By:-DR. VIKRAM SINGH
TANUSHREE SINGH
YEAR OF PUBLICATION-2010
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SAVANT INSTITUTE
TMCLASS XII
PHYSICS
7
RAY OPTICS AND OPTICAL INSTRUMENTS
_____________________ Slide 1 ______________________ Thin Lenses: History
§ Thin lens were first used for practical purposes by a Dutch merchant, Anton van Leeuwenhoek (1632 – 1723). § He used very small pieces of glass (it is easier to have a
flawless small piece of glass than a flawless large one) and polished them so accurately that he could get magnifications of more than 200 without loss of detail.
§ He was able to see blood capillaries, and tiny living animals (protozoa).
_____________________ Slide 2 ______________________ Microscopes
§ Such strong magnifying lenses are microscopes (from Greek words meaning “to see the small”).
§ A microscope, like the one Leeuwenhoek used, made with one lens are called “simple” microscopes.
§ If two lenses are used it is called a “compound” microscope. _____________________ Slide 3 ______________________ Telescopes
§ The word telescope comes from the Greek “to see the distant.
§ The telescope is supposed to have been invented by an apprentice-boy in the shop of the Dutch spectacle maker Hans Lipershey (ca. (1570 – 1619) in about 1608.
§ Galileo Galilei (1564 – 1642), upon hearing rumors of the new device, experimented with lenses until he had built the first practical telescope in 1610.
_____________________ Slide 4 ______________________ Moon from Apollo 16
_____________________ Slide 5 ______________________ Stamp in the honour of Robert Snell
_____________________ Slide 6 ______________________ Introduction:
§ Light is a non-mechanical (requires no medium for propagation) form of energy due to which we have sensation of vision.
§ Broadly optics is divided into three groups. (1) Geometrical optics (2) Wave optics (3) Quantum optics _____________________ Slide 7 ______________________ Geometrical optics
§ In this, light is considered as a ray on which its transfer takes place.
_____________________ Slide 8 ______________________ § Geometrical optics states that for each and every object
there is an image.
§ Geometrical optics works on following assumptions. § Rectilinear propogration of light i.e. light ray travels in in
straight line.
1. Lawsof reflection. 2. Lawsof refraction
3. Physical independence of light rays i.e. two light rays are totally independent from each other.
_____________________ Slide 9 ______________________ The Ray (Particle) Model of Light
§ Evidence suggests that light travels in straight lines under a wide variety of circumstances.
§ We infer the positions of objects by assuming that light moves from the objects to our eyes in straight lines.
§ This is the ray model of light. § Newton used the ray model.
§ Ray model explains reflection, refraction, and the
formation of images by mirrors and lenses.
§ This subject is often referred to as geometrical optics. _____________________ Slide 10 _____________________ Ray Model
Slide 11 What Can Happen to Light?
When it strikes a surface of an object light can be: 1. Reflected
2. Transmitted
3. Absorbed (and transformed to thermal energy)
_____________________ Slide 12 _____________________ Reflected _____________________ Slide 13 _____________________ Transmitted _____________________ Slide 14 _____________________ Absorbed
Slide 15
Reflection and Scattering
_____________________ Slide 16 _____________________ Movie screen scatters light in all directions
_____________________ Slide 17 _____________________ Interactions of Light with Matter
Interactions between light and matter determine the appearance of everything around us: objects reflect some wavelengths, absorb others and emit others.
_____________________ Slide 18 _____________________ The Speed of Light and Index of Refraction
§ The accepted value today for the speed of light in a vacuum is c = 2.99792458 x 108 m/s. We usually round off to c = 3.0 x 108 m/s.
Slide 19 Reflection of Light
_____________________ Slide 20 _____________________ Object
§ Any source of light is called an object and is of two types. Real object
§ When light rays diverse from a light source, object is real.
_____________________ Slide 21 _____________________ Virtual object
§ When light rays converge towards the light source, object is virtual.
§ To find the position of object we should find out the point of intersection of incident rays.
_____________________ Slide 22 _____________________ Image
§ To find the position of Image we should find out the point of intersection of reflected or refracted ray. There are of two types of images.
Real Image
§ If the reflected or refracted ray converges towards the point of intersection the image is real.
Slide 23 Virtual Image
§ If the reflected or refracted rays diverges from a point then the point of intersection is virtual Image.
_____________________ Slide 24 _____________________ Mirror
§ A smooth and shiny surface on which reflection can take place is called a mirror, having two types.
1. Plane mirror. 2. Spherical mirror. _____________________ Slide 25 _____________________ Reflection of light
§ The returing back of light in the same medium from which it has come after striking a surface is called reflection of light. § Reflection may be
§ Diffused reflection – here random reflections take place through irregular surfaces.
§ Specular reflection - here regular reflections take place through plane surfaces.
_____________________ Slide 26 _____________________ Reflection _____________________ Slide 27 _____________________ Diffuse reflection Slide 28 _____________________ Slide 29 _____________________ Image Formation—Plane Mirror
_____________________ Slide 30 _____________________ Image Formation –Plane Mirror
Slide 31
_____________________ Slide 32 _____________________ Conceptual Example
How tall must a full-length mirror be?
A women 1.60 m tall stands in front of a vertical plane mirror. What is the minimum height of the mirror, and how high must it lower edge be above the floor, if she is to see her whole body? (Assume her eyes are 10 cm below the top of her head.)
_____________________ Slide 33 _____________________
_____________________ Slide 34 _____________________
Reflection of light though plane mirror
Laws of Reflection
(i) The incident ray, the reflected ray and the normal to the reflecting surface at the point of incidence all lie in the same plane.
(ii) The angle of reflection r is equal to the angle of incidence i, that is i = r.
_____________________ Slide 35 _____________________ Image of a Point in a Plane Mirror
OA = OA′
_____________________ Slide 36 _____________________ Image of an extended object in a plane
The image will be an inverted one.
_____________________ Slide 37 _____________________ Properties of the image formed by a plane mirror
§ The image is as far behind the mirror as the object is in front of the mirror. Object distance = image distance.
§ The image is unmagnified, virtual and erect. Height of object = height of Image.
§ The image has front-back reversal. i.e. It shows lateral inversion i.e. left potion of object appears as right portion of Image.
_____________________ Slide 38 _____________________ § A plane mirror deviates light through an angle
δ
= 180°−
2i§ Where i is the angle of incidence. The deviation is maximum for normal incidence. δmax = 180°.
§ Glancing angle φ = angle between mirror and reflected rays is called glancing angle φ = 90 – i = 90 – r.
Slide 39 Rotation of Mirror
§ If the direction of the incident ray is kept constant and the mirror is rotated through an angle θ about an axis in the plane of mirror, then the reflected ray rotates through an angle 2θ.
_____________________ Slide 40 _____________________ § If an object moves toward or away from) a plane mirror at
speed v, the image will also approach (or recede) at the same speed v, and the relative velocity of image with respect to object will be 2v, as shown in figure a.
