1.  The near point of a hypermetropic person is 75 cm from the eye. What is d power of  d lens required to enable him to read clearly a book held at 25 cm from the eye?

2.  If the person in the previous problem uses spectacles of power +1.0 D wat is d nearest dist. of distinct vision for him?

1. Defective near vision, v = -75cm

Now, lens formula for this,

1/v + 1/25 = 1/f

=> 1/(-75) + 1/25 = 1/f

=> 2/75 = 1/f

=> f = 75/2 cm

So, power of the lens, P = 1/f

=> P = (100/(75/2)) D

=> P = 2.667 D

2. Given P = 1D

f = 1m = 100cm

Use formula,

1/v + 1/25 = 1/f

to find v = .........?

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1.   Power=1/focal length and in this case values are given in metres so we will convert 1 metre=100 cm.

Now, P=1/f

P=100/75

P=4/3

P=+1.3D

So he will need a lens of +1.3D to read clearly a book.

2. Nearest distinct vision for him is 1 metre as P(given)=1/F.

  • -45

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