To find the approximate value of the focal length of a convex lens, let us do an experiment.
Place a convex some distance away from a light source. At the opposite side of the lens, place a white paper.
Now adjust the position of the lens till a sharp point of light on the white paper is obtained.
Measure the distance between the lens and the white paper. This gives the value of the focal leng
he focal length of convex lens (Graphical method)
PURPOSE: To determine the focal length of converging lens by using a graphical method .
APPARATUS:
1.converging lens .
2.Holder.
3.A plan mirror.
4.Meter scale.
5.Object.
6.Screen.
METHOD;
1. Place an object pin at distance from the lens equal to 2F.
2. Measure the distance between the object and lens which is called "u".
3. Locate the position of its real image on the other side of the lens by using screen ,measure the distance between the image and lens which is called "V".
4. Move the object to other position both nearer to and farther way from the lens, location the new position of the image each time.
READING;
1/V (1/cm)
1/U(1/cm)
Distance of image from lens "v" cm
Distance of object from lens "u" cm
1. Plot a graph of (1/U) against (1/V).
2. Draw the straight line through the pointed and produce it to intersect both axes.
THEORY AND CALCULATION;
1/F=1/U+1/V
1. Straight line included each axis is obtained.
2.The intercept on the (1/v)axis is the numerical value for which (1/u)=0
1/F1=1/V
F1=V
3. Similarly for the intercept on the (1/U) axis
Then F2=U
4. Take the mean value of the two intercept.
F=F1+F2/2
MEDICAL APPLICATION:
Glasses (corrective lenses)help defective vision were among the first prosthetic devices invented in order to discuss the strength of the corrective lens for defective eye we need to review the basic negation of simple lens .
1/F=1/U+1/V
F is measured in meters ,1/F is the strength of the lens in diopter (D)
Now let us discuss defective eyesight due to refractive problems -ametropia.
A metropia affects over half of population of united states.it is often possible to correct it completely with glasses .
There are four general type of ametropia;
1. Myopia (near-sightedness).
2. Hyperopia or Hypermetropia (far-sightedness).
3. Astigmatism (asymmetricalfocusing).
4. Presbyopia (old sight).
The myopia individual has too long an eye ball or too much curvature of the cornea ,distant objects come to a focus in front of the retina ,and the rays diverge to cause a blurred image at the retina .This condition is easily corrected with negative lens.
Focal lengh by displacement method.
[sorry , i can't upload the diagram here as i can't write or upload the image. i hope you can undertstand by explanation.]
First , take the lens and object (like a hole or something, clearly visible). And move the
lens from the object. and a certain length , the image of the object on the screen will be
clearer and equal in the size of the object. This is the nearer focal length. and move the
screen to such a distance such that D=4f. And then do as the same. but at a certain length,
you will see a clear big image and note that length, And an small clear image.take the
Differece of them. that will be S.
Do the same procedure for different values of D (increase D by 10-20cm). and note down the
readings.
Use the formula to calculate the focal length. f=(D2- s2)/4D
and take the mean. That will be your lens focal length.
First and foremost by distance object method we can get approximate focal length
Then base on that value we can assign u values and get required v values and find focal length by u-v method.
Plane mirror method is also there
Two position method would help to find the same
You can place the length in the sunlight, see where the light converges, and measure the distance.
A converging lens is a simple magnifying glass when the object is within one focal length of the lens. The image is then virtual, magnified, and right-side up.
Power is inversely related to the focal length. So convex lens of focal length 20 cm has less power compared to that having focal length 10 cm
If the object is more distant from the lens than the focal length of the lens, a real image is formed.
c virtual,upright,and larger than the object.
Yes converging lens. The power of the lens is given by reciprocal of its focal length. Moreover the power for converging action is +ve. So as we place two converging lenses ie convex lenses, then we have to add the powers. Once again the power becomes +ve. So converging action is definite.
i think it is -0.06m i.e. 6cm
A converging lens is a simple magnifying glass when the object is within one focal length of the lens. The image is then virtual, magnified, and right-side up.
Power is inversely related to the focal length. So convex lens of focal length 20 cm has less power compared to that having focal length 10 cm
If the object is more distant from the lens than the focal length of the lens, a real image is formed.
c virtual,upright,and larger than the object.
Yes converging lens. The power of the lens is given by reciprocal of its focal length. Moreover the power for converging action is +ve. So as we place two converging lenses ie convex lenses, then we have to add the powers. Once again the power becomes +ve. So converging action is definite.
Lenses enable individuals to view objects. A Converging lens has a positive focal length, which facilitates the convergence of the exiting rays. While, diverging lenses have a negative focal length, which facilitates the divergence of the exiting rays.
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farther then f from the lens
C: virtual, upright, and larger than the object
A stamp collector uses a converging lens with focal length 25 cm to view a stamp 13 cm in front of the lens.
if the focal length is greater than the object distance from the lens