Question

(a) A person is suffering from hypermetropia. The near point of the person is \[1.5m\]. Calculate the focal length of the convex lens used in his spectacles. Assume the near point of the normal eye is \[25cm\].
(b)What do you mean by the term range of vision? Also write its value.
(c)What is the limit of accommodation of a healthy human eye?

Answer 1

Hint: As we know Focal length, we can find by using lens maker formula. After finding the focal length we can easily calculate the value of power.People with farsightedness have powers with a positive sign. For example $ +1.50$ or $+2.50$ as their corrective powers.

Formula used:
\[\dfrac{1}{f} = \dfrac{1}{u} + \dfrac{1}{v}\]
Here $f$ is the focal length, $u$ is the object distance and $v$ is the image distance.

Complete step by step answer:
(a) As we know that Hypermetropia is a condition of the eye where the person is not able to see the near things clearly. That person can see the farther objects very clearly. Also the normal eye point of the eye is\[25cm\]. For the normal view point, an object must be at a distance of at least \[25cm\]. Now, apply the formula,
\[\dfrac{1}{f} = \dfrac{1}{u} + \dfrac{1}{v}\]
Here f is the focal length, u is the object distance and v is the image distance.
Here virtual image is formed, so formula will become-
\[\dfrac{1}{f} = \dfrac{1}{u} - \dfrac{1}{v}\]
\[\Rightarrow\dfrac{1}{f} = \dfrac{1}{{25}} - \dfrac{1}{{100}}\]
By solving this, we get value of focal length, that is\[ = 33.3cm\]
Now, for finding the power of the eye,
\[\therefore P = \dfrac{{100}}{f}\]
So, power comes out to be \[3D\].

(b) The range of vision of a person is defined as the distance between the near point and far point from the eyes of that person.
Near point:The point which is at closer distance, by which we can see things clearly.
Far Point: The farther distance at which the object is seen clearly.
For a normal eye, the far point lies at infinity. So, a person with normal vision has an infinite range of vision.

(C) For a healthy human eye, power of accommodation is approximately \[4D\].

Note:Hyperopia is when a person can see nearby objects very clearly and cannot see farther objects. In this question, a virtual image is formed, remembering to substitute the value of $v$ by using sign convention.It is important to remember that Hyperopia is an eye defect and not an eye disease. Therefore, it can be corrected.