Question

(a) What is long-sightedness ? State the two causes of long-sightedness (or hypermetropia). With the help of ray diagrams, show :
(i) the eye-defect long-sightedness.
(ii) correction of long-sightedness by using a lens.
(b) An eye has a near point distance of 0.75 m. What short of lens in spectacles would be needed to reduce the near point distance to 0.25 m ? Also calculate the power of lens required. Is this eye long-sighted or short-sighted ?
(c) An eye has a far point of 2 m. What type of lens in spectacles would be needed to increase the far point to infinity ? Also calculate the power of lens required. Is this eye long-sighted or short-sighted ?

Answer 1

(a)Hypermetropia is also known as long-sightedness. In this defect, a person can see the distant objects clearly but cannot see the nearby objects clearly.

Cause of hypermetropia:
This defect arises due to either of the following two reasons:

• The eyeball becomes too small along its axis so that the distance between the eyelens and the retina is reduced.
• The focal length of the eyelens becomes too large resulting in the low converging power of the eyelens.

(i) The rays coming from an object placed at 25 cm (normal near point) from the eye meet at a point behind the retina. So the object cannot be seen clearly.The object has to be moved away from the eyes to a distance greater than 25 cm inorder to focus the rays again on the retina. Thus, the near point of the eye is not at 25 cm but it has shifted to N' at a distance greater than 25 cm from the eyes.

(ii).A hypermetropic eye is corrected using a convex lens of suitable focal length. This lens diverges the rays such that the rays coming from normal near point N appear to come from near point N' after refraction. That is a virtual image of the object placed at N is formed at N'. Then the eyelens forms a clear image at the retina.

(b)This defect is Hypermetropia (long-sightedness)
v = -75 cm
u = -25cm
f = ?

$\frac{1}{f}=\frac{1}{v}-\frac{1}{u}$

$\frac{1}{f}=\frac{1}{-75}-\frac{1}{-25}$

$\frac{1}{f}=\frac{2}{75}$

$f=\frac{75}{2}+37.5cm=0.375m$

$P=\frac{1}{f}=\frac{1}{f\left(inm\right)}=\frac{1}{0.375}=+2.67D$

The person should use a convex lens.

(c)This defect is Myopia(short-sightedness)
v = -2 m
u = -infinity
f = ?

$\frac{1}{f}=\frac{1}{v}-\frac{1}{u}$

$\frac{1}{f}=\frac{1}{-2}-\frac{1}{\infty }$

Since $\frac{1}{\infty }=0$,

$\frac{1}{f}=\frac{-1}{2}$

$f=-2m$

$P=\frac{1}{f}=\frac{1}{f\left(inm\right)}=\frac{1}{-2}=-0.5D$

The person should use a concave lens.