Notifications
Clear all
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?
0
25/09/2021 7:04 pm
Topic starter
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.
Answer
Add a comment
Add a comment
Topic Tags
1 Answer
0
25/09/2021 7:10 pm
The required diagram is shown below:
Given: Distance of near point of defective eye, i.e.
v = 1 m = 100 cm
Distance of near point of normal eye, i.e.
u = 25 cm
Applying the formula \(\frac{1}{f}\) = \(\frac{1}{v}\)  \(\frac{1}{u}\), we get,
\(\frac{1}{f}\) = \(\frac{1}{100}\)  \(\frac{1}{25}\)
= \(\frac{1}{100}\) + \(\frac{1}{25}\)
= \(\frac{1 + 4}{100}\) = \(\frac{3}{100}\)
or f = \(\frac{100}{3}\) = 33.33 cm
∴ Power, P = \(\frac{100}{f}\) = \(\frac{100}{33.3}\)
= 3.0 D
Hence, the power of the required lens is 3.0 D.
Add a comment
Add a comment
Forum Jump:
Related Topics

Why does the sky appear dark instead of blue to an astronaut?
3 years ago

Why does the sun appear reddish early in the morning?
3 years ago

Explain why the planets do not twinkle.
3 years ago

Why do stars twinkle?
3 years ago

What happens to the image distance in the eye when we increase the distance of an object from the eye?
3 years ago
Topic Tags:
class 10 (8819)
,
cbse (15084)
,
ncert (13840)
,
human eye (135)
,
colourful world (292)
,
Forum Information
 321 Forums
 27.3 K Topics
 53.8 K Posts
 0 Online
 12.4 K Members
Our newest member: Stripchat
Forum Icons:
Forum contains no unread posts
Forum contains unread posts
Topic Icons:
Not Replied
Replied
Active
Hot
Sticky
Unapproved
Solved
Private
Closed