PUMPA - SMART LEARNING
எங்கள் ஆசிரியர்களுடன் 1-ஆன்-1 ஆலோசனை நேரத்தைப் பெறுங்கள். டாப்பர் ஆவதற்கு நாங்கள் பயிற்சி அளிப்போம்
Book Free DemoMake 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.
Hypermetropia:
The causes hypermetropia, also known as .
can be seen clearly, but cannot be seen clearly due to this defect. The distance between the eye lens and the retina is , or the focal length of the eye lens is . As a result, the near point for such eyes will no longer be at \(25\ cm\), and the near point will have moved further. Due to this, the image of nearby objects is formed .
Correction of Hypermetropia:
of suitable power can be used for the correction of hypermetropia. It provide additional focusing power which is required for bringing the image back on the retina.
The given person will be able to clearly see the object which is kept at the near point of the normal eye (that is \(25\ cm\)), the image is formed at his given near point of \(1\ m\).
According to the lens formula,
Answer variants:
u
v
f
Substitute the values,
Focal length \(=\) \(m\)
The power (\(P\)) of a lens of focal length \(f\) is given by the relation,
Power of the lens \(=\) \(D\)
Register for free to see more content