Rules of refraction: Why the rays bend and when they converge? Just the case of a lens made up of glass (denser than air). In this case i > r. Rule 1 : When a light ray travels from a rarer medium to a denser medium, the light ray bend towards the normal.

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But if there’s a lens made up of thin films containing gas of rarer density than air ? Rule 2 : When a light ray travels from a denser medium to a rarer medium, the light ray bends away from the normal. The same case will happen when a glass lens will be placed in a denser medium

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Laws of refraction: The incident ray, the refracted ray and the normal are co-planar.
A ray of light traveling along the normal will not get refracted and will pass un-deviated.
The ratio of the sine of the angle of incidence to that of the sine of the angle of refraction is constant for a given medium. This constant is called the refractive index . This is called the Snell’s Law. Thus Snell’s Law gives-

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Types of convex lenses:- See each convex lens is thicker in the middle.

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Centre of curvature (C1 and C2 ):
The centre of curvature of a lens is defined as the centre of the spherical surfaces from which the lens has been cut. Some definitions regarding convex lens:- Principal axis :
The line joining the centre of curvatures is known as the principal axis of the lens. A ray of light (2 in right fig ) passing along the principle axis will not be refracted.

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Aperture :
The maximum portion of the spherical surfaces from which the lens actions take place is called the aperture of the lens. In the figure, the distance AB is the aperture of the lens. Optical centre :
The intersection of the line joining the lens aperture and the principal axis is called the optical centre of the lens. Any ray of light passing through the optical centre emerges parallel to the direction of the incident ray and for a thin lens the ray become undeviated.The optic centre is denoted by O.

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Focal point:
A convex lens focuses light rays coming from infinite distance (parallel rays)at its focal point. Focal plane :
The plane constructed by all the focal points Second principal focus:
If the incident rays are parallel to the principle axis the focal point will be called second principal focus(f2). Focal length(f2):
The distance between optical centre and principal focus.

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The first principal focus :
Since a lens consists of two surfaces, there are thus first and second focus and first and second focal plane.
On the contrary of the second principal focus the first principal focus f1 of a convex lens is the point on the principal axis from which when the light rays start, on passing through the lens will become parallel. First principal
focus Power of a lens:Power of a lens is the inverse of its focal length measured in meters. The letter P denotes power of lens Standard unit of P is dioptre and is denoted by D.Dioptre=1, means a lens whose focal length is 1 meter. The power of a convex lens is + ve .

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Lensmaker's equation:
f =Focal length of the lens,
n =Refractive index of the lens material,
R1=Radius of curvature of the lens surface closest to the light source,
R2 =Radius of curvature of the lens surface farthest from the light source,
d =Thickness of the lens. ( for thin lens d=0 and ) How the focus of a convex lens depends on its structure & material?

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This can also be put into the newtonian form Where X1=object distance from focus=S1-f and X2=S2-f Relation between object distance, image distance
and focal length:- The general relation for all lens v =image distance
u =object distance Each distance is: Distance from optical centre
Towards source +ve
-ve in opposite direction So for a real image by convex lens as in figure ; the relation become: U=S1 f= -f V=S2 See notes for proof

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Images by convex lenses:
CASE 1 : WHEN THE OBJECT IS PLACED BETWEEN THE OPTICAL CENTRE O AND THE FOCUS F1. Consider two rays of light starting from A.One of them, parallel to the principal axis XY will strike the lens at D. This ray of light on refraction will pass through F2 to become ray DZ. Let the object be a candle AB placed between F1 and O. Explanation: The second light ray from A passing through AO, will go un-deviated and will get refracted along OW.

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S1<f,
S2=-ve Magnification = The rays DZ and OW reach the eye are intersecting at a point A’.Just like A’ rays of light from B extending behind the lens, form an image B’. The refracted image is A’B’ behind the lens. Placed behind the lens and hence is virtual
Erect and not inverted
Magnified Image characteristics: See notes for proof

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CASE 2 : IMAGE WHEN THE OBJECT IS PLACED AT FOCUS F1: Consider two rays of light from point A. AD is parallel to the principal axis XY so passing through F2 being refracted ,become DZ
The second one passing through O become AW.
The two refracted rays are parallel to each other. Thus the image of point A will be at infinity. Image characteristics: Explanation:
Let the candle AB is placed in first focus F1. The image is formed at infinity
The image is inverted and real

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CASE 3 : IMAGE FORMED BY A CONVEX LENS WHEN THE OBJECT IS PLACED BETWEEN THE FOCUS F1 AND 2 F1. Consider two rays of light emanating from point A. AD is parallel to XY so being refracted goes through F2 .
The second ray of light AO cuts the principle axis XY at O so become undeviated.. The image is formed beyond 2 F1
The image is inverted and real
The image is magnified. Image characteristics : Let the candle AB placed between F1 and 2 F1. Explanation: DF2 and AO meet at A’, which is the refracted image of A. Light from B will go un-deflected along XY path. The image B’.

