A person can see objects if they are placed at 1.5 m. What kind of lense whould be rqured to read a book at a distance of 25 cm. What is the defect he is suffering from.calculate the power of lens used to correct this defect
Answers
Hi,
Answer:
The image distance for the person, v = -1.5 m = -150 cm
The object distance for the person, u = -25 cm
Since the person is able to see the object when its image is formed at a distance of 150 cm behind the retina.
This means that the person is suffering from long-sight or Hypermetropia in which the person can see objects far away from them but not close to them.(referring to image (a) & (b) from the figure attached below)
Hypermetropia can be corrected by using a contact lens or prescribed eyeglasses with a convex + powered lens which will help the image to be moved forward and focused correctly onto the retina.(referring to image (c) from the figure attached below)
By using the lens formula for the convex lens in front of the hypermetropic eye, we get
1/f = 1/v – 1/u
⇒ 1/f = 1/(-150) – 1/(-25)
⇒ 1/f = [- 1+6] / [25*6]
⇒ 1/f = 5/[25*6]
⇒ 1/f = [1/30] cm
Thus,
The power of the lens (P) used to correct this defect is,
= [1/f] (in meters)
= [1/30] * 100
= [10/3]
= + 3.33 D
Hope this is helpful!!!!!
Answer:
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