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प्रश्न
When objects at different distances are seen by the eye, which of the following remain constant?
पर्याय
The focal length of the eye-lens.
The object-distance from the eye-lens.
The radii of curvature of the eye-lens.
The image-distance from the eye-lens.
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उत्तर
The image distance from the eye lens
In the human eye, the image is formed on the retina, which is at a fixed distance from the eye lens.
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संबंधित प्रश्न
A virtual image, we always say, cannot be caught on a screen. Yet when we ‘see’ a virtual image, we are obviously bringing it on to the ‘screen’ (i.e., the retina) of our eye. Is there a contradiction?
For a normal eye, the far point is at infinity and the near point of distinct vision is about 25cm in front of the eye. The cornea of the eye provides a converging power of about 40 dioptres, and the least converging power of the eye-lens behind the cornea is about 20 dioptres. From this rough data estimate the range of accommodation (i.e., the range of converging power of the eye-lens) of a normal eye.
Does short-sightedness (myopia) or long-sightedness (hypermetropia) imply necessarily that the eye has partially lost its ability of accommodation? If not, what might cause these defects of vision?
A person looking at a person wearing a shirt with a pattern comprising vertical and horizontal lines is able to see the vertical lines more distinctly than the horizontal ones. What is this defect due to? How is such a defect of vision corrected?
The angle subtended at the eye by an object is equal to the angle subtended at the eye by the virtual image produced by a magnifying glass. In what sense then does a magnifying glass provide angular magnification?
In viewing through a magnifying glass, one usually positions one’s eyes very close to the lens. Does angular magnification change if the eye is moved back?
A Cassegrain telescope uses two mirrors as shown in the figure. Such a telescope is built with the mirrors 20 mm apart. If the radius of curvature of the large mirror is 220 mm and the small mirror is 140 mm, where will the final image of an object at infinity be?
A man wearing glasses of focal length +1 m cannot clearly see beyond 1 m
Mark the correct options.
(a) If the far point goes ahead, the power of the divergent lens should be reduced.
(b) If the near point goes ahead, the power of the convergent lens should be reduced.
(c) If the far point is 1 m away from the eye, divergent lens should be used.
(d) If the near point is 1 m away from the eye, divergent lens should be used.
A person looks at different trees in an open space with the following details. Arrange the trees in decreasing order of their apparent sizes.
| Tree | Height(m) | Distance from the eye(m) |
| A | 2.0 | 50 |
| B | 2.5 | 80 |
| C | 1.8 | 70 |
| D | 2.8 | 100 |
A normal eye has retina 2 cm behind the eye-lens. What is the power of the eye-lens when the eye is (a) fully relaxed, (b) most strained?
A nearsighted person cannot see beyond 25 cm. Assuming that the separation of the glass from the eye is 1 cm, find the power of lens needed to see distant objects.
A myopic adult has a far point at 0.1 m. His power of accomodation is 4 diopters.
- What power lenses are required to see distant objects?
- What is his near point without glasses?
- What is his near point with glasses? (Take the image distance from the lens of the eye to the retina to be 2 cm.)
