Question

5. A person needs a lens of power-5.5 dioptres for correcting his distant vision. For
correcting his near vision he needs a lens of power +1.5 dioptre. What is the focal
length of the lens required for correcting (i) distant vision, and (ii) near vision?
6. The far point of a myopic person is 80 cm in front of the eye. What is the nature and
power of the lens required to correct the problem?
m to show how hypermetropia is corrected. The near point of a​

Answer 1

Explanation:

(i) Distant Vision :

Given, 

    Power =5.5D

    Power, P=f1

    Focal length, f=P1

    ⇒f=−5.51

    ⇒f=−5510

    ⇒f=−112

    ⇒f=−0.181m

    ⇒f=−18.1cm

(ii) Near Vision :

Given,

Power =1.5D

 

    P=f1

    f=P1

    ⇒f=1.51

    ⇒f=1510

    ⇒f=32

    ⇒f=0.667m

    ⇒f=66.7cm

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Answer 2

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Answer:(i). -0.1818 m      (ii). 0.66 m

(i).Power of the lens for correcting the distant vision = -5.5 Dioptres

Power of the lens for correcting the near vision = +1.5 Dioptres

Now,

We know that the Power of the lens is given by,

Therefore, the Focal length of the lens is given by,

Focal length, f = -0.1818 m

(ii).The Power of lens for near vision = +1.5 Dioptres

So,

The focal length of the lens is given by,

Therefore, the Focal length of the lens is given by,

Focal length, f = 0.66m