NCERT Question 7 - Chapter 11 Class 10 - Human Eye and Colourful World
Last updated at Sept. 10, 2019 by Teachoo
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.
NCERT Question 7 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
Near point of the person is 1 m.
Since the hypermetropia.
The type of lens used to correct hypermetropia is a convex lens.
Near point = 1 m means that person can see object placed at 25 cm clearly if the image is formed at 1 m.
To find the power of the lens, we need to find its focal length.
Since the object is in front of the eye, object distance will be negative.
Object distance = u = – 25 cm
Since the image is made in front of the lens, image distance will be negative.
Image distance = v = − 1 m
= –100 cm
We have to find focal length.
Using lens formula,
1/𝑓 = 1/𝑣 − 1/𝑢
1/𝑓 = 1/((−100)) − 1/((−25))
1/𝑓 = (−1)/100 + 1/25
1/𝑓 = (−1 + 4)/100
1/𝑓 = 3/100
f = 100/3 cm
f = 100/3 × 1/100 m
f = 1/3 m
Power of the lens = 1/(𝐹𝑜𝑐𝑎𝑙 𝑙𝑒𝑛𝑔𝑡ℎ (𝑖𝑛 𝑚))
= 1/((1/3) )
= 3 D
Power of the lens is 3 D
and since focal length was positive, it is a convex lens.