On a CD, the pits and lands are covered by a protective acrylic layer with an index of refraction of 1.38. The laser used in a CD has a wavelength equal to 780 nm. At the boundary between pits and lands, the pit depth causes the laser to destructively interfere. (The depth is the minimum distance for this to occur.) How deep are the pits? (Give your answer in nm.)
Ray Optics
Optics is the study of light in the field of physics. It refers to the study and properties of light. Optical phenomena can be classified into three categories: ray optics, wave optics, and quantum optics. Geometrical optics, also known as ray optics, is an optics model that explains light propagation using rays. In an optical device, a ray is a direction along which light energy is transmitted from one point to another. Geometric optics assumes that waves (rays) move in straight lines before they reach a surface. When a ray collides with a surface, it can bounce back (reflect) or bend (refract), but it continues in a straight line. The laws of reflection and refraction are the fundamental laws of geometrical optics. Light is an electromagnetic wave with a wavelength that falls within the visible spectrum.
Converging Lens
Converging lens, also known as a convex lens, is thinner at the upper and lower edges and thicker at the center. The edges are curved outwards. This lens can converge a beam of parallel rays of light that is coming from outside and focus it on a point on the other side of the lens.
Plano-Convex Lens
To understand the topic well we will first break down the name of the topic, ‘Plano Convex lens’ into three separate words and look at them individually.
Lateral Magnification
In very simple terms, the same object can be viewed in enlarged versions of itself, which we call magnification. To rephrase, magnification is the ability to enlarge the image of an object without physically altering its dimensions and structure. This process is mainly done to get an even more detailed view of the object by scaling up the image. A lot of daily life examples for this can be the use of magnifying glasses, projectors, and microscopes in laboratories. This plays a vital role in the fields of research and development and to some extent even our daily lives; our daily activity of magnifying images and texts on our mobile screen for a better look is nothing other than magnification.
On a CD, the pits and lands are covered by a protective acrylic layer with an index of refraction of 1.38. The laser used in a CD has a wavelength equal to 780 nm. At the boundary between pits and lands, the pit depth causes the laser to destructively interfere. (The depth is the minimum distance for this to occur.) How deep are the pits? (Give your answer in nm.)
When a laser is shone onto a CD, it is reflected back to a detector. The pits and lands on the CD's surface affect the reflected light in different ways due to their different physical properties. Specifically, the depth of the pits affects the phase of the reflected wave, and the distance between the pits and lands affects the intensity of the reflected wave.
When the laser hits a pit on the CD's surface, some of the light is reflected back toward the detector, while the rest of the light is scattered in different directions. When the reflected light meets the light reflected from land adjacent to the pit, the two waves interfere with each other. Depending on the depth of the pit, this interference can either be constructive or destructive.
In the case of destructive interference, the reflected waves from the pit and land cancel each other out, resulting in a minimum in the intensity of the reflected light. The condition for destructive interference is given by the equation I mentioned in my previous answer:
2d = (m + 1/2)λ/n
Here, d is the depth of the pit, λ is the wavelength of the laser, n is the index of refraction of the acrylic layer, and m is an integer representing the order of the interference. The factor of 1/2 in the equation accounts for the fact that the reflected wave from the land travels a distance that is half a wavelength longer than the reflected wave from the pit. The index of refraction of the acrylic layer affects the phase of the reflected wave, which is why it is included in the equation.
In this problem, we are given the wavelength of the laser and the index of refraction of the acrylic layer. We are also told that destructive interference occurs at the boundary between pits and lands, which means that m = 0.
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