_____________________ Slide 41 _____________________ § If the mirror is moved toward (or away from) the object with
speed v, the image will also move towards or away from) the object with a speed 2v.
_____________________ Slide 42 _____________________ Number of image by inclined mirrors
§ When θ, the angle between the two mirrors, is an exact sub multiple of 180°
§ The total number of images formed due to successive reflection is equal to either 360°
θ or 360 1 ° − θ . § Accordingly as 360°
θ is odd or even, respectively.
§ In the first case (360°
θ is odd) when the object is placed
exactly midway between the two mirrors, the two images coincide and the total number of images is 360°−1
θ . Slide 43 Solved Example If, θ = 90°, n = 3, θ = 72°, n = 360 72 ° = 5
If the object is placed symmetrically midway between the mirror, then 360 n 1 4. 72 ° = − = _____________________ Slide 44 _____________________ Illustration
Two plane mirrors are inclined at an angle of 60°. An object is placed between the mirrors. What is the total number of images formed by two mirrors?
_____________________ Slide 45 _____________________ Illustration
A man H m tall wishes to see his full-length image in a plane mirror hanging vertically on a wall. Find the length of the shortest mirror which he can see his entire image. If his eyes are H m above the ground, find the position of the mirror.
_____________________ Slide 46 _____________________ Spherical Mirrors
Slide 48
_____________________ Slide 49 _____________________ Rays parallel to the principal axis of the mirror come to a focus at
F, called the focal point, as long as the mirror is small in
diameter, d, as compared to its radius of curvature, r. In that case q will be small and the rays will cross each other at very nearly the same point.
_____________________ Slide 50 _____________________ Reflection from Curved Surfaces
Important Terms
(i) Centre of Curvature (C)
It is the centre of sphere of which the mirror is a part. (ii) Radius of Curvature (R)
It is the radius of the spherical reflecting surface.
Slide 51 (iii) Pole (P)
It is the geometrical centre of the spherical reflecting surface.
(iv) Principal Axis
It is the straight line joining the centre of curvature to the pole.
_____________________ Slide 52 _____________________ (v) Focus (F)
When a narrow beam of rays of light, parallel to the principal axis and close to it, is incident on the surface of a mirror, the reflected beam is found to converage or appear to diverge from a point on the principal axis. This point is called the focus.
(vi) Focal length (f)
It is the distance between the pole and the principal focus. For spherical mirrors f = R/2
_____________________ Slide 53 _____________________ Reflection of light from spherical (curved) sign convention § Pole is taken to be the origin and the principal axis as the
x-axis.
§ The quantities u, v, R and f is positive if the corresponding point lies on the positive side of the origin (in the direction of incident light) and negative if it is on the negative side. (opposite to the direction of incident light).
_____________________ Slide 54 _____________________ § The distances measured in the upward direction,
perpendicular to the principal axis of the mirror are taken as positive and those measured in downward direction is taken as negative.
_____________________ Slide 55 _____________________ Rules for ray diagrams
§ The position, size and nature of an image formed by mirrors are conventionally expressed as ray diagrams. We can locate the image of any extended object graphically by drawing any two of the following four special rays:
(a) A ray, initially parallel to the principal axis is reflected through the focus of the mirror.
Slide 56
(b) A ray, initially passing through the focus is reflected parallel to the principal axis.
(c) A ray passing through the centre of curvature is reflected back along itself.
(d) A ray incident at the pole is reflected symmetrically. _____________________ Slide 57 _____________________ Mirror Formula
In terms of Cartesian Sign convention mirror formula may be expressed as:
1 1 1 v+ =u f
Where, u = object distance from the pole v = image distance from the pole, f = focal length of the mirror.
_____________________ Slide 58 _____________________ Relation between u, v and R for spherical mirrors
§ Consider the situation shown in figure. A point object is placed at the point O of the principal axis of a concave mirror.
§ The line CA is the normal at A.
_____________________ Slide 59 _____________________ § Thus, by the laws of reflection. ∠OCA = ∠CAI. Let α, β, γ
and θ denote the angles AOP, ACP, AIP and OAC respectively.
§ As the exterior angle in a triangle equals the sum of the opposite interior angles, we have,
From triangle OAC
β
=α
+θ
….(i) And from triangle OAI γ = α + 2θ ….(ii) _____________________ Slide 60 _____________________ § Eliminating θ are from (i) and (ii)2β = α + γ …..(iii) § If the point A is close to P, the angles α, β and γ are small
and we can write.
§ AP, AP and AP PO PC PI α = β = γ ≈ Fig (i) § Hence , 2AP AP AP PC=PO+ PI or, 1 1 2 ... iv
( )
PO+PI= PC _____________________ Slide 61 _____________________ § The pole P is taken as the origin and the principal axis asthe X – axis.
§ As, the distances PO, PI and PC are positive, PO = –u, PI = – V and PC = – R. Putting in (iv)
1 1 2 –u+–v =–R 1 1 2 1 u v R f = + = = Note:
§ While using the mirror formula always write only known quantities with sign and unknown quantities without sign. _____________________ Slide 62 _____________________ Magnification
§ Magnification – It is defined as the ratio of dimension of Image to the dimension of object (m).
§ It is a pure number.
_____________________ Slide 63 _____________________ Lateral magnification or transverse magnification
§ It is given by, i 0 h v m h u − = =
§ Where hi = height of image, h0 = height of object.
§ If magnification m is positive, the image is erect w.r.t. respect to the object;
§ If m is negative image is inverted with respect to the object. § For a real image by spherical mirror, m is negative. § For a virtual image by spherical mirror, m is positive.
_____________________ Slide 64 _____________________ Application of mirror formula
§ The mirror formulae may be used to find the position, nature and size of the image formed by a spherical mirror.
Limitation
§ The limitation of the formulae is that it is applicable only for paraxial rays (the rays which make very small angle with the principal axis).
Slide 65 Relation between f and R –
§ From mirror formula we have derived 1 1 2
u+ =v R, now if the object is at infinity the image is formed at focus.
§ Hence, when u = ∞ v = f, Now from mirror formula 1 1 2
f+∞=Ror 1 2
f= R or, f = R/2 i.e. focal length of a mirror is half of its radius of curvature.
_____________________ Slide 66 _____________________ Object Not at Infinity
(a) Ray 1 goes out from O′ parallel to the axis and reflects through F.
_____________________ Slide 67 _____________________
(b) Ray 2 goes through F and then reflects back Parallel to the axis.
_____________________ Slide 68 _____________________
(c) Ray 3 heads out perpendicular to mirror and the reflects back on itself and goes through C center of curvature
Slide 69
_____________________ Slide 70 _____________________
_____________________ Slide 71 _____________________ Convex Mirrors
Slide 73 Illustration
An object of length 3 cm is placed at a distance of (5/4)f from a concave mirror where f is the magnitude of the focal length of the mirror. The length of the object is parallel to the principal axis. Find the length of the image, is the image erect or inverted?
