Physics Department College of Science General Physics
Code: 4031101-4 Chapter 9
Reflection and Refraction
of Light
Units of Chapter 9
REFLECTION AND REFRACTION AT PLANE SURFACE
IMAGE FORMAITON BY PLANE MIRRORS
TOTAL INTERNAL REFRLECTION
SPHERICAL MIRROS
THIN LENSES
Learning goals of this chapter
□ On completing this chapter, the student will be able to :
□ Define the following terms:
□ The incident ray, The refracted ray, The reflected ray, The angle of incidence, The refractive index, The law of reflection, The law of refraction (Snell’s law), The total internal reflection.
□ Define the following terms:
□ The focal length of a spherical mirror, The focal length of a lens, The equation of mirrors, The equation of lenses, Concave and convex mirrors, Concave and convex lenses, The image formed by mirrors and lenses.
□ Calculate the focal length of the spherical mirrors and lenses.
□ Determine the properties of the images formed by the spherical mirrors and lenses
Geometrical Optics and Wave Optics
Geometrical optics, describes light propagation in terms of rays to treat reflection and refraction. It does not consider optical effects such as diffraction and interference
A light ray is a line that is perpendicular to the light's wave fronts
There are two sections of optics studies
which are geometrical optics and wave
Optics.
Geometrical Optics and Wave Optics
Wave optics studies interference, diffraction, polarization, and other phenomena for which geometric optics is not valid.
The figure shows how light is diffracted while passing
through narrow slit, and in case of decreasing the width of
the slit.
Electromagnetic wave
Reflection and Refraction
□ If light travel from a medium to another; two
phenomenal may happen: Reflection and Refraction
□ All angles lie in the same plane
□ The phenomenon by which the incident light falling on a surface is sent back into the same medium is reflection
□ Laws of Reflection 1. =
□
Incident ray
Reflected ray Normal
Medium 1 Medium 2
Refracted ray
A B
C
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2�
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Reflection and Refraction
□ Laws of Refraction:.
1. Snell’s Law
are refractive index of medium 1 and medium 2
□ Snell’s Law can be written as
Where & is called the relative refractive index from medium 1 to medium 2.
If the medium 1 is air (or vacuum), then , and the last equation becomes
□
Incident ray
Reflected ray Normal
Medium 1 Medium 2
Refracted ray
A B
C
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2�
1�
2�1′
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Reflection and Refraction
□ The refractive index of the medium is defined by
Since the speed of light in vacuum is higher than that in any medium, then the refractive index is always greater than 1.
□
Incident ray
Reflected ray Normal
Medium 1 Medium 2
Refracted ray
A B
C
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2�
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Problem 1
□ A light beam in air is incident on one face of a glass prism.
The angle is chosen so that the emerging ray also makes an angle with the normal to the other face. Derive an
expression for the index of refraction of the prism material,
taking n=1 for air.
Problem 1
Note that , and , where is the prism angle.
Therefore : (1)
The deviation angle is the sum of the two opposite interior angles in the triangle , i.e.,
Substituting for and solve for yield, (2)
At point , is the angle of incidence and is the angle of refraction. The refraction index of the material of the prism is given by
Substituting from 1, 2, we get
□
Image Formation by Plane Mirrors
□ In the case of plane mirrors, the image is Virtual.
□ Virtual images are images that are formed in locations where light does not actually reach.
□ Consider only two rays, The first is incident perpendicular and strikes the mirror at , and the other ray incident with an angle and strikes the mirror at a. Then:
□ From the similarity of the triangles aOv and , it follows that
□
(the –ve sign means the image is in the opposite side on the mirror)
□
Image Formation by Plane Mirrors
□ If an object O is placed a distance o in front of a plane mirror, rays out of that object fall on the mirror.
□ Each ray strikes the mirror and is then reflected back.
□ At the points of contact between each ray and the mirror, we construct a
normal and draw the reflected ray.
□ If we extend the reflected rays backward, they intersect at the point I which is the image.
