REFRACTION OF LIGHT & OPTICAL INSTRUMENTS Chapter 14.

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CHAPTER - 10 LIGHT : REFLECTION AND REFRACTION
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CHAPTER - 10 LIGHT : REFLECTION AND REFRACTION
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REFRACTION OF LIGHT & OPTICAL INSTRUMENTS Chapter 14

SNELL’S LAW According to Snell’s law "The ratio of the sine of the angle of incidence to the sine of the angle of refraction is always constant. " Mathematically, Sine <i/sine <r = constant or sin< i/sine< r = mew where mew = Refractive index of the material of medium.

TOTAL INTERNAL REFLECTION When light rays enter from one medium to the other, they are refracted. If we increase the angle of incidence, angle of refraction will also increase. At certain angle of incidence light rays are reflected back to the first medium instead of refraction. This condition or phenomenon is called Total Internal Reflection.

CRITICAL ANGLE The angle of incidence at which the angle of refraction will become 90 o is called Critical Angle. If angle of incidence further increased then instead of refraction, reflection will take place.

DEFECTS OF VISION There are four common defects of vision: 1. SHORT SIGHTEDNESS OR MYOPIA 2. LONG SIGHTEDNESS OR HYPER METROPIA 3. ASTIGMATISM 4. PRESBYOPIA

SHORT SIGHTEDNESS OR MYOPIA SYMPTOMS In Myopia, a person can not see distant objects clearly, but he can see clearly the objects near to him. REASON The reason for Myopia is either the focal length of lens of eye is too short or the eyeball is very much elongated. WHAT HAPPENS IN MYOPIA In Myopia, light rays from a distant object are focused in front of the Retina.

CORRECTION OF DEFECT This defect can be corrected by using a concave lens of suitable focal length

ASTIGMATISM - PRESBYOPIA ASTIGMATISM If the cornea or the surface of eye is not perfectly spherical. In this situation the eye has different focal points in different planes and an object is not focused clearly on the retina. CORRECTION OF DEFECT ASTIGMATISM is corrected by using asymmetrical lenses which have different radii of curvature in different planes PRESBYOPIA or lack of accommodating At old age, the eye lens loses its natural elasticity and ability to change its shape and the ciliary muscles weaken resulting in a lack of accommodation. This type of long sightedness is called "PRESBYOPIA". CORRECTION OF DEFECT This defect can be corrected by using convex lens for long sighted person and concave lens for short sighted person.

LONG SIGHTEDNESS OR HYPER METROPIA SYMPTOMS In HYPER METROPIA, a person can not see objects clearly which are near to him, but he can see clearly distant objects REASON The reason for HYPER METROPIA is that either the focal length of the lens of eye is too long or the eyeball is too short. WHAT HAPPENS IN HYPER METROPIA In HYPER METROPIA, light rays from a near object are focused behind the Retina.

CORRECTION OF DEFECT This defect can be corrected by using a convex lens of suitable focal length

POWER OF LENS Power of lens is defined as the reciprocal of the focal length of the lens in meters. FORMULA: Power = 1/f(in meter) Unit of power of lens is Dioptre. DIOPTRE Dioptre is defined as the power of lens whose focal length is one meter if f =1 meter then the power of the lens = 1 dioptre.

Image Formation by convex lens POSITION OF OBJECT When the object is placed at infinity NATURE AND POSITION OF IMAGE 1. The image will form at the principal focus (F). 2. The image will be real and inverted. 3. The image will be very small in size.

POSITION OF OBJECT When the object is placed beyond 2F Nature and position of image 1. The image will form between F and 2F. 2. The image will be real and inverted. 3. The image will be smaller in size.

POSITION OF OBJECT When the object is placed at 2F NATURE AND POSITION OF IMAGE 1. The image will form at 2F. 2. The image will be real and inverted. 3. The image will be equal in the size of object.

POSITION OF OBJECT When the object is placed between F and 2F NATURE AND POSITION OF IMAGE 1. The image will form beyond 2F. 2. The image will be real and inverted. 3. The image will be magnified.

POSITION OF OBJECT When the object is placed at F NATURE AND POSITION OF IMAGE 1. The image will form at infinity. 2. The image will be real and inverted. 3. The image will be highly magnified.

POSITION OF OBJECT When the object is placed between the pole (P) and F NATURE AND POSITION OF IMAGE 1. The image will form on the same side of object. 2. The image will be virtual and erect. 3. The image will be magnified.