_____________________ Slide 74 _____________________ Newton’s formula
If the object and image distances are measured from the focus instead of the pole of the mirror. Then, the mirror formula reduces to a simple form called the Newton’s formula.
x0 xi = f2
Where x0 is the object distance from the focus.
xi is the image distance from the focus.
_____________________ Slide 75 _____________________ Illustration
A short linear object is placed at a distance ‘u’ along the axis of a spherical mirror of focal length f.
(i) Obtain an expression for the longitudinal magnification. (ii) Also, obtain an expression for the ratio of the velocity of
image (v) to the velocity of object (u).
_____________________ Slide 76 _____________________ Illustration
A small strip of plane mirror A is set with its plane normal to the principal axis of a convex mirror B and placed 15 cm in front of B which it partly covers. An object is placed 30 cm from A and the two virtual images formed by reflection in A and B coincide without parallax. Find the radius of curvature of B.
_____________________ Slide 77 _____________________ Illustration
A concave mirror of focal length 15 cm) and a convex mirror (focal length 10 cm) are placed co-axially 70 cm apart facing each other. A 2 cm tall object is placed perpendicular to the common axis 20 cm from the concave mirror. Find the position, size and nature of the final image formed by two reflections, first at concave mirror and then at convex mirror.
Slide 78 Introduction:
Refraction through plane surface refraction of light
§ Refraction is a phenomena of light due to which it bends while travelling from one medium to another.
§ When light goes from one medium to another medium. ü Its velocity changes
ü Its wavelength changes
ü Its path may or may not change. ü Its frequency remains unchanged.
_____________________ Slide 79 _____________________ § During, the propagation of light from one medium to another
medium changes its velocity and wavelength and the path also changes (except ∠i = 0).
§ When light goes from an optically rarer to an optically dense medium, it bends towards normal provided ∠i ≠ 0 as shown in the fig.
_____________________ Slide 80 _____________________ When light goes from an optically dense to an optically rarer medium it bends away from normal provided.
Slide 81 _____________________ Slide 82 _____________________ _____________________ Slide 83 _____________________ Slide 84 _____________________ Slide 85 _____________________ Refractive index
§ The degree of bending of light depends upon the medium quality called refraction Index (µ), a pure number.
§ Refractive index (µ) is of two types 1. Relative Refractive Index (µ) 2. Absolute Refractive Index
_____________________ Slide 86 _____________________ § The absolute refractive index (µ) of a medium is defined as
the ratio of the speed of light in the vacuum (c) to the speed of light in the medium (Vm).
m c V
µ =
_____________________ Slide 87 _____________________ Relative refractive index
§ The relative refractive index of two media is equal to the ratio of their absolute refractive indices.
2 2 1 21 1 1 2 c v v c v v µ µ = = = µ
§ This is read as refractive index of 2nd medium with respect 1st medium and written as 1µ2.
§ The refractive indices of glass and water are 3
(
1.5)
2 = and(
)
4 1.33 3 = . _____________________ Slide 88 _____________________ Note :§ Absolute Refractive index µ is the refractive index of a medium with respect to air or vacuum, i.e.
air m medium c V λ µ = =
Slide 89 Refraction of Light (Snell's Law)
§ As light travels from one medium to another, its frequency
does not change.
_____________________ Slide 90 _____________________ ü Both the wave speed and the wavelength do change. ü The ratio of the indices of refraction of the two media
can be expressed as various ratios
1 2 1 1 2 1 2 2 sin n v 1n2 sin n v θ λ = = = = θ λ _____________________ Slide 91 _____________________ Refraction of Light Critical Angle
§ When light attempts to move from a medium having a given index of refraction to one having a lower index of refraction.
Slide 92
§ As the incidence angle (θ1) is increased until a particular
angle (θc), as shown in figure, the angle of refraction will be
90º and the refracted ray would skim the surface of the glass.
§ The incidence angle at which this occurs is called the critical angle, denoted as
_____________________ Slide 93 _____________________ Total internal reflection
“Total internal reflection is an effect occurs only when light
attempts to move from a medium of given index of refraction to a medium of lower index of refraction and the incidence angle is equal or greater than the critical angle of the higher refractive index medium .”
_____________________ Slide 94 _____________________ Total internal reflection
§ WWeeccaannuusseeSSnneellll’’ssllaawwooffrreeffrraaccttiioonnttooffiinnddtthheeccrriittiiccaallaannggllee. . § WWhheennθ = θ1 c, θ =2 90ºaannddSSnneellll’’ssllaawwggiivveess O 1 c 2 2 1 2 2 c c 1 1 n sin n sin90 n n n sin sin n n − θ = = θ = ⇒ θ = F Foorr nn11>>nn22
Slide 95 Example
Calculate the critical angle of a diamond (n = 2.42)
c D c c 1 sin n 1 sin 2.42 24.4º θ = θ = θ = _____________________ Slide 96 _____________________ _____________________ Slide 97 _____________________ Fiber Optics
Application of total internal reflection
§ Such a “light pipe” is flexible if thin fibers are used rather than thick rods. If a bundle of parallel fibers is used to construct an optical transmission line, images can be transferred from one point to another.
Slide 98
§ Light, signals or other forms of communication can travel a long distance without losing much intensity.
§ Applications include medical use of fiber optic cables for diagnosis and correction of medical problems Telecommunications
_____________________ Slide 99 _____________________ Application of total internal reflection
§ TToottaalliinntteerrnnaallrreefflleeccttiioonnoofflliigghhttbbyypprriissmmssiinnbbiinnooccuullaarrss. .
_____________________ Slide 100 ____________________ Application of total internal reflection
_____________________ Slide 101 ____________________ Diamonds achieve their brilliance from a combination of dispersion and total internal reflection. Because diamonds have a very high index of refraction of about 2.4, the critical angle for total internal reflections only 25 degree. Incident light therefore strikes many of the internal surfaces before it strikes one at less than 25 degree and emerges. After many such reflections, the light has traveled far enough that the
colors have become sufficiently separated to be seen individually and brilliantly by the eye after leaving the crystal.
Slide 102 Some Indices of Refraction
_____________________ Slide 103 ____________________ Snell’s law
§ Incident ray, Refracted ray and Normal lie in the same plane.
§ When light propagates through a series of layers of different medium as shown in the figure, then according to Snell’s law we may write,
µ1 sin θ1 = µ2 sin θ2 = µ3 sin θ3 = … = constant
_____________________ Slide 104 ____________________ § In general, µ sin θ = constant.
i.e. for given any two media where light ray strikes the interface
µ1 × sin i = µ2 × sin r
_____________________ Slide 105 ____________________ Conclusion from Snell’s law
§ When light passes from rarer to denser medium, it bends
towards the normal as shown in the figure.
§ Using Snell’s Law at point P
µ1 sin θ1 = µ2 sin θ2 1 2 1 sin or, 1 sin 2 θ =µ > θ µ
§ Thus, if µ2 > µ1 the θ2 < θ1. i.e light ray bends towards the
normal.
