Science, asked by mauryayash86gmailcom, 1 year ago

limit of revolution eye 1 minute formula for comitiraion of lens 1/f=1/f+1/f2

Answers

Answered by Sneha037
0
I don't know bye bye

mauryayash86gmailcom: bye
Answered by sarabjotkaur707
1

Thin lenses are used in many optical systems, and understanding how they work involves many basic

concepts in geometrical optics. These include:

1. the assumption of “rectilinear propagation ”(light as “rays”),

2. the mapping from object space to image space

3. the variation in quality of images of “on-axis ”and “off-axis ”objects due to aberrations present

in the lens (primarily chromatic aberration, spherical aberration and coma),

4. the effect of stops and the lens “shape factor ”on image quality.

Consider the imaging equation for a single thin lens:

1

so

+

1

si

= 1

f (1)

where so and si are object and image distances from the lens, respectively, and f is the focal length

of the lens. By simple rearrangement, we can express si as a function of so and f:

si = sof

so − f (2)

Note that this relationship is extremely nonlinear: the image distance is definitely NOT proportional

to the object distance. This relationship will be demonstrated by imaging the white light source at

various distances from the lens.

fo

so si

fi

Ray diagram for a single thin lens: an input ray parallel to the optical axis passes through the rear

focal point, an input ray through the center of the lens emerges at the same angle, and an input ray

that passes through the front focal point emerges parallel to the optical axis.

1

The transverse magnification of the image (compared to the object) is the ratio of the image

distance to the object distance

MT = − si

so

(3)

(one can easily derive this result by analyzing similar triangles in Figure 1), where the negative

sign indicates that the image is inverted if both s0 and si are positive.

For combinations of two thin lenses in contact, we have seen that the focal length of the combination

is given by an expression involving the individual focal lengths:

1

f = 1

f1

+

1

f2

(4)

Combinations of two thin lenses can be used to make an extremely important optical device: the

refracting telescope. Although it is not totally clear who invented the telescope, records in The Hague

(in Holland) demonstrate that Hans Lippershey (1587-1619) applied for a patent on the device in

1608. Soon after, Galileo built his first telescope using one positive lens with convex surfaces and

one negative lens with concave surfaces. Kepler also used a telescope, but favored an arrangement

with two convex lenses. In both cases, the lens where the light enters is called the “objective lens ”,

while and the lens where the user places their eye is the “eye lens.” Both the Galilean and Keplerian

designs are shown in Figure 2. They differ in that the position of the eye lens as well as the type of

lens. The Keplerian telescope forms a real intermediate image, while the Galilean does not. Note

that both designs are “afocal”; the systems do not form accessible “images”. Rays that enter at

an angle θ measured relative to the optical axis emerge at the same angle, so parallel entering rays

emerge in parallel. A third lens (usually your eye lens or a camera) forms the real image. The

telescope is formed when the two lenses are separated by a distance equal to the sum of their focal

lengths: d = f1 + f2. The focal length of the resulting system is obtained by substitution into the

equation:

1

f = 1

f1

+

1

f2

− d

f1f2

= 1

f1

+

1

f2

− f1 + f2

f1f2

= 0 mm−1 =⇒ f = ∞mm, for a telescope (5)

:The “power” of the system (reciprocal of the system focal length) is:

1

f = ϕ = ϕ1 + ϕ2 − ϕ1ϕ2d = 0 mm−1, for a telescope (6)

l1 (f1 > 0) l2 (f2 < 0)

(f2)i

(f2)o

(f1) (f1)o i

Optical layout of Galilean telescope: the “image space” (rear) focal point of the positive objective

lens coincides with the “object space” focal of the negative eye lens. Light rays that are incident

parallel to the axis emerge parallel to the axis. Parallel rays that enter the object at an angle to the

optical axis emerge parallel but at a “steeper” angle to the axis.

2

l1 (f1 > 0) l2 (f2 > 0)

(f2)o

(f1)i (f1)o (f2)i

Optical layout of Keplerian telescope: the “image space” (rear) focal point of the positive objective

lens coincides with the “object space” focal of the positive eye lens. Light rays that are incident

parallel to the axis cross the axis at the focal point of the objective lens and emerge parallel to the

axis and on the “other” side. Parallel rays into the object at an angle relative to the optical axis

emerge parallel and at a “steeper” angle.


Similar questions