KPK Physics Class 10 Notes Chapter 12 Geometrical Optics Conceptual Questions Comprehensive Questions, and Numerical Questions, the best notes geometrical optics questions and answers of 2021.
Table of Contents
Physics Class 10 Notes Chapter 12 Geometrical Optics Conceptual Questions
Q.1) Which type of lens would you use to start fire from light from sun concave or convex, would work best? At what distance from the lens should the paper be held for best results?
Answer: A convex lens would be the best option to start a fire from sunlight. The reason is that a convex lens converging the incoming rays of light to a focal point. The rays of light are concentrated to a point (focal point), as a result, the intensity of light at that point becomes greater. So we place the paper at its focal point to start the fire.
Q.2) If a concave mirror produces a real image, is the image necessarily inverted? Explain.
Answer: Yes, if a concave mirror produces a real image, it necessarily has to be inverted. The incident light on the mirror reflects back with the same angle. Due to the shape of the concave mirror, it bounces back at the centre and below the principal axis, hence the image formed is inverted.
Q.3) Are rearview mirrors used in cars concave or convex?
Answer: Convex mirrors are used as rearview mirrors. The reasons are as follows:
Image is magnified and upright, so the orientation is the same as the object.
Produces a reduced size image, due to which it can cover a large area.
Q.4) A magician during a show makes a glass lens with n = 1.47 disappear in a trough of liquid. What is the refractive index of the liquid? Could the liquid be water?
Answer: The magician put the glass in a liquid having the refractive index equal to that of the glass, i.e. 1.47. The phenomenon behind this trick lies in refraction. As the refractive indices of both medium (glass and liquid) are same, no refraction or reflection occurs. Hence, as a result, no light is bend or reflected from the glass, so it looks disappeared.
The liquid the magician used was either turpentine oil or glycerine.
Q.5) Suppose that you were handed a lens and a ruler and told to determine the focal length of the lens. How would you proceed?
Answer: Following are the steps to find the focal length of a lens:
Grab the lens from its sides and stand near a wall. Try to focus the background image to that wall.
At first, it would view diminished or unclear to view. A point comes when the background image on the wall becomes clear and visible.
At this point grab the ruler and measure the distance from the centre of the lens and the wall. This measured value is the focal length of the lens.
Q.6) Can we achieve total internal reflection from optically rare medium to optically dense medium?
Answer: No, for total internal reflection to occur it is compulsory for a light wave to travel from a denser medium to a rarer medium. When light travels from a denser medium to a rarer medium, it bends away from the normal and above critical angles it reflects instead of refracting. As a result, total internal reflection occurs.
If the light wave enters from the rarer medium to the denser medium it bends towards the normal. And it can not be reflected back to the same medium. So no total internal reflection phenomenon takes place.
Q.7) Will a nearsighted person who wears corrective lenses in her glasses be able to see clearly underwater when wearing those glasses?
Answer: No, the glasses that she wears were prescribed to use in the “air” medium. The refractive index of water is different as compared to the air. So the light will diverge differently underwater as compared to in the air, and it will not focus accurately on the cornea. Hence, she will not be able to see clearly underwater.
Q.8) When you use a simple magnifying glass, does it matter whether you hold the object to be examined closer to the lens than its focal length or farther away? Explain.
Answer: Yes, it depends on the position of the object to be placed. A magnifying glass uses a convex lens. To view an object magnified, it is necessary to place it in between the focal point and the lens. In this case, an upright, magnified and virtual image is formed. Whereas, in any other case, an inverted image will be formed which is not required result.
Q.9) In blind turns on hilly roads, mirrors are used to help drivers. Are these mirrors plane mirrors, concave mirrors or convex mirrors? Explain.
Answer: Convex mirrors are used on the blind turns on hilly roads. They are used to diverge the incoming light waves which give a larger field view. Due to this phenomenon, we are able to view cars from the other blind end and so as they can see us too. These mirrors are used to avoid accidents due to ill visibility on the blind turns.
Comprehensive Questions Physics 10 Class Notes Chapter 12
Q.1) What is meant by reflection of light? State and explain laws of reflection with diagrams.
Answer: Reflection of Light: Consider a beam of light travelling through a homogeneous medium in a straight line falls on the boundary of another medium. When the beam falls then a part of this beam will bounce back by striking the surface of that other medium. This phenomenon of bouncing back of the light is known as a reflection of light. Amount of reflected light depends on the nature and smoothness of the surface of the medium from which light gets reflected. Some mediums reflect almost 100 percent of the falling light such as mirrors. These materials must have a very smooth polished surface.
REad more: Biology Class 10 Notes Chapter 17 Biotechnology Short Questions Laws of Reflection of Light: There are two laws used to explain reflection of light and are known as laws of reflection of light. These laws are given as follows: 1. First law of Reflection of Light: This law states that: “The incident ray, reflected ray and the normal to the point of incidence are all lie in the same plane” This can be seen in the figure below where ray if incidence, ray of reflection and normal lie in the same plane.
2. Second Law of Reflection of Light: The second law of reflection of light states that: “The angle of incidence is always equal to the angle of reflection” This can be understood from the figure above and is mathematically expressed as,
Q.2) Derive spherical mirror formula.