_____________________ Slide 106 ____________________ § When a light ray passes from denser to rarer medium it
bends away from the normal as shown in the figure. § From Snell’s Law, we know that
1 2 1 sin 1P sin 2 θ =µ > θ µ
§ Thus, if µ2 < µ1, θ2 > θ.i.e. refracted ray bends away form the
normal.
_____________________ Slide 107 ____________________ Refraction: Principle of Least
1 2 1 2 1 2 1 1 2 2 Time, s s t t t v v n s n s c v , t n c c = + = + = = + 2 2 1/2 2 2 1/2 1 1 2 2 1 dt t = [n(y + x ) + n ( y + ( a - x ) ) ] = 0 c dx
Prove Snell’s law, n1sin?1= n2sin ?2. Demonstrate law of
reflection.
_____________________ Slide 108 ____________________ Relation between object and image distance
§ An object O placed in medium 1 (refractive index µ1) is
viewed from the medium 2 (refractive index µ2).
§ It is important to note that the object and image both are formed on the same side of the boundary.
§ The image distance y and the object distance x are related as 2 y x 1 µ = µ
Slide 109
Here, x = Real depth/height y = Apparent depth/height
_____________________ Slide 110 ____________________ § If µ2 > µ1, that is, when the object is observed from a denser
medium, it appears to be farther away from the interface, i.e., y > x. or, Apparent height > Real height.
§ If µ2 < µ1, that is, when the object is observed from a rarer
medium, it appears to be closer to the interface, i.e. y < x. i.e. Apparent depth < Real depth.
Note:
§ The above formula is only applicable for normal view or
paraxial ray assumption.
_____________________ Slide 111 ____________________ Total Internal Reflection
§ The phenomenon of total internal reflection occurs when light travels from a medium of high refractive index to a medium of lower refractive index.
§ At the critical angle (θc), the refracted ray just grazes the
boundary between two media.
_____________________ Slide 112 ____________________ § Using Snell’s law, we get
§ µ1 sin θc = µ2 sin 90°or , sin−1 2 1 µ θ = µ § Here, θ1 < θc < θ2 Slide 113
§ For an angle of incidence greater than θc, the light is totally
reflected back into the medium of higher refractive index.
This phenomenon is called total internal reflection.
_____________________ Slide 114 ____________________ Optical fibers:
§ The optical fibres can transmit light beam from one end to the other due to the repeated total internal reflections even if the fiber is bent or twisted.
_____________________ Slide 115 ____________________ Mirage:
§ The optical illusion that water is present at some distance place is called inferior mirage.
§ This generally occurs on very hot summer days . This is due to total internal reflection.
Looming:
§ The optical illusion of object floating in air is called superior mirage. It is also known as looming.
§ This occurs in very cold regions due to total internal reflection.
_____________________ Slide 116 ____________________ Total Internal Reflection; Fiber Optics
Slide 117
_____________________ Slide 118 ____________________
_____________________ Slide 119 ____________________ Deviation ( δ )
§ The figure shows a light ray traveling from a denser to a rarer medium at an angle α lesser than the critical angle θc.
§ The deviation δ of the light ray is given by
_____________________ Slide 120 ____________________ Conceptual Example
View up from under water
Describe what a person would see who looked up at the world from beneath the perfectly smooth surface of a lake or swimming pool.
_____________________ Slide 121 ____________________
Slide 122 Refraction through glass slab
§ The refracting surfaces of a glass slab are parallel to each other. When a light ray travels through a glass slab, it is refracted twice at the two parallel faces and finally emerges out parallel to its incident direction.
§ The light ray undergoes zero deviation, δ = O. § Angle of emergence = Angle of incidence
Or, e = i
_____________________ Slide 123 ____________________ § The lateral displacement of the ray is the perpendicular
distance between the incident and the emergent ray and is given by
(
)
sin = − t d i r cos r _____________________ Slide 124 ____________________ _____________________ Slide 125 ____________________ Apparent shiftWhen a glass slab of thickness t and refractive index µ is placed in the path of a convergent beam as shown in the figure, then the point of convergence is shifted by
1 1 = − µ s t _____________________ Slide 126 ____________________ § When the same glass slab is placed in the path of a
diverging beam, the point of divergence is 1 1 = − µ s t
§ It is important to note that the shift (s) is always on the direction of light.
_____________________ Slide 127 ____________________ § If there are n number of slabs with different refractive indices
are placed between the observer and the object, then the total apparent shift is equal to the summation of the individual shifts. s = s1 + s2 + ……… + sn 1 2 n 1 2 n 1 1 1 or, s = t 1- + t 1- +...t 1-µ µ µ _____________________ Slide 128 ____________________ Illustration
In the figure, determine the apparent shift in the position of the coin. Also, find the effective refractive index of the combination of the glass and water slab.
Slide 129 Prism
§ Prism is a transparent medium with two refracting surfaces through which refraction takes place.
§ The two faces are not parallel but are inclined to each other. § Any geometrical figure with two refracting surfaces are not
parallel in a prism.
§ The above figure is an example of prism. § In general, we take a triangular prism.
_____________________ Slide 130 ____________________ Visible Spectrum and Dispersion
_____________________ Slide 131 ____________________ n Dependence on Wavelength
Slide 132 _____________________ Slide 133 ____________________ Dispersion _____________________ Slide 134 ____________________ _____________________ Slide 135 ____________________ Slide 136 Newton’s Analysis of Light
1 2 2 1 n n λ = λ _____________________ Slide 137 ____________________ Basic Terms
Angle of prism or reflecting angle (A)
§ The angle between the refracting faces is calledangle of prism.
§ The adjoining figures shows various terms related to prism. _____________________ Slide 138 ____________________ Here, i = angle of incidence
r1 = angle of refraction through 1st surface.
r2 = angle of refraction through 2nd surface.
e = angle of emergence
δ = angle of deviation
_____________________ Slide 139 ____________________ Angle of deviation (δ)
§ It is the angle between the emergent and the incident ray. § It other words , it is the angle through which incident ray
turns in passing through a prism. § Now, the angle of deviation is given by
δ
= (i – r1) + (e + r2)or,
δ
= i + e – (r1 + r2)Also, ∠A = r1 + r2
Slide 140 Condition of minimum deviation
§ At the minimum deviation the angle of incidence is equal to the angle of emergence i.e., i = e
Also, r1 = r2 = r A = 2r
§ The refracted ray inside the prism becomes parallel to the base of the prism.
§ The deviation of the prism is given by
δ
= ( i + e – A). or,δ
m = ( 2i– A).§ Where δm is the minimum deviation produced.
m A i 2 δ + = _____________________ Slide 141 ____________________ § Using Snell’s law
min A sin sin i 2 A sin r sin 2 δ + µ = = _____________________ Slide 142 ____________________ Graphical representation of minimum deviation
§ The graph shows the minimum deviation produced by the prism.
_____________________ Slide 143 ____________________ Thin prisms
§ In thin prisms, the distance between the refracting surfaces is negligible and the angle of prism (A) is very small.