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CASE 4 : IMAGE WHEN THE OBJECT IS PLACED AT 2 F1. Two rays emanating from A; one AD is parallel to XY so passes through F2.
The second ray is passing through O, and remain undeviated.
Df2 and AO meet at A’, which is the refracted Forms at 2 F2 on other side of the lens
Inverted and real
Same in size as that of the object. Let the object AB placed at 2 F1. Image characteristics : Explanation: image of A.Ray from B goes undeflected along XY path forms B’.

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CASE 5 : IMAGE FORMED BY A CONVEX LENS WHEN THE OBJECT IS PLACED AT A DISTANCE BEYOND 2 F1. The image is inverted and real
The image is reduced in size.
. Let the candle AB placed at a distance beyond 2 F1. consider two rays from A; one is AD parallel to XY so passes through F2.
The second ray passes through O and remain undeviated.
DF2 and AO meet at A’, which is the refracted image Image characteristics : Explanation: of A. Light from B will go un-deflected along XY path & forms B’

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CASE 6 : IMAGE FORMED WHEN THE OBJECT IS PLACED AT A DISTANCE FAR OFF (OR INFINITE DISTANCE) BEYOND 2 F1. Rays emanating from A are striking the lens in parallel fashion. They refract and meet at A’ straight from F2
The image is thus highly diminished.
Rays from B go undeflected and forms image A’B’. The image is formed at F2
The image is inverted and real
The image is highly reduced in size Let the object i.e. Candle AB placed at infinity. B is still on line XY. Image characteristics : Explanation:

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Example:
An object 5 cm high(h0) is placed 35 cm(p) in front of a converging lens of 10 cm focal length(f). A) where is the image located(q)? B) what is the magnification(m)? C) what is the size of the image(hi)?
Solution:
A) let the image is real,so according to the formula Q=14 cm. (So we have assumed properly that it’s a real image )
B) m=v/u=-q/p=-14 cm/35 cm=-0.4. Since M is negative, the image is real and inverted.
C) m=h0/hi Or, hi=h0m=(5 cm)(-0.4)=-2 cm. The image is smaller than the object and inverted.

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Combinations of Lenses:-
Most optical systems consist of combination of two or more lenses. Fig. shows the combination of two convex lenses separated by a distance of d. The image distance of the first and second lens is found from equation and

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The magnification of the lens combination:- The magnification of the first lens is M1=hi1/h01 (h01=hi1/M1) and from the second lens M2=hi2/h02=hi2/hi1 (hi2=hi1M2). Substituting the equations in brackets into equation , we get M= hi1M2/( hi1/M1) or M=M1M2 Ratio of the height of the final image to the height of the initial object. Equivalent focal length of two lenses placed co-axially:- CASE 1. PLACED CLOSELY CASE2.PLACED IN DISTANCE D

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Example:-
A combination of lenses. An object 5 cm high (h01) is placed 15 cm (=p1) in front of a convex lens of 10 cm (f1)focal length. A second convex lens, also of 10 cm focal length (=f2) is placed 48 cm (=d) behind the first lens. Find (a) the image of the lens combination, (b) the magnification of the system, and (c) the height of the final image. Solution:
(A) with equation we get Q1=30 cm.The object distance for the second lens p2=d-q1=48 cm-30cm =18 cm. Hence, the final image Q2=22.5 cm Thus, the final image of the lens combination is 22.5 cm behind the second lens, or 70.5 cm behind the first lens.
(B) the magnification of each lens is found as m1=-q1/p1=-30 cm/15cm=-2 and m2=-q2/p2=-22.5 cm/18 cm=-1.25.
The final magnification is M=M1M2=(-2.00 )(-1.25)=2.50.
(C) the final height of the image is hi2=h01m=(5.00 cm)(2.50)=12.5 cm.

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When there is a variation of the refractive indexes as well as materials in the lens: Dose a lens always give the perfect image ? No ,generally there are two types of problems occur often.

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When spherical surfaces are not in the ideal shape: This problem may happen if it is shaped incorrectly while making it.
Long use of a lens may change its shape and this problem may appear. Solution:
Aberrations can be compensated for to a great extent by using a combination of simple lenses with complementary aberrations.

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Use of a convex lens in human eye:- In time of normal vision we see through a convex lens which is working in our eye from the very beginning of our life in this earth. Convex lens

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When the internal lens become poor:-

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Use of another convex lens to rectify our vision:-

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How the total image forms? Retina Convex
Lens Light

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By: sohasaeed (40 month(s) ago)

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