make a video to explain this question and how to solve it---**Question Stem:**
2. Make a diagram to show how hypermetropia is corrected. The near point of a hypermetropic eye is 1 m. What is the power of the lens required to correct this defect? Assume that the near point of the normal eye is 25 cm.
**Other Relevant Text (Answer Introduction):**
Ans: The diagram showing the correction of hypermetropia is drawn below.
**Chart/Diagram Description:**
* **Type:** Ray diagram illustrating the correction of hypermetropia (farsightedness) using a convex (converging) lens.
* **Main Elements:**
* **Eye:** Represented by an outline of an eyeball with an internal biconvex lens (eye lens) and a red point on the retina (back of the eye).
* **Corrective Lens:** A biconvex lens is placed to the left of the eye.
* **Points:**
* A red point labeled 'N' is located to the far left. This represents the object (the normal near point).
* A red point labeled 'N'' is located between the corrective lens and the eye. This represents the virtual image formed by the corrective lens, which acts as the object for the eye.
* A red point on the retina indicates where the light rays converge after passing through the eye's lens.
* **Lines and Rays:**
* Three solid blue lines with arrows represent light rays originating from point 'N'.
* These rays travel towards the corrective lens.
* After passing through the corrective lens, the rays converge towards point 'N'' (indicated by dashed lines extending backwards from the refracted rays).
* These refracted rays (now appearing to come from N') then enter the eye and pass through the eye's lens.
* Inside the eye, the rays converge precisely onto the red point on the retina.
* Dashed blue lines are used to show the virtual image formation by the corrective lens, appearing to originate from N'.
* **Relative Position and Direction:** Light rays travel from left to right. Point N is to the left, followed by the corrective lens, then point N', and finally the eye. The rays successfully converge on the retina after passing through the corrective lens and the eye's lens.
**Other Relevant Text (Solution Explanation):**
Near point of a hypermetropic eye is 1 m and that of a normal eye is 25 cm.
Given,
$\Rightarrow v = -1 \text{ m} = -100 \text{ cm}$
$\Rightarrow u = -25 \text{ cm}$
From the lens formula, we have:
$\Rightarrow \frac{1}{f} = \frac{1}{v} - \frac{1}{u}$
$\Rightarrow \frac{1}{f} = \frac{1}{-100} - \frac{1}{-25}$
$\Rightarrow f = \frac{100}{3} \text{ cm} = \frac{1}{3} \text{ m}$
The focal length is given by,
$\Rightarrow f = \frac{1}{P}$
Near point of a hypermetropic eye is 1 m and that of a normal eye is 25 cm.
Given,
⇒ v = -1 m = -100 cm
⇒ u = -25 cm
From the lens formula, we have:
⇒ 1/f = 1/v - 1/u
⇒ 1/f = 1/(-100) - 1/(-25)
⇒ f = 100/3 cm = 1/3 m
The focal length is given by,
⇒ f = 1/P
⇒ P = 1/(1/3) D
⇒ P = +3.0 D
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Hypermetropia, also known as farsightedness, is a common vision defect where distant objects are seen clearly, but near objects appear blurred. In a normal eye, light rays from objects at the near point of 25 centimeters converge precisely on the retina. However, in a hypermetropic eye, these same light rays converge behind the retina, causing the image to be out of focus. For our problem, we have a hypermetropic eye with a near point of 1 meter instead of the normal 25 centimeters.
To correct hypermetropia, we use a convex or converging lens placed in front of the eye. This corrective lens creates a virtual image of objects at the normal near point of 25 centimeters, positioning this virtual image at the hypermetropic eye's near point of 1 meter. The lens essentially moves the apparent position of near objects farther away, to a distance where the hypermetropic eye can focus them properly on the retina. Light rays from point N pass through the convex lens and appear to come from point N-prime, which the eye can then focus clearly.
Now let's set up the mathematical framework for solving this problem. We have the given information: the near point of the hypermetropic eye is 1 meter, and the near point of a normal eye is 25 centimeters. For the lens formula, we need to identify the object distance u and image distance v. The object is at the normal near point, so u equals negative 25 centimeters. The virtual image must be formed at the hypermetropic near point, so v equals negative 100 centimeters. Both distances are negative according to sign conventions: u is negative because it's a real object, and v is negative because it's a virtual image.
Now let's apply the lens formula step by step to calculate the focal length. We start with the lens formula: one over f equals one over v minus one over u. Substituting our values: one over f equals one over negative 100 minus one over negative 25. This simplifies to negative one over 100 plus one over 25. Converting to a common denominator: negative one over 100 plus four over 100 equals three over 100. Therefore, f equals 100 over 3 centimeters, which converts to one-third meter. This positive focal length confirms we need a convex lens for correction.
Hypermetropia, also known as farsightedness, is a common vision defect where distant objects are seen clearly but near objects appear blurred. This occurs because light rays from nearby objects converge behind the retina instead of directly on it. In this problem, we need to find the power of a corrective lens for an eye with a near point of 1 meter, compared to the normal near point of 25 centimeters.
Let's analyze the problem. A normal eye can focus on objects as close as 25 centimeters, but this hypermetropic eye can only focus clearly on objects that are 1 meter or farther away. Our goal is to find a corrective lens that will allow this eye to see objects at the normal near point of 25 centimeters.
To solve this problem, we use the lens formula. The corrective lens must create a virtual image at 1 meter when an object is placed at 25 centimeters from the lens. Using sign conventions: object distance u equals negative 25 centimeters, and image distance v equals negative 100 centimeters, since both are on the same side as the object.
Now let's solve the problem step by step. Given that v equals negative 1 meter or negative 100 centimeters, and u equals negative 25 centimeters. Using the lens formula: one over f equals one over v minus one over u. Substituting our values: one over f equals one over negative 100 minus one over negative 25. This simplifies to negative one over 100 plus one over 25, which equals negative one over 100 plus four over 100, giving us three over 100. Therefore, f equals 100 over 3 centimeters, or one-third meter.
Finally, we calculate the power of the lens using the power formula: P equals one over f. Substituting our focal length of one-third meter: P equals one divided by one-third, which gives us positive 3.0 diopters. The positive sign confirms this is a convex lens, and the value of 3.0 diopters indicates a strong converging lens is needed. This completes our solution: a hypermetropic eye with a near point of 1 meter requires a corrective lens with a power of positive 3.0 diopters to restore normal near vision at 25 centimeters.