplitude is big, no matter what you do with the freq/wavelength, the wave won't go in the hole right? But if you lower the amplitude, it can fit. That's my confusion on visualizing the light wave reflecting. Now I know it's not like that exactly, I know how light propogates through space with the em field and photon and all that jazz but basically, it's a wave. So if you shorten the wavelength of your right hand wave, will it go into a small left hand hole? Why?

Answer 1

Light as you know can be treated as a wave, so it can cancel itself out if you reverse it's phase, it also varies in amplitude over time.

i.e. if a wave has an amplitude of 1 at one particular point in time and it has a frequency of 1 Hz i.e. 1 cycle per second then 1 second latter it will again have an amplitude of 1 after half a second however it will have an amplitude of minus 1 (or an amplitude of 1 with reversed phase) after 0.5 seconds and 0.75 seconds it's amplitude will be zero. (All these numbers can be offset by any number by adding the same number to all of them. They can also be scaled by multiplying all of them by a constant)

Apart from that it's all to do with how you extract information from the wave, how you store data on a disk and resolution limits.

In a CD player a laser is shone onto the CD the laser is bounced off the disc to a detector. On the CD there are two possible outcomes, the laser could bounce off a pit or a ridge.

The depth of the pit is such that if the laser travels the extra distance into the pit and back out again after reflecting the phase of the light is reversed and the wave is canceled out to zero.

So by measuring the strength of the returned light it is possible to tell whether the laser is pointing at a pit or a ridge.

If the disk is stationary it's as simple as that. When you spin the disk the laser will only be resting on a pit or a trough for a certain amount of time.

Due to the nature of a wave there are times when the magnitude of the wave is zero (twice per wavelength) if the disk was spinning so fast that the "bit" of data (pit or ridge) passed by the point of contact with the laser so fast that only one or less than one cycle of wavelength was reflected from the CD then the data will not be able to be resolved, at a time when the natural amplitude of light is zero, due to it's oscillation, the reading at the detector will be zero no matter what the laser is reflecting off.

In practice several wavelengths for each bit of data are required.

So if you up the frequency, you can can, spin your disk faster i.e. more bits per second. Or you can make your pits and ridges smaller in physical size i.e. more bits per second or you can slow the disk down, reduce the physical size of your pits and store more data on a disk the same size.

I'm not sure I've explained myself too well, please do ask me any thing you would like clarifying

The answers above are wrong in the case of CDs their explanations work for sending data down a line or through space with radio waves microwaves etc. but this is not the same situation you find yourself in when considering a CD player

I'm editing my post because I'm afraid the website you refear to does not show how a cd player works, it claims that the cd player reflects or does not reflect the light, this is a common misconception, it is an interference effect.

see

http://www.pctechguide.com/32CD-ROM_Operation.htm

This is a full explaination.

Answer 2

It's simple really. Shorter means smaller waves right? Smaller waves can read smaller bumps etched on an optical disk. The smaller the bumps, the more bumps (and the more data) you can put onto an optical disk. Bigger bumps take up more space on a disk so the smaller you can make them, the more you can fit onto the disk. You need shorter wavelengths to read the smaller bombs since the longer waves would be too big.

Answer 3

Lf = c; where L is the wavelength (meters), f is the frequency (sec^-1), and c is the speed of light (m/sec).

A bit of data can be based on on or off (0 or 1), such as digital computers use. So more bits mean more information.

If each cycle carries just one bit (0 or 1), then the total number of bits (n) = f del(t) = c/L del(t); where del(t) = the transmitting or, in the case of an optical disc, recording time. As you can see, n --> inf. as L --> 0, which clearly shows why shorter wavelengths "allow more data on optical discs."

Answer 4

Yes short wavelength means its a tiny wave,whose distance between two succesive nodes is small.and frequency means number of waves passing thru a point .so that tell us why it is high frequency,because many short wave passes thru a point at a span of time. so more data can be allowed in .

Answer 5

You got it when you said higher frequency.

If you were align magnets that were lying on x axis. They should be aligned south-north-north-south. Eave wavelength allows one south and north combination.

In one sec you could align many magnets with higher frequency than with lower frequency