§ For thin prism, δm is very small. i.e.,
m m
A A A A
sin( ) and sin
2 2 2 2
+ δ ; + δ ;
_____________________ Slide 144 ____________________ § Now, the refractive index for thin prism is given by
min A sin 2 A sin 2 δ + µ = min A 2 A 2 δ + = min or, δ = µ −( 1)A. _____________________ Slide 145 ____________________ Illustration
A rectangular block of refractive index µ is placed on a printed page lying on a horizontal surface as shown in the figure. Find the minimum value of µ so that the letter L on the page is not visible from any of the vertical sides.
_____________________ Slide 146 ____________________ Illustration
A ray of light PQ is incident at an angle i on face ML of a prism and is refracted along OR (figure). This ray, after refraction at face MN, travels along RN at grazing emergence.
If µ is the refractive index and A refracting angle of the prism, show that 1 2 2 sini cosA 1 sinA + µ = + _____________________ Slide 147 ____________________ Dispersion
§ It explains the blue colour of the sky at day time.
§ When while light passes through a prism it is spilled into its seven constituent colours which is known as dispersion of white light.
§ It is because for different constituent colours of light there are different values of Refractive index (µ).
Slide 148
§ We get a continuous bends of seven colours on screen, which is called spectrum of light as shown in the fig.
_____________________ Slide 149 ____________________ Scattering of light
§ When sunlight travels through the earth’s atmosphere, it gets scattered by the atmospheric particles.
§ The amount of scattering is given by Rayleigh scattering. _____________________ Slide 150 ____________________ Rayleigh scattering
§ The amount of scattering S is inversely proportional to the fourth power of the wavelength (λ).
4 1 S=
λ
§ The law is applicable only for small particles which have relative size with the wavelength of the light.
_____________________ Slide 151 ____________________ Spherical surface
§ A spherical refracting surface is a part of a sphere of refracting material.
§ It is of two types. Convex refracting surface
§ A refracting surface which is convex towards the medium where the object is present.
_____________________ Slide 152 ____________________ Refraction at spherical surface:
Assumptions
§ Object is a point object lying on the principle axis.
§ The incident and the refracted rays make small angles with the principal axis.
§ The aperture is small.
Slide 153
Mathematical expression of Refraction at spherical surface § Two media of refractive index µ1 and µ2, when separated by
a transparent curved surface, can be regarded as the case of refraction at spherical surface.
Refraction at a spherical surface separating two media _____________________ Slide 154 ____________________ § The figure shows how light refracts at the interface of two
curved media.
§ C is the centre of curvature of medium ‘2’. § CN is the normal to the curved surface and § O is the point where the object lies.
§ After refraction, let the image be formed at I.
§ Let ‘u’ be the object distance, and ’v’ be the image distance.
_____________________ Slide 155 ____________________ § From figure, for small angles, tanq q;
MN
tan NOM NOM
OM Ð = ;Ð MN and tan NCM NCM MC Ð = ;Ð MN
tan NIM NIM
MI
Ð = ;Ð
§ From plane geometry
( )
MN MNi i
OM MC
= + − − − (exterior angle = sum of interior
Slide 156
§ Similarly, r MN–MN
( )
ii MC MI= − − −
(Q r = ∠ NCM – ∠ NIC) § Now from Snell’s law, µ1 sin i = µ2 sin r and for small angles
sinθ≈θ. § Hence, µ1 i = µ2 r
_____________________ Slide 157 ____________________ § From (i) and (ii)
( )
1 2 2– 1 iii
OM MI MC
µ +µ =µ µ − − −
§ By using sign convention, OM = – u, MI = + v, MC= + R § Hence, from eqn. (iii)
2 – 1 2 – 1
v u R
µ µ =µ µ
_____________________ Slide 158 ____________________ Important terms
§ The medium in which the incident ray travels is called the 1st medium (µ1).
§ The medium in which the refracted ray travels is called the 2nd medium (µ2).
§ The above equation relates object and image distance in terms of refractive index and radius of curvature.
§ This formula is applicable for both convex and concave surfaces with proper sign conventions.
_____________________ Slide 159 ____________________ Solved example
A convex refracting surface of radius of curvature 15 cm separates two media of refractive indices 4/3 and 1.5. An object is kept in the first medium at a distance of 240 cm from the refracting surface. Calculate the position of the image.
_____________________ Slide 160 ____________________ Solution
As the object is in the rarer medium (i.e. the incident ray travels in the rarer medium), we have
1 2 2 – 1 –u v R µ +µ =µ µ Here, u = – 240 cm: v = ? ; R = + 15 cm, µ1 = 4/3 ; µ2 = 1.5 Slide 161 ∴ 4 / 3 1.5 1.5– 4 / 3
(
)
240+ v = 15 ∴ v = 270 cmAs v is positive, the image is formed in the second medium at a distance of 270 cm from the refracting surface.
The image is real.
_____________________ Slide 162 ____________________ Illustration
An empty spherical flask of diameter 30 cm is Placed in water of refractive index. A parallel beam of light strikes the flask. Where does it get focused, when observed from within the flask?
_____________________ Slide 163 ____________________ Illustration
What curvature must be given to the bounding surface of a refracting medium (µ = 1.5) for the virtual image of an object in the adjacent medium
(µ = 1) at 10 cm to be formed at a distance of 40 cm?
_____________________ Slide 164 ____________________ Refraction by a lens
Lens
§ A combination of two refracting surfaces, at least one of which is curved, is called a lens.
§ There are two types of lens ü Concave lens
ü Convex lens.
_____________________ Slide 165 ____________________ Concave lens
§ Concave lens is a lens which is thinner at middle and wider at its ends and diverge light rays.
§ Concave lens is of three types
_____________________ Slide 166 ____________________ § A concave lens can be somewhat supposed to be made up
of two prisms placed on each other, vertex to vertex as shown in figure
§ Concave lens is also called a diverging lens because it diverges parallel beam of light after refraction through it.
_____________________ Slide 167 ____________________ Convex lens
§ Convex lens is a lens which is thicker at middle and thinner at its ends and converges light rays.
§ Convex lens is of three types
_____________________ Slide 168 ____________________ § A convex lens is supposed to be made up of two prisms
placed on each other, base in to base contact, as shown in fig.
§ Convex lens is also called converging lens because it converges parallel beam of light after refraction through it as shown in fig.
Note
In general, focal length of a concave lens is taken negative and that of convex lens is taken positive.
Slide 169
_____________________ Slide 170 ____________________
_____________________ Slide 171 ____________________
Slide 173
_____________________ Slide 174 ____________________ Lens maker’s formula
_____________________ Slide 175 ____________________
_____________________ Slide 176 ____________________
Slide 177
§ The geometry of image formation by a double convex lens. § The first refracting surface forms the image I1 of the object O.
§ The image I1 acts as a virtual object for the second surface
that forms the image at I.