Answer: Spherical Mirror Formula: The formula which gives a relation between image distance, object distance and focal length of a mirror is known as mirror formula or mirror equation. In order to derive this equation, consider an object ‘AB’ is placed between the principle focus ‘F’ and centre of curvature ‘C’ as shown in the figure.
Two rays ‘AP’ and ‘AE’ are incident to the mirror. Obeying the law of reflection after striking the mirror, ray A’P will reflect with the same angle along PA. From the figure above, considering two triangles A’PB’ and APB, we can deduce following relation,
As AB = h = h0, A‘B‘ = h’ = hi, PB = p PB‘ = q So we can also write it as,
Similarly, the ray AE which is passing through the focal point F will become parallel to the principal axis PB. From the figure we can see that triangles ABF and FPE are identical, therefore we can write
As, EP = A’B’ = hi, FP = f and BF = p – f, therefore we can write,
Comparing equation (1) and (2), we get
By cross multiplication, we can write pf = q (p – f) = qp – qf Dividing both sides by ‘pqf’, we get
This equation is known as mirror equation or spherical mirror formula. This equation gives us the relation between image distance, object distance and focal length of a spherical mirror.
Q.3) What is meant by refraction of light? What is the index of refraction?
Answer: Refraction of Light: “The change in velocity and direction of light when it passes from one medium to another is called refraction of light” It is a property of light that it travels in a straight line but that can only be true if the light is travelling only in one homogenous medium. However, if the light has to pass obliquely from one medium to another medium then the light rays will change their direction at the junction of these two mediums. For example, when light that is travelling in air enters to the glass then it will change its path. This can be understood from the figure (a).
In this diagram, PQ is the surface that separates two mediums from each other. The medium that is above from that surface is air and below is glass. Consider a ray CO is falling at a point O of the surface PQ. When this ray will enter to the second medium i.e. glass then it will change its path. Normally it will have to travel along OD but after entering it will move along OE. This OE ray is known as refracted ray. Now if we draw a normal NON’ then angle CON will be the angle of incident and angle EON’ will be the angle of refraction and we can see that refracted angle s a little bend towards normal i.e. angle of refraction is smaller than the angle of incidence.
But we consider the case where the light ray is emerging from glass and entering to air then it will bend away from the normal as shown in figure (b). So, in general, we can say that when light enters from a rear medium to a denser medium then it will bend towards the normal but if it enters from denser medium to a rear medium then it will bend away from the normal.
index of refraction: “The ratio of the speed of light in vacuum divided by the speed of light in the material is called the refractive index of the material”
It is given by: n = c/v
Q.4) State and explain laws of refraction with diagrams.
Answer: Laws of Refraction: Refraction of light obeys two laws, known as laws of refraction.
First Law of Refraction: The first law of refraction states that, “The incident ray, refracted ray and normal at the point of the incident, all lie in the same plane” This can be understood from the figure (c), where incident ray AO, refracted ray OB and normal at the point of the incident, all lie in the same plane.
Second Law of Refraction: The second law of refraction gives us a relation between the angle of incident and angle of refraction and this law is also known as Snell’s law as Snell was the first scientist who develops this relationship between them. This law states that: “The ratio of the sine of the angle of incidence and sine of the angle of refraction is constant for a given pair of media” Its mathematical expression is as follows:
Q.5) What is total internal reflection? How we can calculate the critical angle for total internal reflection? What are the conditions for total internal reflection?
Answer: Total Internal Reflection: We know that when light enters from an optical denser medium to an optical rare medium then it will bend away from the normal. And in this process, the angle of refraction is always greater than the angle of incidence. It means that if we increase the angle of incidence slightly then the angle of reflection will also increase. And if we keep on increasing the angle of incidence then at a certain angle of incidence, the angle of reflection will become 90o. In a denser medium, when the angle of incidence becomes greater than critical angle then there will be no refracted beam at all. Therefore, are rays will reflect internally back in the optical denser medium. Such a reflection is known as total internal reflection. For total internal reflection, the following two conditions must be satisfied: i) The incident light must pass from a denser medium into a rarer medium. ii) The angle of incidence in the denser medium must be greater than the critical angle of the medium.
Critical Angle: That angle of incidence, for which the angle of reflection becomes 90o is known as the critical angle. It is denoted by ‘C’ and for glass, the value of the critical angle is 42o. The sin of the critical angle of a medium has an inverse relation with the refractive index of that medium. We know that the refractive index is the ratio of the sine of an incident angle to the sine of refracted angle and if the incident angle is 90o then the refracted angle will be the critical angle. Consider a light ray emerging from a denser medium and entering to a rare medium say from glass to air as shown in the figure.
At a critical angle C, the refractive index is given by,
As, Sin 90 = 1, therefore
Q.6) What are optical fibres? Give some applications of optical fibers.
