Explain with a drawing how hypermetropia is corrected. The near point of a hypermetropic eye is 75 cm. What is the focal length of the lens required to correct this defect? The near point of a normal eye is 25 cm.
Show Hint
For correcting hypermetropia, a convex lens with a focal length equal to the difference between the near points of the defective and normal eye is used.
Step 1: Understand hypermetropia.
- Hypermetropia, or farsightedness, is a defect in the eye where the near point is farther away than normal. In this case, the near point is 75 cm, whereas for a normal eye, the near point is 25 cm.
- The defect occurs when the eye's focal point falls behind the retina, making it difficult to focus on nearby objects.
Step 2: Correction of hypermetropia.
Hypermetropia can be corrected by using a convex lens, which converges the light before it enters the eye. The lens brings the image of nearby objects to the normal near point (25 cm).
Drawing the diagram of correction Step 3: Calculation of focal length.
To calculate the focal length of the lens required to correct the defect, we can use the lens formula:
\[
\frac{1}{f} = \frac{1}{v} - \frac{1}{u}
\]
where:
- \( f \) is the focal length of the lens,
- \( v \) is the image distance (which, for correction, is the normal near point of 25 cm),
- \( u \) is the object distance (which is the near point of the hypermetropic eye, 75 cm).
\[
\frac{1}{f} = \frac{1}{25} - \frac{1}{75}
\]
Calculating the right-hand side:
\[
\frac{1}{f} = \frac{3 - 1}{75} = \frac{2}{75}
\]
Thus, the focal length \( f \) is:
\[
f = \frac{75}{2} = 37.5 \, \text{cm}
\]
Step 4: Conclusion.
The focal length of the lens required to correct the hypermetropic defect is 37.5 cm.
