An eyepiece is made of two positive thin lenses, each of focal length f = 20 mm, separated by a distance of 16 mm. (a) Where must a small object viewed by the eyepiece be placed so that the eye receives parallel light from the eyepiece? (b) Does the eye see an erect image relative to the object? Is it magnified? (c) Use a ray-trace diagram to answer these questions by inspection.
An eyepiece is made of two positive thin lenses, each of focal length f = 20 mm,
separated by a distance of 16 mm. (a) Where must a small object viewed by the
eyepiece be placed so that the eye receives parallel light from the eyepiece? (b) Does
the eye see an erect image relative to the object? Is it magnified? (c) Use a ray-trace
diagram to answer these questions by inspection.
pls do this
To answer these questions, we need to understand the basic principles of thin lenses and how they work. A thin lens is a transparent medium with one or two curved surfaces that refracts light rays, causing them to converge or diverge. These lenses have a property called focal length (f), which is the distance from the lens at which parallel light rays converge or appear to diverge from.
(a) To determine where the small object should be placed to receive parallel light from the eyepiece, we need to consider the individual lenses in the eyepiece and their arrangement. The eyepiece consists of two thin lenses, each with a focal length of 20 mm. The lenses are separated by a distance of 16 mm.
To obtain parallel light, the image formed by the first lens should be at the focus of the second lens. Since the eyepiece is made of two positive lenses (convex lenses), the real image formed by the first lens will be inverted. The second lens will then produce a virtual image that is erect and located at the near point of the eye, which is typically about 25 cm away.
Using the lens formula (1/f = 1/v - 1/u), where f is the focal length, v is the image distance, and u is the object distance, we can determine the object distance (u) for the first lens:
1/f1 = 1/v1 - 1/u
As the first lens forms a real and inverted image, the image distance (v1) will be negative. Let's assume the image formed by the first lens is at a distance v1 from the first lens.
1/20 = -1/v1 - 1/u
Now, let's consider the second lens. The object distance (u2) for the second lens will be the negative of the image distance (v1) of the first lens, as it acts as an object for the second lens.
u2 = -v1
Using the lens formula for the second lens:
1/f2 = 1/v2 - 1/u2
Since the second lens produces a virtual image, the image distance (v2) will be positive and equal to the near point of the eye (25 cm or 250 mm).
1/20 = 1/v2 - 1/(-v1)
Simplifying the expressions, we have:
1/20 = 1/250 + 1/v1
To obtain parallel light, the image formed by the first lens should be at the focus of the second lens. Therefore, v1 should be the negative of the focal length of the second lens (f2 = 20 mm).
1/20 = 1/250 + 1/-20
Now, we can solve for u:
1/20 - 1/250 = -1/20
u = -25 mm
Therefore, the small object viewed by the eyepiece should be placed 25 mm in front of the first lens.
(b) To determine if the eye sees an erect image relative to the object and whether it is magnified, we can use the magnification formula (m = -v/u), where m is the magnification, v is the image distance, and u is the object distance.
For the first lens:
m1 = -v1/u
Since the image formed by the first lens is real and inverted (v1 is negative), the magnification will also be negative, indicating an inverted image.
For the second lens:
m2 = -v2/u2
Since the second lens produces a virtual image (v2 is positive), the magnification will be positive, indicating an erect image.
Therefore, the eye sees an inverted image relative to the object after the first lens, but then the image is corrected and appears erect after the second lens. The overall magnification can be calculated by multiplying the individual magnifications of the two lenses (m1 * m2). In this case, since m1 and m2 have opposite signs, the overall magnification will be negative, indicating a reduced, inverted image compared to the object.
(c) To verify these conclusions using a ray-trace diagram, we can draw the path of two representative rays from the object through the eyepiece. The first ray is drawn parallel to the optical axis, and the second ray passes through the center of the lens.
When you draw the ray-trace diagram, you'll notice that the first lens forms a real and inverted image, while the second lens produces a virtual and erect image. The rays converge to the near point of the eye, indicating that the eye receives parallel light.
By analyzing the ray-trace diagram, you can verify that the image formed by the eyepiece is inverted after the first lens but then appears erect after the second lens. The diagram also confirms the position of the object in front of the first lens.
In summary:
(a) The small object should be placed 25 mm in front of the first lens to receive parallel light.
(b) The eye sees an inverted image relative to the object after the first lens, but it appears erect after the second lens. The overall magnification is negative, indicating a reduced, inverted image compared to the object.
(c) By drawing a ray-trace diagram, you can visually confirm these conclusions.