Advertisements
Advertisements
प्रश्न
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.
Advertisements
उत्तर
Given:
As the person wears spectacles at a distance of 1 cm from the eyes.
Distance of the image from the glass, v = Distance of the image from the eye − Separation between the glass and the eye
v = 25 − 1 = 24 cm = 0.24 m
For the near-sighted person, u = ∞.
The lens formula is given by
`1/v -1/u = 1/f`
On substituting the values in the above formula, we get:
`1/f = 1/((-0.24))-1/∞ = -42`
Power of the lens, P = `1/f` = − 4.2 D
Thus, the power of the lens required to see distant objects is − 4.2 D.
APPEARS IN
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.
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?
What should be the distance between the object and the magnifying glass if the virtual image of each square in the figure is to have an area of 6.25 mm2. Would you be able to see the squares distinctly with your eyes very close to the magnifier?
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?
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?

When objects at different distances are seen by the eye, which of the following remain constant?
A person A can clearly see objects between 25 cm and 200 cm. Which of the following may represent the range of clear vision for a person B having muscles stronger than A, but all other parameters of eye identical to that of A?
The distance of the eye-lens from the retina is x. For a normal eye, the maximum focal length of the eye-lens
A man wearing glasses of focal length +1 m cannot clearly see beyond 1 m
When we see an object, the image formed on the retina is
(a) real
(b) virtual
(c) erect
(d) inverted
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 |
Can virtual image be formed on the retina in a seeing process?
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 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.)