□ The image is at the same distance behind
the mirror that the object O is in front of
it.
critical angle and Total Internal Reflection
If the light source is set inside the glass as shown; and the angle of incidence is increased as shown in rays a, b, c, d.
Rays will be refracted near to the surface, till it becomes parallel to the surface as in ray e, at which the angle of refraction is 90° . This is called the
“critical angle”
critical angle and Total Internal Reflection
□ We find the critical angle by putting in Snell’s law
□ Or
□ If we increase the angle of incident further, as in ray f, then the ray is reflected back internally in the glass, which is called “total internal reflection”
□
Problem 2
□ A triangular prism of glass is shown. If a ray incident normal to one face and is totally reflected. If is 45°, what can you conclude about the index of refraction of the glass?.
□ Solution:
□ The angle must be equal to or greater than the critical angle , where is given by
□ Then
□ In which the index of refraction of air (.
□ Since total internal reflection occurs, then must be less than 45°, and so
□
Problems
3- A ray of light passing through a glass prism of refracting angle 60º, undergoes a minimum deviation of 30º. Calculate
a) Refracted index of prism
b) The velocity of light in glass if the velocity of light in air is 3×108 m /s 4- The speed of light in a piece of glass is measured to be 2.2x108 m /s.
a- What is the index of refraction for this glass?
b- If a beam of light traveling in air of index of refraction n1 = 1 makes an angle of 8o with the normal to a surface of a piece of glass What is the angle of refraction of the beam that passes into the glass?
5- What is the critical angle for light traveling from diamond into glass, given that the refractive index of diamond is 2.4 and the refractive index of glass is 1.5?
6 -The diagram shows the passage of a ray of light from
air into a substance X. What is the index of refraction of X ?
7- The index of refraction of benzene is 1.80. What is the critical angle for total internal reflection, at a benzene to air?
SPHERICAL MIRROS
□ Spherical Mirrors can be formed as a part of hollow sphere.
They are either Concave or Convex mirrors.
□ Some Basic Definitions
1. The center of curvature C: In a spherical mirror the center of curvature is the center point of the hollow sphere of a mirror. In concave mirror, center of curvature is in front of it but in convex mirror it is behind the mirror.
2. The pole of the mirror P: is The center point on the spherical mirror.
3. The radius of curvature r: is the radius of the hollow
sphere from which the mirror is made, or it is the distance between center of curvature and pole is called radius of curvature.
4. The principle axis : is the straight line passing through center of curvature and pole.
5. The focus F: is the point on the principal axis to which all the light rays which are parallel to the principle axis converge after reflection from the concave mirror.
6. The focal length f: is the distance between the pole of the mirror to the focus.
SPHERICAL MIRRORS EQUATION
□ Convex Mirrors □ Concave Mirrors
□ Mirrors Equation is:
o is the distance from the object to the mirror
i is the distance from the image to the mirror
f is the focal length.
The focal length , where is the radius of curvature.
i
o
i
o
SPHERICAL MIRROS
□ Magnification
M > 0 , if image is upright with respect to the object
M < 0 , if image is inverted with respect to the object
IMI = 1 the image is at the same size of the object
IMI < 1 the image is smaller in size than the object
IMI > 1 the image is larger in size than the object
□ Sign Convention
o is always positive (always in front of the mirror)
i is positive if the image is real, and negative if the image is virtual.
f is positive for concave mirrors and negative for convex mirrors.
SPHERICAL MIRROS
□ Magnification
M > 0 , if image is upright with respect to the object
M < 0 , if image is inverted with respect to the object
□ Real Image : A real image is formed at the point where the light rays originating from a particular object converge by reflection (by mirror) or by refraction (by lens). It can be obtained on the screen when the screen is set in the plane of the image.
□ Virtual Image : A virtual image is formed by extensions of diverging rays. It cannot be obtained on the screen. In diagrams of optical systems, virtual rays are
conventionally represented by dotted lines
□
SPHERICAL MIRROS
Ray Diagrams
To predict the image formed, you can draw three rays from the tip of the object as shown.
