If the near point of a far-sighted eye is 1 m, what is the power of th – InterviewSolution

1.	If the near point of a far-sighted eye is 1 m, what is the power of the lens required to correct this defect? (Assuming that the near point of a normal eye is 25 cm)
Answer» If the near point of a far-sighted eye is $1 m$ , what is the power of the lens required to correct this defect? (Assuming that the near point of a normal eye is $25 c m$ )

Discussion

No Comment Found

Related InterviewSolutions

Explain the sign convention for the spherical mirror.
The magnitude of the force acting on a current carrying wire placed in a magnetic field in the direction perpendicular to its length depends on ___.
COM OF ROD is at2L/3 then in problems why we take it at L/2
What is responsible for the release of nuclear energy?
what is relation between self induction in solenoid with number of turns
The magnitude of gravitational potential energy of the earth-satellite system is U with zero potential energy at infinite separation.The kinetic energy of satellite is K. Mass of satellite << Mass of earth. Then
If in a plano-convex lens, radius of curvature of convex surface is 10 cm and the focal length of lens is 30 cm, the refractive index of the material of the lens will be
a. Identify the words in the advertisement that gave away the fact that it was placed by John Garrideb. b. Make a list of words which are spelt differently in American and British English.
(a) Explain the functions of the following parts of the eye : (a) cornea (b) iris (c) pupil (d) ciliary muscles (e) eye-lens (f) retina (g) optic nerve (b) If you walk from a dark room into sunlight and back again into dark room, how would your pupils after in size ? What makes this happen ? (c) Explain why, we cannot see our seats first when we enter a darkened cinema hall from bright light but gradually they become visible.
Hypermetropia defect arises due to: