Lenses are transparent materials that refract light. There are two main types: convex and concave. Convex lenses converge light rays to a focal point, while concave lenses diverge light rays. Eye defects like myopia, hyperopia, astigmatism and presbyopia can be corrected using lenses with different optical properties. Myopia is nearsightedness corrected with concave lenses. Hyperopia is farsightedness corrected with convex lenses. Astigmatism involves non-uniform refraction corrected with cylindrical lenses. Presbyopia is age-related stiffening of the lens corrected with reading glasses. More severe conditions like cataracts and strabismus may require surgery.

1. LENSE
S

2. What is Lenses?
Lenses are smoothly
curved transparent
materials, usually
made of glass that
refract light.

3. LENSES

4. The degree of
bending depends
on the curvature
and index of
refraction of the
lens.

5. Two Main Types of Lenses
CONVEX
( Converging Lens)
CONCAVE
( Diverging Lens)

6. Also known
as
Converging
Lens
CONVEX

7. Convex Lens
It is thick in middle and
thin at the edges.
 Lens such that a beam
of light passing through it
is brought to a point or
focus.

8. Also
known as
Diverging
Lens.
CONCAVE

9. Concave Lens
It is thin at the middle
and thick at the edges.
Lens such that a parallel
beam of light passing
through it is caused to
diverge or spread out.

10. Different Kinds
of Eye Defects

11. Myopia
 (nearsightedness) This is a defect of
vision in which far objects appear
blurred but near objects are seen
clearly. The image is focused in front
of the retina rather than on it usually
because the eyeball is too long or the
refractive power of the eye’s lens too
strong. Myopia can be corrected by
wearing glasses/contacts with
concave lenses these help to focus
the image on the retina.

12. Hyperopia
 (farsightedness) This is a defect of
vision in which there is difficulty with
near vision but far objects can be seen
easily. The image is focused behind
the retina rather than upon it. This
occurs when the eyeball is too short or
the refractive power of the lens is too
weak. Hyperopia can be corrected by
wearing glasses/contacts that contain
convex lenses.

13. Astigmatism
 This defect is when the light rays do not all
come to a single focal point on the
retina, instead some focus on the retina and
some focus in front of or behind it. This is
usually caused by a non-uniform curvature of
the cornea. A typical symptom of astigmatism
is if you are looking at a pattern of lines
placed at various angles and the lines
running in one direction appear sharp whilst
those in other directions appear blurred.
Astigmatism can usually be corrected by
using a special spherical cylindrical lens; this
is placed in the out-of-focus axis.

14. Presbyopia
 Presbyopia is part of aging. When
someone gets into their forties, and is
caused by the hardening of the
crystalline lenses in the eye. People
with this will find that their arms are
“getting shorter” as they have to hold
reading material farther and farther
away.

15. Strabismus
 Strabismus, more commonly known
as crossed-eyes, is when the two eyes
do not look toward the same object
together. One eye is normal, while the
other one points in some other
direction. Strabismus can be corrected
with prisms. The prisms bend the light
so the light goes to the correct spot.

16. Cataract
 A cataract is when the lens of the eye
or its capsule is opaque. It is the
leading cause of blindness in the
world. The word “cataract” means
waterfall. For someone with a serious
cataract, it's like seeing through a
waterfall to them.

17.  Sample Problem #1
A 4.00-cm tall light bulb is placed
a distance of 45.7 cm from a
double convex lens having a focal
length of 15.2 cm. Determine the
image distance and the image
size. Like all problems in physics,
begin by the identification of the
unknown information.

18. ho = 4.00 cmdo = 45.7 cmf = 15.2
cmNext identify the unknown
quantities that you wish to solve for.
di = ???hi = ???To determine the image
distance, the lens equation must be
used. The following lines represent the
solution to the image distance;
substitutions and algebraic steps are
shown.

19. 1/f = 1/do + 1/di1/(15.2 cm) = 1/(45.7
cm) + 1/di
0.0658 cm-1 = 0.0219 cm-1 +
1/di
0.0439 cm-1 = 1/di
di = 22.8 cm