1- The first ray incident parallel to the principle axis, will be reflected towards the focus.
2- The second ray that passes through the focus will be reflected back parallel to the principle axis.
3- The third ray (if is possible) that passes through the center of curvature, will incident normal on the surface of the mirror and reflected back to its origin.
4- The crossing of the reflected rays is the image.
Principal axis Object
Real image
Focal point
Concave mirror
Pole Principal axis
Object
Virtual image
Focal point
Concave mirror Pole
SPHERICAL MIRROS
Ray Diagrams
Object beyond C
1. The image is located between C and F.
2. It is a real, 3. inverted image
4. that is smaller in size than the object
Object between C and F
1. The image is located beyond C.
2. It is a real,
3. inverted image
4. that is larger in size than the object.
To predict the image formed, draw two rays from the tip of the object as shown. The crossing of the reflected rays is the image.
SPHERICAL MIRROS
Ray Diagrams
In
convex mirrors,
the image of an object is always virtual, upright, and smaller in size.
Object at C
1. The image will be formed at C also, 2. but it will be inverted.
3. It will be real
4. and the same size as the object.
Object in front of F
1. The image is located behind the mirror.
2. It is a virtual, 3. upright image
4. that is larger in size than the object.
Problem 1
In the figure, suppose f =12 cm and o = 30 cm. Find the position of the image and the magnification.
Solution
Problem 2
Solution
A convex mirror has a radius of curvature of 22 cm. An object is placed 14 cm from the mirror. Locate and describe the image using (a) graphical and (b) algebraic methods.
(a) The image is virtual, upright and smaller.
(b)
Image is virtual (negative) and smaller
Image is upright (m>0)
THIN LENSES
Consider a lens having an index of refraction n and two spherical surfaces with radii of curvature r1 and r2
The lens formula is
where
Converging lens Diverging lens
THIN LENSES
□ Sign Convention
□ For each surface of the lens
r is positive if the center of curvature is on the R-side
r is negative if the center of curvature is on the V-side
□ The object distance o is positive if the object is real and lies on the V-side of the lens
□ The image distance i is positive if the image is on the R-side
□ The image distance i is negative if the image is on the V-side
□ Magnification is negative when both i and o are positive.
THIN LENSES
Ray Diagrams
To predict the image formed, draw three rays from the tip of the object as shown.
The crossing of the three rays after passing through the lens is the image.
THIN LENSES
Ray Diagrams
Special cases will occur if parallel light rays are incident on the lenses, or goes of the focus of the lens as shown.
Problem 3
The lenses in the figures have radii of curvature of 42 cm and are made of glass with n= 1.65. Compute their focal lengths.
Solution
r1 is positive, r2 is negative r1 is negative, r2 is positive
Problem 4
Solution
An object is 38 cm in front of a diverging lens of focal length -24 cm.
Find the location and magnification of the image using (a) graphical and (b) algebraic techniques.
Problems
5. A concave mirror has focal length 10 cm , an object is placed 16 cm from the mirror. Find the location of the image and describe the image ?
6. An object is 25 cm in front of a converging lens of focal length 13 cm. describe the image ?
7. An object is placed 45 cm in front of a convex mirror whose radius of curvature is 30 cm. Describe the image?
8. A mirror has a focal length of 15 cm. An object is located 8 cm from the surface of the mirror. describe the image ?
9. A light bulb is placed a distance of 20 cm from a diverging lens
having a focal length of 10 cm. describe the image ?
Homework
1. A prism with angle 60o , if the deviation angle is 16o , calculate the refractive index of the prism.
2. The glass has the refractive index 1.52, if the speed of light in air is 3×108 m/s, calculate the speed of light within the glass.
3- The index of refraction is 1.52. What is the critical angle for total internal reflection, at a glass to air?
4. A lens made of glass with refractive index 1.5, with radii of curvature , A -calculate the focal length and the type of the lens.?
B- if an object is placed at a distance 40 cm, calculate the position and the magnification of the image. ?