ASTRONOMICAL TELESCOPE Introduction It is an optical instrument used to view heavenly bodies such as moon,stars, planets and distant object. Construction Astronomical telescope consists of two convex lenses: 1:Objective 2:Eye piece Objective The objective is a convex lens of large focal length and large aperture. It usually made of two convex lenses in contact with each other to reduce the chromatic and spherical aberrations.

Eye piece The eye piece is also a convex lens.Its focal length is smaller than that of objective. It is also a combination of two lenses. The objective is mounted on a wide metallic tube while the eye piece is mounted on a small tube.The distance b/w the eye piece and the objective can be changed by moving tubes.

WORKING The rays coming from a distant object falls on objective as parallel beam at some angle say "a" and these rays after refraction and passing through the objective converge at its focus and make an inverted & real image AB. This image acts as an object for the eye piece. The distance of the eye piece is so adjusted that the image AB lies within the focal length of the eye piece. The eye piece forms the final image.The final image is magnified,virtual and inverted with respect to object. The final image is formed at infinity.

WORKING

MAGNIFYING POWER The magnifying power (M) of astronomical telescope is given by: It is because the object is at infinite distance and hence the angle subtended by the object at eye may be taken as the angle subtended by the object at objective. M =  since  and are small angles, therefore we can take:  = tan  and  = tan 

MAGNIFYING POWER In right angled triangles DAOB & DAEB This expression shows that in order to obtain high magnification, focal length of object must be large and that of eye piece is small.

LENGTH OF TELESCOPE The distance b/w objective lens and the eye piece is equal to the length of the telescope. From figure: OE = length of telescope =L But OE = OB + BE OB = Fo & BE = Fe OR L = focal length of objective + focal length of eye piece

THIN LENS FORMULA FOR CONCAVE LENS FOR CONCAVE LENS Consider an object placed in front of a concave lens of focal length " f " on the principle axis of the lens. Concave lens forms a virtual and erect image at a distance of " q " from the optical centre of the lens as shown in the diagram below.

FORMULA FOR CONCAVE LENS Consider similar triangles and Similarly in triangles and

FORMULA FOR CONCAVE LENS Comparing equation (1) and (2) p (f - q) = fq pf - pq = fq Dividing both sides by "pqf" 1/f - 1/p = 1/q

COMPOUND MICROSCOPE Compound microscope is an optical instrument which is used to obtain high magnification. Construction It consists of two converging lenses: Objective Eye piece Objective The lens in front of object is called objective. Its focal length f 1 = f o is taken to be very small.The objective forms a real, inverted, and magnified image of the object placed just beyond the focus of objective. Eye piece The lens towards the observer's eye is called piece.Focal length of eye piece is greater than the focal length of objective. Eye piece works as a magnifying glass.

Working The objective is so adjusted that the object is very closed to its focus. The objective forms a real, inverted and magnified image of the abject beyond 2fo on the right hand side. The eye piece is so adjusted that it forms a virtual image at the least distance of distinct vision "d".The final image is highly magnified.

Magnifying power In order to determine the magnifying power of a compound microscope,we consider an object oo' placed in front of objective at a distance p 1. Objective forms an inverted image II' at a distance of q 1 from objective. Magnification produced by :M o = size of image / size of object M o = q 1 / p (1) Eye piece works as a magnifying glass. It further magnifies the first image formed by objective. Magnification produced by the eye piece is given by: M e = size of image / size of object M e = q 2 / p 2 the objective is given by:

Magnifying power We know that the eye piece behaves as a magnifying glass therefore the final image will be formed at least distance of distinct vision i.e at 25 cm from the eye. Hence q 2 = d M e = d / p (2) Using thin lens formula for eye piece :

Magnifying power 1/f 2 = 1/q 2 + 1/p 2 Here f 2 = f e, q 2 = - d and p = p 2 1/f e = 1/-d + 1/p 2 1/f e = -1/d + 1/p 2 Multiplying both sides by "d" d/f e = -d/d + d/p 2 d/f e = -1 + d/p d/f e = d/p 2 d/p 2 = 1 + d/f e (3) Comparing equation (2) and (3) M e = 1 + d/f e (4)

Magnifying Power Total magnification is equal to the product of the magnification produced by the objective and the eye piece. M =M o X M e M = (q 1 /p 1 )(1 + d/f e ) In order to get maximum magnification, we must decrease p 1 and increase q 1.Thus maximum possible value of p 1 is fo i.e p = fo and maximum possible value of q 1 is the length of microscope i.e q 1 = L Therefore the magnification produced by a compound d microscope is given by: M = (L/fO)(1 + d/f e )