_____________________ Slide 178 ____________________ Applying to the first interface ABC, we get 1 2 2 1
1 1
n n n – n OB+BI = BC § To the second interface ADC,
2 1 2 1
1 2
n n n – n –
DI + =DI DC § For a thin lens, BI1 = DI1.
_____________________ Slide 179 ____________________ § Adding and, we get
(
)
1 1 2 1 1 1 2 n n 1 1 n – n OB DI BC DC æ ö÷ ç ÷ + = çççè + ÷÷ ø§ Suppose the object is at infinity, i.e,, OB →∞ and DI1 = f,
gives
§ By the sign convention, BC1 = + R1, DC2 = – R2
§ So, it can be written as
2 21 21 1 2 1 n 1 1 1 (n –1) – n f R R n æ ö æ÷ ö÷ ç ÷ ç ÷ = ççç ÷÷ ççç = ÷÷ è ø èQ ø
Where n21 is the refraction index of the medium 2 w.r.t
medium 1.
§ Equation is known as the lens maker’s formula.
_____________________ Slide 180 ____________________ Important Points
§ Len’s maker’s formula is used to find the lens of desired focal length.
§ While using this formula we should use proper sign conventions.
§ A lens has two focii, two radius of curvature, one optical centre (same function as that of pole in mirror)
§ The focus on the side of the original source of light is called the 1st focal point, while the other is called 2nd focal point. _____________________ Slide 181 ____________________ Relation between object distance (u), Image distance (v) and focal length, (f) of a thin lens
§ The formula for object distance(u), Image distance (v) and focal length (f) of a lens is given by
1 1 1 – v u= f
§ This formula is applicable for both convex and concave lenses with proper sign convention.
Slide 182 Rules for image formation through a lens
§ To find out the position of image through a lens we should find out the point of intersection of refracted rays.
§ A ray parallel to principal axis of a lens after refraction passes through or appears to diverge from the focus of lens. § A ray through optical centre of lens passes undedicated
along its original path after refraction through lens.
§ A ray through focus of lens becomes parallel to principal axis, after passing through the lens.
_____________________ Slide 183 ____________________ Image formation by a concave lens
_____________________ Slide 184 ____________________
_____________________ Slide 185 ____________________ Magnification
§ Magnification (m) produced by a lens is defined as the ratio of the size of the image to that of the object.
§ It is denoted by m.
§ m is negative for real image which are inverted w.r.t. to object.
§ m is positive for virtual image which are erect w.r.t. object. § Mathematically, magnification is given by
v f – v f 1 1 1 m using, – u f f u v u f = = = = + _____________________ Slide 186 ____________________ Solved Example
A convex lens is to be used to throw on a screen 10 cm from the lens, a magnified image of an object. If the magnification is to be 19, find the focal length of the lens.
_____________________ Slide 187 ____________________ Solution
Given: magnification, m = −19, Image distance, v = 10 cm Now, magnification is given by
f – v m , f f – 1 0 or, –19 f or, f 0.5 cm = = = _____________________ Slide 188 ____________________ Power of lens
§ The degree of converging or diverging of light rays through a lens is called its power.
§ The power of a lens is the reciprocal of the focal length of the lens.
where f is measured in metre.
_____________________ Slide 189 ____________________ § The power of a lens is the tangent of the angle by which it
converges or diverges a beam of light falling at unit distance from optical centre.
§ The SI unit of power of lens is called Dioptre (D).
§ The power of convex lens is positive and that of concave is negative.
_____________________ Slide 190 ____________________ Combination of thin lenses in contact
Image formation by a combination of two thin lenses in contact
Slide 191
§ Consider two thin convex lenses A and B having focal length f1 and f2 are placed in contact with each other as shown in fig.
§ Image formed by the 1st lens ‘A’. is given by
( )
1 1
1 1 1
– i
v u =f − − −
§ For image formed by second lens B, we get
( )
1 2 1 1 1 – = - - - i i v v f _____________________ Slide 192 ____________________ § Adding (i) and (ii) we get1 2 1 1 1 1
–
v u= f +f
§ If the two lens system is regarded as a single lens, then the equivalent focal length fe is related as
e 1 1 1 – v u=f § Hence, e 1 2 1 1 1 f = +f f _____________________ Slide 193 ____________________ § If several thin lenses are placed in contact then the effective
focal length of their combination is given by
e 1 2 3
1 1 1 1
f = + + + − − − − −f f f § In terms of power, we can write
P = P1 + P2 + P3 +- - -
_____________________ Slide 194 ____________________ § The total magnification m of the combination is a product of
magnification of individual lenses m = m1 m2 m3 - - -
§ Such a system of combination of lenses is used in designing lenses for cameras, microscopes, telescopes and other optical instruments.
_____________________ Slide 195 ____________________ The rainbow
§ It is a beautiful patterns of colours seen in the sky after a shower
§ The rainbow is an example of the dispersion of sunlight by the water drops in the atmosphere.
§ The conditions for observing a rainbow are that the sun should be shining in one part of the sky while it is raining in the opposite part of the sky.
Slide 196 Primary rainbow
§ The primary rainbow has violet colour on the inner edge and the red colour on the outer edge of the rainbow.
§ Primary rainbow is formed due to two refractions and one total internal reflection of the light incident on the droplets. _____________________ Slide 197 ____________________ Secondary rainbow
§ It has red colour on the inner edge and violet colour on the outer edge of the rainbow.
§ It is formed due to refraction and two total internal reflections of light incident on the droplets.
§ Order of colour of the secondary rainbow is just reverse of that of the primary rainbow.
_____________________ Slide 198 ____________________ _____________________ Slide 199 ____________________ Scattering of light § § WWhheennssuunnlliigghhttttrraavveellss tthhrroouugghhtthheeeeaarrtthh’’ss aattmmoosspphheerree,,iittss d diirreeccttiioonncchhaannggeessbbyytthhee aattmmoosspphheerriiccppaarrttiicclleess aanndd tthhiissiiss c caalllleeddssccaatttteerriinnggoofflliigghhtt.. § § TThhee aammoouunntt ooffssccaatttteerriinnggiiss iinnvveerrsseellyypprrooppoorrttiioonnaall ttoo tthhee f foouurrtthhppoowweerroofftthheewwaavveelleennggtthh,,ccaalllleeddRRaayylleeiigghhSSccaatttteerriinngg.. Slide 200
§ Blue light being shorter wavelength, the sky look blue during daylight due to scattering.
§ Red colour being least scattered light reaching our eyes, therefore, the sun looks reddish at sunrise and at sunset.
Slide 201 Human eye
_____________________ Slide 202 ____________________ How the Eye Works:
_____________________ Slide 203 ____________________
Slide 204 Vision Defects
Nearsightedness:
(a) Light focuses before the retina Cannot see distant objects Gets worse as the body grows
(b) Fix: concave lens which will focus the light back on the retina
_____________________ Slide 205 ____________________ Vision Defects (cont)
Farsightedness:
(c) Light focuses behind the retina Cannot see close objects Gets worse as the body ages (40+): lens is less flexible (d) Fix: convex lens which will focus the light back on the retina
_____________________ Slide 206 ____________________ Contact Lenses
§ Rests on a layer of tears between it and the cornea § Produces the same result as eyeglasses
§ Most refraction occurs at air-lens surface where change in refractive index is greatest.
