Find the radius of curvature of a biconcave lens symmetrically ground from a glass with index of refractive 1.55 so that its focal length in air is 20 cm (for a biconcave lens, both surfaces have the same radius of curvature).
Strategy
Use the thin-lens form of the lens maker’s equation:
Solving for
R and inserting
$f=\mathrm{-20}\phantom{\rule{0.2em}{0ex}}\text{cm},{n}_{2}=1.55,\phantom{\rule{0.2em}{0ex}}\text{and}\phantom{\rule{0.2em}{0ex}}{n}_{1}=1.00$ gives
Find the location, orientation, and magnification of the image for an 3.0 cm high object at each of the following positions in front of a convex lens of focal length 10.0 cm.
$\text{(a)}\phantom{\rule{0.2em}{0ex}}{d}_{\text{o}}=50.0\phantom{\rule{0.2em}{0ex}}\text{cm}$ ,
$\text{(b)}\phantom{\rule{0.2em}{0ex}}{d}_{\text{o}}=5.00\phantom{\rule{0.2em}{0ex}}\text{cm}$ , and
$\text{(c)}\phantom{\rule{0.2em}{0ex}}{d}_{\text{o}}=20.0\phantom{\rule{0.2em}{0ex}}\text{cm}$ .
Strategy
We start with the thin-lens equation
$\frac{1}{{d}_{\text{i}}}+\frac{1}{{d}_{\text{o}}}=\frac{1}{f}$ . Solve this for the image distance
${d}_{\text{i}}$ and insert the given object distance and focal length.
Solution
For
${d}_{\text{o}}=50\phantom{\rule{0.2em}{0ex}}\text{cm,}\phantom{\rule{0.2em}{0ex}}f\phantom{\rule{0.2em}{0ex}}=+10\phantom{\rule{0.2em}{0ex}}\text{cm}$ , this gives
The image is positive, so the image, is real, is on the opposite side of the lens from the object, and is 12.6 cm from the lens. To find the magnification and orientation of the image, use
The negative magnification means that the image is inverted. Since
$\left|m\right|<1$ , the image is smaller than the object. The size of the image is given by
For
${d}_{\text{o}}=5.00\phantom{\rule{0.2em}{0ex}}\text{cm,}\phantom{\rule{0.2em}{0ex}}f\phantom{\rule{0.2em}{0ex}}=+10.0\phantom{\rule{0.2em}{0ex}}\text{cm}$
The image distance is negative, so the image is virtual, is on the same side of the lens as the object, and is 10 cm from the lens. The magnification and orientation of the image are found from
The positive magnification means that the image is upright (i.e., it has the same orientation as the object). Since
$\left|m\right|>0$ , the image is larger than the object. The size of the image is
For
${d}_{\text{o}}=20\phantom{\rule{0.2em}{0ex}}\text{cm,}\phantom{\rule{0.2em}{0ex}}f\phantom{\rule{0.2em}{0ex}}=+10\phantom{\rule{0.2em}{0ex}}\text{cm}$
The image distance is positive, so the image is real, is on the opposite side of the lens from the object, and is 20.0 cm from the lens. The magnification is
When solving problems in geometric optics, we often need to combine ray tracing and the lens equations. The following example demonstrates this approach.
Choosing the focal length and type of lens
To project an image of a light bulb on a screen 1.50 m away, you need to choose what type of lens to use (converging or diverging) and its focal length (
[link] ). The distance between the lens and the lightbulb is fixed at 0.75 m. Also, what is the magnification and orientation of the image?
Strategy
The image must be real, so you choose to use a converging lens. The focal length can be found by using the thin-lens equation and solving for the focal length. The object distance is
${d}_{\text{o}}=0.75\phantom{\rule{0.2em}{0ex}}\text{m}$ and the image distance is
${d}_{\text{i}}=1.5\phantom{\rule{0.2em}{0ex}}\text{m}$ .
Solution
Solve the thin lens for the focal length and insert the desired object and image distances:
The minus sign for the magnification means that the image is inverted. The focal length is positive, as expected for a converging lens. Ray tracing can be used to check the calculation (see
[link] ). As expected, the image is inverted, is real, and is larger than the object.
every matter made up of particles and particles are also subdivided which are themselves subdivided and so on ,and the basic and smallest smallest smallest division is energy which vibrates to become particles and thats why particles have wave nature
according to de Broglie any matter particles by attaining the higher velocity as compared to light'ill show the wave nature and equation of wave will applicable on it but in practical life people see it is impossible however it is practicaly true and possible while looking at the earth matter at far
Manikant
a centeral part of theory of quantum mechanics example:just like a beam of light or a water wave
Swagatika
Mathematical expression of principle of relativity
given that the velocity v of wave depends on the tension f in the spring, it's length 'I' and it's mass 'm'. derive using dimension the equation of the wave