1. The answer is best understood by this simple analogy. If someone were to stand behind you and say, "Hands up and look back", there are two options ahead of you. One is the normal one where you take the easiest path and turn right to left. The other is to go down on your head and look back. Which one would you take? ..... That's exactly what light does too. It takes the easiest path, that of least TIME.
In physicists jargon, you would say that a mirror carries out a 'lateral inversion' that is, there is only a left-right reversal of the newspaper. It does not perform a longitudinal reversal or an up-down reversal although you may personally like to somersault! TOM in English is reversed, but TOM written in CHINESE style is not.
Moreover, what is your right hand is on the left hand side of the IMAGE in the mirror. Imagine yourself in the mirror. To which side is the your right hand up now INSIDE the mirror?
2. You're mixing issues here. You're right, light does not require a medium to travel. If you're asking like sound is there a range only which you can hear, yes there is a range of frequencies you can see. Roughly between 700 nm (red) to 400 nm (violet) is all we can see. Below the red in energy and frequency is the infrared, (microwave) and radio waves. Beyond the violet in energy and frequency you'll find the ultraviolet, x-rays and gamma rays (and cosmic rays beyond).
Approximately below 20 Hz you can't hear and it is called infrasonic (heard by whales, dolphins and so on) and beyond 20,000 Hz you can't hear too and the region is called ultrasonic (heard by bats, dogs and so on).
Sound always requires a medium whereas light does not need a medium at all. In fact that's one good way to distinguish between the two.
3. The light from a lens can refract parallel to the principal axis (line passing through the centre and focus) if it starts at the focus, somewhat like the adjustments you can make with a torch light to either spread or focus the light.
Since light follows the principle of reversibility (that is it doesn't matter if it approaches from the left or right), it follows from above that if the light is parallel, then it must gather at the focus after refraction.
Now to answer the question about what is meant by "infinity", in physics a dot can mean infinity if the distance you are comparing to is also small, say the size of amoeba or paramecium.
In the case of a lens, all distances are compared to the radius of curvature (that is the radius of the sphere of which the lens surface is a part. If you consider a tpyical lens it is of the order of 10-20 centimeters. Even ten metres is considered to be infinity in this situation.
You can make an order of magnitude calculation of the radius if you know the refractive index, focal length and type of lens using the lens makers' formula.
4. In case of mirrors, the focal length (f) is related to the radius of curvature (R) through the relation
f = R/2
In case of lenses however, the focal length is related to the radius of curvature through what's popularly called the lens makers' formula. You must be careful with the sign conventions here. The formula is
1/f = (mu - 1)[1/R1 - 1/R2]
For a double convex lens R1 is positive and R2 is negative. 'mu' as you're aware is the refractive index of the material of the lens.
The focal length in general can be defined as the image distance when the object is at infinity.
If you're talking about Autofocus cameras, then they work by using different techniques. One such technique is to use an infrared signal and is great for subjects within twenty feet or so of the camera. Infrared systems use a variety of techniques to judge the distance. Typical systems might use:
Triangulation
5. One can explain your question using the so called 'magnification equation' for any lens. The linear magnification of a lens is defined as follows:
Magnification = height of image/height of object ...(i)
If the object is placed such that the light comes from infinity (if we consider light from sun), it will form a point size image at its focal length, because the denominator in equation (ii) above makes the magnification infinity. This is confirmed by our common sense idea also, isn't it. An object far away looks very small and appears bigger as it gets closer. The lens in the eye too follows the same logic.
If the object is nearer to the lens(using your example of the image formed on the lens of a camera)it forms an image of some particular size because the object distance is now a finite value in the denominator of equation (ii) above. Would you agree that this follows from our common experience with the human eye?
To answer your final question on what distance is considered to be infinity for a lens of particular focal length, you can view my answer to a similar question posed by another anonymous member. That should give you some idea of what I think about this subject.
6. This had me fascinated for a while on my cars too. You've actually got the answer in your question itself. In any case you'll find a simple explanation to this and several other "How" questions by clicking on the URL below:
http://www.howstuffworks.com/question20.htm
If you are unable to see the URL I've suggested by clicking on link above,then copy and paste the full URL above into your browser.
7. Such questions always have multile answers that one can think of. At this moment what comes to my mind is in all of these it that there are DIFFERENT physical processes taking place in their working although they all use light waves for their operations.
A camera uses a lens which REFRACTS or BENDS light to form images on a film.
