The lens shown in FIGURE CP is called an achromatic doublet, meaning that it has no chromatic aberration. The left side is flat, and all other surfaces have radii of curvature R.
a. For parallel light rays coming from the left, show that the effective focal length of this two-lens system is , where and are, respectively, the indices of refraction of the diverging and the converging lenses. Don’t forget to make the thin-lens approximation.
b. Because of dispersion, either lens alone would focus red rays and blue rays at different points. Define and as for the two lenses. What value of the ratio makes for the two-lens system? That is, the two-lens system does not exhibit chromatic aberration.
c. Indices of refraction for two types of glass are given in the table. To make an achromatic doublet, which glass should you use for the converging lens and which for the diverging lens? Explain
Crown glass 1.525 1.517
Flint glass 1.632 1.616
d. What value of R gives a focal length of ?
a. The statement is proved below.
b. The value of the ratio is .c. The required statement is explained below.
d. The value of is .
We have to find out effective focal length of the two lens.
Here, is focal length, is refractive index and is radius of curvature of lens
we have to find what value of the ratio .
Here, is focal length of blue and is refractive index.
We have to find the for each type of glass.
Here, is the refractive index for respective names.
We have to find focal length in expression for R.
Here, is refractive index and is the focal length of lens.
Once dark adapted, the pupil of your eye is approximately 7 mm in diameter. The headlights of an oncoming car are 120 cm apart. If the lens of your eye is diffraction limited, at what distance are the two headlights marginally resolved? Assume a wavelength of 600 nm and that the index of refraction inside the eye is 1.33. (Your eye is not really good enough to resolve headlights at this distance, due both to aberrations in the lens and to the size of the receptors in your retina, but it comes reasonably close.)
The resolution of a digital camera is limited by two factors:
diffraction by the lens, a limit of any optical system, and the fact
that the sensor is divided into discrete pixels. Consider a typical
point-and-shoot camera that has a 20-mm-focal-length lens and
a sensor with 2.5@mm@wide pixels.
a. First,ass ume an ideal, diffractionless lens. At a distance of
100 m, what is the smallest distance, in cm, between two
point sources of light that the camera can barely resolve? In
answering this question, consider what has to happen on the
sensor to show two image points rather than one. You can use
s′ = f because s W f.
b. You can achieve the pixel-limited resolution of part a only if
the diffraction width of each image point is no greater than
1 pixel in diameter. For what lens diameter is the minimum
spot size equal to the width of a pixel? Use 600 nm for the
wavelength of light.
c. What is the f-number of the lens for the diameter you found in
part b? Your answer is a quite realistic value of the f-number
at which a camera transitions from being pixel limited to
being diffraction limited. For f-numbers smaller than this
(larger-diameter apertures), the resolution is limited by the
pixel size and does not change as you change the aperture. For
f-numbers larger than this (smaller-diameter apertures), the
resolution is limited by diffraction, and it gets worse as you
“stop down” to smaller apertures
A 2.0-cm-tall object is to the left of a lens with a focal length of A second lens with a focal length of is to the right of the first lens.
a. Use ray tracing to find the position and height of the image. Do this accurately using a ruler or paper with a grid, then make measurements on your diagram.
b. Calculate the image position and height. Compare with your ray-tracing answers in part a.
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