_____________________ Slide 207 ____________________ § The light from an object enters the eye through the
transparent cornea.
§ It passes through a transparent lens held in place by ciliary muscles and is focused on the retina.
§ The retina is sensitive to light and sends messages to the brain by way of optic nerve.
§ An iris in front of the lens changes in size to regulate the amount of light that enters the eye.
_____________________ Slide 208 ____________________ § It may be noted that the image of the object formed on the
retina is real and inverted w.r.t. object.
§ However, our brain interprets it as an erect image w.r.t. object.
§ The human eye is about 2.5 cm in diameter and its neat spherical shape is maintained by the pressure of the fluid within it.
§ A normal eye can focus on object located anywhere from about 25 cm to hundreds of kilometers away.
_____________________ Slide 209 ____________________ Power of accommodation of the eye
§ The ability of the eye to focus on objects at different distances is called power of accommodation of the eye. Far point of the eye
§ The most distant point that the eye can see clearly is called the far point of the eye.
_____________________ Slide 210 ____________________ Near point of the eye
§ The closest point at which an object is seen most clearly without strain is called the near point of the eye/ least distance of distinct vision (D)
§ For an adult with normal eye, this distance is taken to be 25 cm (by convention).
Slide 211 Defects of vision
Short sightedness (or myopia)
§ A person who can see the near objects clearly but cannot focus on distant objects is short sighted.
§ This defect occurs if a person’s eyeball is larger than the usual diameter.
§ In such a case, the image of a distant object is formed in front of the retina as shown in fig.
_____________________ Slide 212 ____________________ Correction
To correct short-sighted vision, a diverging lens (concave lens) of suitable focal length is placed in front of the eye.
_____________________ Slide 213 ____________________ Long sightedness (or hypermetropia)
§ A person who can see distant objects clearly but cannot focus on near objects clearly is farsighted.
§ This defect may occur if the diameter of person’s eyeball is smaller than usual.
§ In such a case, for an object placed at the normal near point (i.e., 25 cm from eye), the image of the object is formed behind the retina as shown in fig.
Slide 214 Correction
To correct for long sighted vision a converging lens of suitable focal length is placed in front of the eye.
_____________________ Slide 215 ____________________ Presbyopia
§ It is the defect in which a person can see nearby objects due to the increase of the near point of the eye.
§ In this, the near point increases to 200 m.
§ It is caused due to the decreasing effectiveness of the ciliary muscle and the loss of flexibility of the lens.
§ It is corrected by using a converging lens for reading. _____________________ Slide 216 ____________________ Astigmatism
§ It is a defect in which eye lens looks at a wire eye mesh or a grid of lines, focusing in either the vertical or the horizontal plane which may not be as sharp as in the other plane. § ‘It is caused due to the non-uniformity in the spherical shape
of the cornea.
§ It can be corrected using a cylindrical lens of desired radius of curvature with an appropriately directed axis.
Note
§ Astigmatism can occur along with myopia or hypermetropia. _____________________ Slide 217 ____________________ Solved Example
Far point of a short-sighted person is 100 cm. What lens should he use to see distant objects clearly?
Slide 218 Solution
Since the far point of the short-sighted person is 100 cm away, he can see objects situated up to a distance of 100 cm. so, the lens used should be such that it forms the image (virtual) of the distant object at a distance of 100 cm.
∴ u = – ∞, v = – 100 cm _____________________ Slide 219 ____________________ Now, 1 1 – 1 f=v u 1 1 1 – f –100 – ∴ = ∞ 1 1 or, f –100 or, f –100 cm = =
So, a concave lens of 100 cm focal length should be used. _____________________ Slide 220 ____________________ Illustration 1
Where an object should be placed from a converging lens of focal length 20 cm so to obtain a real image of magnification 2?
_____________________ Slide 221 ____________________ Photometry
§ Photometry is the science of measurement of light in terms of its perceived brightness to the human eye.
Important terms Luminous flux
§ It is the amount of light energy emitted per second by a source.
§ The unit of luminous flux is lumen
§ One lumen is the luminous flux emitted per unit solid angle by a uniform point source of luminous intensity 1 candela. _____________________ Slide 222 ____________________ Luminous intensity
§ Luminous intensity of source in a given direction is defined as the luminous flux per unit solid angle in that direction. § The unit of luminous intensity is candela (cd).
§ Candela is defined as the luminous intensity, in a given direction, of a source that emits monochromatic radiation of frequency 5.40 × 1014 Hz and that has a radiant intens ity in the direction of watt per steradian.
_____________________ Slide 223 ____________________ Illuminance or Intensity of illumination
§ Illuminance or Intensity of illumination of a surface at any point may be defined as the luminous flux incident normally on a unit area of the surface held at that point.
§ It is generally denoted by E.
§ If luminous flux ∆φ falls normally on an area ∆A of a surface, then illuminance is given by
_____________________ Slide 224 ____________________ § The unit of illuminance is lux.
§ Lux or metre candle is the illuminance produced at the inner surface of a sphere of one metre radius when a source of one candela is placed at its centre.
§ Radiant flux is measured in watt.
_____________________ Slide 225 ____________________ Luminous efficiency
§ The luminous efficiency of a light source is the ratio of the luminous flux emitted and the input electrical power.
§ The unit of luminous efficiency is lumen/watt.
_____________________ Slide 226 ____________________ Simple Microscope
It is an optical instrument used to see small and minute particles.
Slide 227 Magnifying power of a simple microscope
The angular magnification or magnifying power of a simple microscope is the ratio of the angle subtended at the eye by the image at the near point and the angle subtended at the unaided eye by the object at the near point.
_____________________ Slide 228 ____________________ § The linear magnification is given by m v
u = 1 or, m v u 1 1 or, m v – v f = = v or, m 1– f =
§ Here, in accordance with sign Convention v = − D[ far point]
D
m 1
f
∴ = +
_____________________ Slide 229 ____________________ Magnification when the image is at infinity
§ Consider an object having height h when it is at the near point D of the eye.
0 h tan D ∴ θ = § Since θ0 is small
( )
0 h ... 1 D;q _____________________ Slide 230 ____________________§ Now, magnification when the object is at infinity.
( )
I 0 I 0 h v m h u v or, h h .... 2 u = = =Where hI = image distance, h0 = object distance.
_____________________ Slide 231 ____________________ § The angle subtended by the image
I i h tan – v q =
( )
N 0 h v . using eq 2 – v u é ù = êë\ úû0 h – u = i 0 i i i forsmall –h tan u é\ q ù ê ú \ q ê q q ú ë û ; ;
§ The angle subtended by the object, when u = −f is given
by 0
i h
f ; q
_____________________ Slide 232 ____________________ § Now, the angular magnification is given by
i 0 0 0 m h D f h D m f q =q = ´ \ = _____________________ Slide 233 ____________________ § So, smaller the focal length of the lens, greater will be the
magnifying power.
