What would be the position of the image with respect to the retina for a hypermetropic eye, when the object is placed at (i) Infinity (ii) at 25 cm from the eye
Answers
Answer:
NCERT IN-TEXT ACTIVITIES SOLVED
Q1. What is meant by power of accommodation of the eye?Ans. The ability of the eye to focus the distant objects as well as the nearby objects on the retina by changing the focal length of the eye lens is called power of accommodation.
Q2. A person with a myopic eye cannot see objects beyond 1.2 m distincti What should be the type of the corrective lens used to restore proper vision?Ans. A person with a myopic eye can use concave lens to restore proper vision.
Q3. What is the far point and near point of the human eye with normal vision?Ans. The far point is infinity and the near point is 25 cm of the human eye with normal vision.
Q4. A student has difficulty reading the blackboard while sitting in the It row. What could be the defect the child is suffering from? How can it be corrected?Ans. A student is suffering with the eye defect named myopia, in this defect person can see nearby objects clearly but cannot see far off objects distinctly. It can be corrected by using concave lens.
Explanation:
Answer:
Hypermetropia can be corrected by using a convex lens. A convex lens converges the incoming light such that the image is formed on the retina.
An object at 25 cm forms an image at the near point of the hypermetropic eye. Here, near point is 1 m.
Given,
Object distance,u=−25 cm
Image distance, v=−100 cm
From lens formula, v1−u1=f1
−1001−−251=f1
Focal length,f=100/3 cm=1/3 m
Power, P=f1=1/31=3 D
Explanation: