REFRACTION OF LIGTHT

When light crosses boundary between two media it changes its direction of travel or it is bent, this process is called refraction. This is a result of light changing speed as travels from one medium to another.

Refractive indexThe ability of a material to refract light is measured by the ratio of speed of light in air to the

speed of light in the material, known a refractive index (n). medium

air

cc

n = , where cair is the speed of

light in air, cmedium is the speed of light in the medium. For vacuum n =1, for air also we assume n=1 for most practical purposes.

Other measures of n: n =ri

sinsin

, where i is angle of incidence, and r is angle of refraction; n =

pthapparentderealdepth

,

Refraction of light makes objects look shallower than their actual depth.Diagram

Exercise1:1. A light ray from air (n = 1) is incident on to a piece of fused quartz (n = 1.46) and refracted at an angle of 37. Calculate the angle of incidence: Solution

2. Explain, with the help of a diagram, why an object under water appears shallower than it really is.

Solution: diagram same as in figure 1(N.B. indicate the direction in which the light rays travel)Rays from the object under the water get refracted as the reach the water/air boundary. As their direction of travel changes, they appear to come from a different position from that of the original object. In this case, the new position is shallower than the original position.Lenses

Figure 1

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47.61)37sin*46.1(sin

)sin(sinsinsin

sinsin

1

1

=

=

=

=

=

ii

rnirni

rin

Lenses have the ability to refract light from illuminated objects such that they form images of those objects. Lenses are characterized by the distance from the lens at which they would focus the parallel (beam) light rays; this distance is called the focal length of the lens (f)

Types of lenses

Generally, there are two types of lenses, Convex or positive or converging lens and lens Concave or negative or diverging lens

(i) Convex lens (ii) Concave lensf is in front of lens; i.e. positive f is in front of lens i.e. negative.

Exercise 3:1. State and/or explain the main difference between positive and negative lenses.

2. A spectacle lens held 75 cm from a white wall forms a real, inverted image of distant objects on the wall. What type of lens is it and what is its focal length?

The lens equation

io xxf111

+= Where f is the focal length of the lens, xo is the object distance from the center of

the lens; xi is the image distance from the center of the lens.

Linear Magnification

Magnification of a lens (M) is given by; o

i

hh

M = , where hi is the image height and ho is the

object height or o

i

xx

M

= , where xi is the image distance and xo is the object distance

Exercise 3: 1. A thin converging lens has a focal length of + 24 cm. An object is placed 9.0 cm from the lens. (i) What is the lens strength or power in dioptres?

ff

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(ii) Find the distance of the image formed from the lens. (iii) State whether the image is real or virtual. (iv) What is the magnification of the image? (v) State whether the image is upright (erect) or inverted.

Refractive power

The refractive power, dioptric power or refractive power(s) is the degree to which a lens or mirror converges or diverges light. It is equal to the reciprocal of the focal length.

That is, the refractive power of a lens is given by; fs 1= diopters, where f is the focal length of

the lens measured in meters. The unit for measuring the refractive power is diopter. 1diopter =1 m-1.Associated questions:1. A magnifying lens is held facing sunlight and is observed to focus the sun's rays to a bright point on a sheet of paper when the lens is held 5 cm above the paper. What is the power or strength of the lens?

The eye

In the eye, refraction of light by the cornea and the crystalline lens enables the eye to focus images of object on the retina. N.B In simple the refracting system of cornea and the crystalline lens is often represented by the lens alone.

Diagram

Distant vision Close vision

.

Diagram Diagram

Associated questions: 1. Explain clearly the changes taking place when the human eye views objects at different distances

Normal visionWhen the eye is fully relaxed it should focus images of distant object at its focal length. A normal eye, when relaxed, the cilliary muscles are relaxed, but cilliary fibers are taut. In this condition, the lens is thin and has maximum focal length. Thus, the maximum focal length of a normal eye should be equal to distance of the retina from the center of the lens (image distance). Since the

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distance of the retina from lens (image distance) is fixed. To focus objects near than infinity, the eye has to adjust the focal length of the lens. This process is called accommodation the process by which the eye increases optical power to maintain a clear image (focus) on an object as it draws near the eye, or accommodation is the process by which the eye changes its focal length to form clearer retinal images of objects at different distances from the eye. For objects, ranging from infinity to within several meters very little adjustment of the focal length is required; however, for objects within a meter from the eye a significant adjustment is required, up to approximately 20% change of focal length. (Images of a very distant object are focused at roughly the focal length of a convex lens. For distant objects, the image is at exactly the focal length. Therefore, in a relaxed condition images of distant object are focused at the retina.More details of an object are seen when it is about 25 cm from the eye. This is called distance of most distinct vision. For kids, the distance of most distinct vision is below this average value. It naturally increases with age, at an age of 70, being about 4 meters! And one has difficulty reading a newspaper just at an arms length.

Common vision defects

Myopia, near-sightedness or short-sightedness: This is the condition where the focal length of the eye is too short; the image of a distant object will form in front of the retina. Thus, very far objects cannot be clearly seen (blurred), near objects are clearer. This condition is corrected with a diverging/concave/negative lens of appropriate diopter strength, in front of the eye.

DiagramHyperopia, hypermetropia, far-sightedness or long-sightedness. This is the condition where the focal length of the eye is too long; the image of a distant object will form behind the retina. The suffers may have difficulty focusing both nearer and far objects. However far objects may be seen clearer than near objects. This condition is corrected with a converging/convex/positive lens of appropriate diopter strength.

Diagram

The focal length of the necessary correcting lens is given by the following equation.

correctedlenseye fff111

=+ Or in refractive power, correctedlenseye sss =+

Associated questions:

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1. Define hyperopia (hypermetropia) and explain how this eye defect can be corrected. Draw a sketch to illustrate your answer.

2. A farsighted person has a near point of 200 cm. What kind of glasses does this person need to be able to read a book held at 25 cm from the eye?

3. The far point of a certain eye is 1 m in front of the eye. Explain the term far point of an eye. What lens should be used for the eye to see clearly an object at infinity?

Magnifying glass

This is the situation where a convex lens is used to form magnified and virtual images of objects. The object must be placed within the focal length of the lens.

Diagram

Optical Fibers

Thin glass fibers that can transmit light even when they are bent, without losing it. Optical fibers use the principle of total internal reflection to pipe light. Gastro scopes use optical fibers for producing internal images. Diagram

Total internal reflectionTotal internal reflection is the process where light is reflected within a medium of higher optical density. This happens when the light rays from an optically denser material are incident on its surface at an angle grater than a special angle for that particular material called the Critical Angle. Critical angle is the angle of incidence that corresponds to the angle of reflection of 90. When the angle of incidence increases beyond this angle, total internal reflection occurs.Diagram

Associated questions:

1. Explain the terms critical angle and total internal reflection. Draw a sketch to illustrate your answer.

2. Give two clinical applications of total internal reflection.

1. An optical fiber is made of a material with a refractive index 1.3 and bound by air (na =1). (i)For what angles with the surface will light, remain contained in the optical fiber.

Magnifying glass

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Compound Microscope

In a microscope, images of objects can be magnified several hundred times. This is achieved by, a set of convex lenses of shorter and longer focal lengths called the objective and the eye piece, respectively. The array of the lenses is such that the objective is near to the object to be viewed. The object is placed just outside the focal length of the objective, resulting in a magnified, real image of the object. This magnified image now serves as an object for the eyepiece. The distance of eye piece is adjusted such that the position of the later image coincides with its focal length. The image, seen through the eye piece is perceived as that of an object at infinity. In this setting, the microscope is in its normal adjustment. This adjustment does not strain our eyes that much, when viewing images for longer times. The resultant magnification of the microscope is the product of magnification of the two lenses. M=me x mo, where me is the magnification of the eyepiece and mo is the magnification of the objective

Associated questions:1. A compound optical microscope consists of two converging lenses: an objective lens of focal length fo = 4 cm and an eyepiece of focal length fe = 10 cm. The separation between the lenses is L = 20 cm, and if an object is placed 6 cm in front of the objective lens,(i) Locate the position of the image formed by the objective lens. (ii) Locate the final image formed by the eyepiece, and seen by the observer. (iii) Hence, calculate the total magnification of the microscope. (iv) Draw the ray diagram of the microscope showing the image locations.

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