A rainbow is formed when light waves from the sun is TOTALLY INTERNALLY REFLECTED by the water droplets in the rain.
A mirror usually REFLECTS or BOUNCES light waves of itself to form images.
Water surfaces usually are colored green or blue depending on the depth, incident light, presence of impurities and so on by SCATTERING the light waves.
So to summarize this answer, all of them have used light waves but have used FOUR DIFFERENT physics principles Refraction, Total Internal Reflection, (Simple) Reflection, and Scattering to form images. Of course as you may have observed it is not only one process which takes place but there is always a combination of processes taking place in all these cases. What I have listed is the PREDOMINANT process. The other processes are only minor.
8. This is a classic problem, that involves what's called the conjugate positions of a convex lens in forming a magnified and diminished images.
You have to set up two equations and solve them simultaneously. The best way forward in such questions is to draw a quick sketch of the situation. If you did that you'll find something like this.
<----u--- = 20 cm ---> (lens) <---- v --- = v cm ---> !( the screen)
You can use the lens equation to write the following equation.
1/20 + 1/v = 1/f
After you move the lens towards the screen the new diagram would be
<----u--- = 40 cm ---> (lens) <---- v --- = v - 20 cm ---> !( the screen)
Writing the second equation
1/40 + 1/v-20 = 1/f
Solving the two equations simultaneously you'll find v = 40 cm. Substituting in any one of the equations you'll find
f = 40/3 cm.
9. These problems can be worked easily if you know the lens equation and sign conventions properly. What I'm going to use below is called the REAL IS POSITIVE sign convention. According to this, all real objects and images have a positive sign and all objects and images which are imaginary have a negative sign. Let me know if you need more details on this sign convention.
Here's how you go about solving the problem. First make a rough sketch of the arrangement. You are given the object (lamp) and image (screen) distances. You can thererore find the focal length of the converging lens using the lens equation.
In general from the lens equation
1/u + 1/v = 1/f
Here
1/40 + 1/40 = 1/f (since lens is midway)
Solving this you'll find f = +20 cm
When you place the concave lens between the lens and the lamp, you consider this as a problem one with a combination of lenses. In such a case, you would use the image formed by the first lens as the object for the second lens. This image formed by the second lens is the final image of the combination.
Consider only the CONCAVE lens. Look at your sketch. What is the object distance? You've moved the lamp 30 cm away. Therefore the object distance is 20 + 30 = 50 cm. Using the lens equation, we have for the concave lens
1/50 + 1/v = 1/F
where F is the focal length of the concave lens. You should expect a negative value for the image distance of the concave lens.
Obviously you have two unknowns in the above equation and cannot solve it without writing down another equation. To do so, now consider the CONVEX lens. The object distance for the convex lens is positive and 20 (midway) + v.
The image distance and focal length of the convex lens are known. Therefore
1/(v+20) + 1/40 = 1/20
Solving this you'll find 20 cm for the value of 'v' as you may have predicted since the screen has not been moved.
Substituting this value of v, but with a negative sign (since it is to the left of the lens) following the convention, you'll have
for the CONCAVE LENS
1/50 + 1/(-20) = 1/F
Solving you'd find the value of the focal length of the concave lens is
F = -100/3 cm
(Note the negative sign is what you'd have expected for an answer for the concave lens)
10. The Raman effect involves the interaction of light with matter. Sir C V Raman found in 1928 that a change in frequency (or wavelength) of light is observed when light is scattered in a transparent material. When monochromatic light, such as that obtained from a laser, is passed through a transparent gas, liquid, or solid and then observed with the spectroscope, the spectral line ordinarily produced by the light has associated with it lines of longer and of shorter wavelength, called the Raman spectrum.
Analysis of the scattered light, as a Raman spectrum, reveals vital information about a sample's chemical structure and physical state.
Sometimes it is convenient to explain the blue of the sky and green of water surfaces in terms of the scattering of molecules and refer to it as the Raman effect. While this did bring Sir Raman the coveted Nobel Prize in 1930, there is a strong lobby which still believe his student Krishnan must have got more credit for first discovering the effect. I read last year some other articles showing the other side of Raman too. For details on this you can go to the URL:
http://expressindia.com/ie/daily/19970708/18950683.html
Scientists are also human like you and me and have their good and bad sides!
Amount of infrared light reflected from the subject
Time, and so on
= image distance/object distance ...(ii)