Answer: Optical fibres: ” Light can be trapped by total internal reflection inside a bent glass rod and ‘piped’ along a curved path called optical fibre “. Various glasses and plastics can be used to make optical fibres. Optical fibre transmits a beam of light by means of total internal reflection. Total internal reflection can occur in any transparent medium that has a higher index of refraction than the surrounding medium. An optical fibre cable has a cylindrical shape and consists of three concentric sections: the core, the cladding and the jacket. The core is the innermost section and consists of one or more very thin strands, made of glass or plastic. Each strand is surrounded by its own cladding, a glass or plastic coating that has optical properties different from the core.
Applications of optical fibres: 1. Surgical techniques have been revolutionized by the use of optical fibres. In arthroscopic surgery, a small surgical instrument (several millimetres in diameter) is mounted at the end of an optical fibre cable which allows light to be shone on the internal area and the reflected light to be viewed by the surgeon. The surgeon can insert the instrument and cable into a joint, such as the knee, with only a tiny incision, and minimal damage to the surrounding tissues. 2. An endoscopy is a procedure where the inside of our body is examined using an instrument called an endoscope. An endoscope is a long, thin, flexible tube has a light source and camera at one end. Images of the inside of our body are displayed on the computer monitor. An endoscope can be inserted into the body through a natural opening, such as the mouth and down the throat, or through the bottom. An endoscope can also be inserted through a small cut (incision) made in the skin when keyhole surgery is being carried out. An endoscope can also be used by engineers to light up some awkward spot for inspection.
Q.7) Describe the behavior for a ray of light after passing through a prism.
Answer: 1. Deviation of light through Prism: In the prism, the incident light ray travels through the air and enters the left side of the glass. The light bends toward the normal in the glass, because glass has a higher index of refraction (optical density) than the air. When the light leaves the glass and emerges into the air on the other side of the prism, the light is refracted away from the normal. The change in direction of the light as it passes through the glass is known as its deviation.
2. Dispersion of light through Prism: When sunlight (white light) falls on a triangular prism a band of colours called a spectrum is obtained as shown in the figure below. The effect is termed as dispersion. It arises because white light is a mixture of many colours, the prism separates the colours because the refractive index of glass is different for each colour (it is greatest for violet light).
Q.9) Describe power of lens and its resolving power. What are its units?
Answer: Power of lens and its Unit: The degree of convergence or divergence of light rays falling on lens is called power of lens. Instrument makers often quote the power of a lens rather than its focal length. The Power of a lens D, in diopters, a dimensionless number, is given by the equation D = 1/f where f is the focal length of the lens expressed in metres. Eyeglass lenses are typically characterized in terms of diopters. The power of a lens in diopters should not be confused with the familiar concept of power in watts. It is an unfortunate fact that the word “power” is used for two completely different concepts. If you examine a prescription for eyeglasses, you will note lens powers given in diopters. If you examine the label on a motor, you will note energy consumption rate given as a power in watts. Resolving power: Resolving power is the capacity of an instrument to distinctively separate two points which are close together. The sharpness of vision-in particular, the ability to visually separate closely spaced objects is referred to as resolution.
Q.10) How the human eye works? How the defects in the eye like short-sightedness and long-sightedness be corrected by using lenses.
Answer: Human Eye: The human eye is like a camera in its basic structure but is more complex. The interior of the eye is filled with a transparent gel-like substance called the vitreous humor with index of refraction n = 1.337. Light enters in this region through the cornea and lens. Between the cornea and lens is a watery fluid the aqueous humor with n = 1.336. The iris adjusts automatically to control the amount of light entering the eye, similar to a camera. The retina which plays the role of the film or sensor in a camera, is on the curved back surface of the eye. The retina is composed of a many nerves which act to change light energy into electrical signals that travel along the nerves to brain for interpretation. The lens of the eye ( n = 1.386 to 1.406) does little of the bending of the light rays. Most of the refraction is done at the front surface of the cornea at its interface with air. The lens acts as a fine adjustment for focusing at different distances. This is accomplished by the ciliary muscles, which change the curvature of the lens so that its focal length is changed. To focus on a distant object, the ciliary muscles of the eye are relaxed and the lens is thin , as shown in figure below and parallel rays focus at the focal point (on the retina). To focus on a nearby object, the muscles contract, causing the centre of the lens to thicken, see figure below thus shortening the focal length so that images of nearby objects can be focused on the retina, behind the new focal point. This focusing adjustment is called accommodation.
Short or nearsightedness: Some people cannot see distant objects clearly without the aid of spectacles. This defect of vision is known as short sight or nearsightedness. Correctness: The near-sighted eye can be corrected with glass or contact lenses that use diverging lenses (concave lens) Light rays from the distant objects are now diverged by this lens before entering the eye. To the observer, these light rays appear to come from far point and are therefore focused on the retina, thus forming a sharp image.
Long or Farsightedness: The disability of the eye to form distinct images of nearby objects on its retina is known as Long or farsightedness. Correctness: This defect can be corrected with the aid of suitable converging lens (convex lens). The lens refracts the light rays and they converge to form an image on the retina. To an observer, these rays appear to come from near point to form a sharp virtual image on the retina. When we show up to the present moment with all of our senses, we invite the world to fill us with joy. The pains of the past are behind us. The future has yet to unfold. But the now is full of beauty simply waiting for our attention.