How does this spectroscope work?

[Phys12]

I build the following spectroscope:

However, I'm not sure how it works. I thought that the light entered the slit, diffracted from the CD (because the tracks in them are comparable to the wavelength of visible light) and then we observe the different components in the camera.(https://www.exploratorium.edu/snacks/cd-spectroscope)

The explanation here (https://www.livescience.com/41548-spectroscopy-science-fair-project.html) says that light gets diffracted from the slit by interfering itself forming the interference pattern (the rainbow) and then reflects off the CD and then that's what we see.

Which one is true?

P.S. When I look at light from one angle, I see the following:
https://ibb.co/mSvo1x
But when I rotate it by 90 degrees, I see:
https://ibb.co/j6UsTc
Why is it the case of the orientation of the spectroscope changes the spectrum seen? If the orientation of the slit and the CD are the same, shouldn't it stay the same?

Also, I don't see the same two distinct spectrum when I look at the spectroscope with my eyes, I see multiple in different directions. Why is it different for a camera vs eyes?

 

[Charles Link]

Usually the slit widths of a spectrometer are chosen wider than what would cause appreciable diffraction. The light source whose spectrum is to be measured is focused onto the entrance slit. $\\$ Inside of the spectrometer is a lens or concave mirror whose focal point is on the entrance slit. This way the light comes off of the first optic as parallel rays that are incident onto either a prism or diffraction grating. The light emerges from the prism or diffraction grating as parallel rays that have a direction that is wavelength dependent, i.e. parallel rays of wavelength $\lambda_1$ will emerge at angle $\theta_1$, parallel rays of wavelength $\lambda_2$ will emerge at angle $\theta_2$, etc. In general, for a broad spectrum source, parallel rays emerge at a wide variety of angles $\theta$, where $\theta$ is a function of wavelength $\lambda$, i.e. $\theta=\theta(\lambda)$. $\\$ The parallel rays are then brought to focus in the plane of the exit slit by a second lens or concave mirror.(The spectrum will appear across the focal plane of this second lens or concave mirror as the colors of the rainbow). The exit slit is located at a position in the focal plane. It only gets light from one specific angle, i.e. the (focused) colors of the rainbow are spread out on both sides of the exit slit, and only one color will be focused on the exit slit at a time, so there will only be light of one wavelength emerging from the exit slit. To change the wavelength that comes out of the exit slit, the prism or diffraction grating is rotated (about a vertical axis), causing the whole pattern of the colors of the rainbow to rotate horizontally. (Note:The spectrometer slits are in the vertical direction). $\\$ Note: If you start with a monochromatic source, the prism or diffraction grating needs to be rotated to the proper angle to get the light coming off of the prism or diffraction grating to focus onto the exit slit. (Otherwise, it will come to a focus in the focal plane, but at a different location than right on the exit slit). When it does, the light that is focused onto the exit slit will have an image whose shape is just like that of the illuminated entrance slit. It's not completely necessary, but often times, the exit slit is chosen to be the same width as the entrance slit.

 

[Phys12]

Usually the slit widths of a spectrometer are chosen wider than what would cause appreciable diffraction. The light source whose spectrum is to be measured is focused onto the entrance slit. $\\$ Inside of the spectrometer is a lens or concave mirror whose focal point is on the entrance slit. This way the light comes off of the first optic as parallel rays that are incident onto either a prism or diffraction grating. The light emerges from the prism or diffraction grating as parallel rays that have a direction that is wavelength dependent, i.e. parallel rays of wavelength $\lambda_1$ will emerge at angle $\theta_1$, parallel rays of wavelength $\lambda_2$ will emerge at angle $\theta_2$, etc. In general, for a broad spectrum source, parallel rays emerge at a wide variety of angles $\theta$, where $\theta$ is a function of wavelength $\lambda$, i.e. $\theta=\theta(\lambda)$. $\\$ The parallel rays are then brought to focus in the plane of the exit slit by a second lens or concave mirror.(The spectrum will appear across the focal plane of this second lens or concave mirror as the colors of the rainbow). The exit slit is located at a position in the focal plane. It only gets light from one specific angle, i.e. the (focused) colors of the rainbow are spread out on both sides of the exit slit, and only one color will be focused on the exit slit at a time, so there will only be light of one wavelength emerging from the exit slit. To change the wavelength that comes out of the exit slit, the prism or diffraction grating is rotated (about a vertical axis), causing the whole pattern of the colors of the rainbow to rotate horizontally. (Note:The spectrometer slits are in the vertical direction). $\\$ Note: If you start with a monochromatic source, the prism or diffraction grating needs to be rotated to the proper angle to get the light coming off of the prism or diffraction grating to focus onto the exit slit. (Otherwise, it will come to a focus in the focal plane, but at a different location than right on the exit slit). When it does, the light that is focused onto the exit slit will have an image whose shape is just like that of the illuminated entrance slit. It's not completely necessary, but often times, the exit slit is chosen to be the same width as the entrance slit.

I don't think that answers my question. In the setup that I have, is the spectrum caused by diffraction of light when it passes through the slit or when it passes through the gap between the tracks on the CD?

 

[Charles Link]

@Phys12 My fault for not looking more carefully at your post. I see it is a video which I now watched. It's very interesting. The CD is a transmission type grating. $\\$ Your first photo looks like the correct spectrum. $\\$ The spectrum contains a diffraction pattern of primary maxima of the form $m \lambda=d \sin(\theta)$ where $m=$ integer, and $d$ is the distance between the grooves on the CD. If you would aim it at a faraway screen and/or view from far away, the angle $\theta$ would be the azimuth angle measured from center. (You don't have sufficient light to do this but instead focus the far field diffraction pattern with your eyes or with a camera.)The camera or your eye puts the far-field diffraction pattern in the focal plane of the camera (need to focus at $+\infty$), or on the retina of your eye (and again for best results, focus your eye to faraway when viewing the spectrum with your eye). $\\$ The explanation for the spectrum is the white light is $m=0$. That is the central maximum at $\theta=0$, which will be independent of wavelength. For $m=1$, violet and blue will have smaller angles $\theta$ than greens and yellows and reds because the wavelengths $\lambda$ are shorter. Then you have $m=2$ violet and blue and then $m=2$ green, and yellow and red going from center to the right.(Note: You just have $m= 2$ violet. You didn't get a wide enough angle to see the m=2 blue, green, yellow and red). The same pattern appears on the left because of $m=-1$ and $m=-2$. (If $d$ is small enough, you could perhaps get a set of $m=3$ maxima, and the $m=3$ violet and blue would start to overlap the $m=2$ yellow and red. For this CD, $d$ was apparently too large to get that to appear). $\\$ Note: The horizontal position $x$ of the focused rays at angle $\theta$ on the camera focal plane (pixels of photo-receptors), and also on your retina, is determined by $x=f \theta$, where $f$ is the focal length of the camera lens or the lens of your eye. It then gets enlarged when you see the photo in the display screen of your camera. $\\$ Additional item: I think the slit you have may be sufficiently narrow that it causes a spreading of the incident light as a single slit diffraction pattern. This however does not create the spectrum. The speading of the light by the slit will cause the incident light to expand somewhat. The longer wavelengths will get spread more by this diffraction effect but it has minimal effect on the operation of this spectrometer. This single-slit diffraction effect helps to illuminate a wider region on the CD. Alternatively, oftentimes spectrometers have a collimating lens or mirror between the entrance slit and the diffraction grating. By having a very narrow slit, and using single-(narrow) slit diffraction to expand the beam, the collimating lens or mirror is then unnecessary.