§ The simple microscope may be used in such a way that the image is formed at infinity.
Note
§ The maximum angular magnification is obtained when the image is the near point.
§ The minimum angular magnification is obtained when the image is at infinity.
_____________________ Slide 234 ____________________ Light and Telescopes
Slide 235 Telescopes § Light collectors § Two types: ü Reflectors mirrors) ü Refractors (Lenses) _____________________ Slide 236 ____________________ Yerkes 40” Refractor- The World’s Largest
Slide 237 Reflecting Telescopes
_____________________ Slide 238 ____________________ Reflection
§ Light that hits a mirror is reflected at the same angle it was incident from
§ Proper design of a mirror (the shape of a parabola) can focus all rays incident on the mirror to a single place
_____________________ Slide 239 ____________________ Whoops - The Hubble needs a contact lens
§ Both types of telescope can suffer from a defect called spherical aberration so that not all of the light is focused to the same point.
§ This can happen if the mirror is not curved enough (shaped like part of a sphere instead of a paraboloid) or the glass lens is not shaped correctly.
Slide 240
§ The Hubble Space Telescope objective suffers from this (it is too flat by 2 microns, about 1/50 the width of a human hair) so it uses corrective optics to compensate.
§ The corrective optics intercept the light beams from the secondary mirror before they reach the cameras and spectrographs. Fortunately, the Hubble Space Telescope's spherical aberration is so perfect, that it is easy to correct for!
_____________________ Slide 241 ____________________
Before Costar After Costar _____________________ Slide 242 ____________________ Famous Telescopes: Galileo
§ First telescope: 3x magnification § Last one: 32 x
_____________________ Slide 243 ____________________ Famous Telescopes: Newton
§ First reflector ever § Built around 1670
Slide 244 Famous Telescopes: Lord Ross
_____________________ Slide 245 ____________________ Famous Telescopes: Mt Palomar
5 Meter Telescope – Huge and heavy mirror _____________________ Slide 246 ____________________ Famous Telescopes: Arecibo Radio Telescope
_____________________ Slide 247 ____________________
Slide 248
_____________________ Slide 249 ____________________ Telescope Size
§ A larger telescope gathers more light (more collecting area). § Angular resolution is limited by diffraction of light waves; this
also improves with larger telescope size
_____________________ Slide 250 ____________________
_____________________ Slide 251 ____________________ Largest Earth-Based Telescopes
§ Hobby-Eberly Telescope, Davis Mountains, TX ü 11 m diameter
ü Cannot see all parts of the sky § Keck I and II, Mauna Kea, HI
ü 36 ×1.8 m hexagonal mirrors; equivalent to 10 m ü Above most of atmosphere (almost 14,000 ft ASL) ü Operating since 1993
_____________________ Slide 252 ____________________ Resolving Power of Telescopes
Andromeda Galaxy
Telescope 1 Telescope 2 of double size _____________________ Slide 253 ____________________ Resolving Power of Telescopes (II)
Andromeda Galaxy Resolution: (a) 10’ (b) 1’ (c) 5” (d) 1” _____________________ Slide 254 ____________________ _____________________ Slide 255 ____________________ Not to be outdone, the European Union has proposed building the 100 meter OWL (Overwhelmingly Large Telescope)
_____________________ Slide 256 ____________________ Compound Microscope
§ A compound microscope is one which has much larger magnification to see very small objects.
Slide 257
Magnifying power of a compound microscope
_____________________ Slide 258 ____________________ § Angular magnification or magnifying power of a compound
microscope is defined as the ratio of the angle subtended by the final image at the eye to the angle subtended by the object seen directly, when both are placed at the least distance of distinct vision.
_____________________ Slide 259 ____________________ § The magnification produced by the compound microscope is
the product of the magnifications produced by the eyepiece and objective.
M = Me × Mo
Where Me and Mo are the magnifying powers of the
eyepiece and objective respectively. § Now, e e D M 1 f = +
Where fe is the focal length of the eyepiece.
_____________________ Slide 260 ____________________ § Also, o o o v M u
= , Where vo is the distance A’B’ from the
objective and uo is the distance of the object from the
objective.
§ Now, the magnification is given by o
o e v D M 1 u f = + _____________________ Slide 261 ____________________ § Since the object is placed very close to the principle focus of
the objective therefore uo is nearly equal to fe. i.e. u0≈ f0
§ vo is nearly equal to the length L of the microscope tube. i.e.
v0≈ L
§ L is the separation between the two lenses.
o e L D M – 1 f f = +
§ It is clear from the above equation that the smaller the focal length of the objective and eyepiece, larger is the magnifying power. _____________________ Slide 262 ____________________ Again, o o o 1 1 1 – v u = f or , o o o o o o v v v – v u = f o o o o o o o o v v or, – –1 u f v v or, 1 – u f = + = o o e v D M 1– 1 f f ∴ = + _____________________ Slide 263 ____________________ § When the final image is formed at infinity, then
o e
o e L D M = M × M = –
f ×f
§ In this case, the microscope is said to be in normal adjustment.
_____________________ Slide 264 ____________________ Historical Optical Microscopes
Slide 266
_____________________ Slide 267 ____________________
_____________________ Slide 268 ____________________ Current Optical Microscopes
Upright Inverted
Slide 269 Solved Example
A person with a normal near point (25 cm) using compound microscope with an objective of focal length 8.0 mm and an eyepiece of focal length of focal length 2.5 cm can bring an object placed 9.0 mm from the objective in sharp focus. What is the separation between the two lenses? What is the magnifying power of the microscope?
_____________________ Slide 270 ____________________ Solution
Given: D = 25 cm; fo = 0.80 cm ; u0 = – 0.9 cm
Objective:
Using lens formula for the objective, we have,
0 0 0 1 1 1 – v u = f 0 1 1 1 or v +0.9=0.8 ∴ v0 = 7.2 cm Eyepiece:
Using lens formula for the eyepiece, we have, e e e 1 1 1 – v u = f _____________________ Slide 271 ____________________
(
e)
e e 1 1 1 or – – v D – 2 5 c m D u = f Q = = e 1 1 1 or – – 25 u =2.5 ∴ ue = – 2.27 cmSeparation between the two lenses = v0 + |ue| = 7.2 + 2.27 = 9.
47 cm
Magnifying power of microscope, M = M0 × Me 0 0 e v D 7.2 25 1 1 88 u f 0.9 2.5 = + = + = _____________________ Slide 272 ____________________ Solved Example
The total magnification produced by a compound microscope is 20. The magnification produced by the eyepiece alone is 5. The microscope is focused on a certain object. The distance between the objective and eyepiece is observed to be 14 cm. If least distance of distinct vision is 20 cm, calculate the focal lengths of the objective and eyepiece.
