# NCERT Solutions Class 10 Science Chapter 10

## NCERT Solutions for Class 10 Science Chapter 10: Light Reflection and Refraction

Class 10 Science Chapter 10-Light Reflection and Refraction is one of the key chapters for the students to learn light phenomena such as reflection and refraction. Concepts like refraction and reflection using straight-line propagation of light, a reflection of light by spherical mirrors, mirror formula, magnification, etc., are illustrated with detailed explanations and written with easy-to-understand language  NCERT Solutions for Class 10 Science Chapter 10 provided by Extramarks. These solutions help students build a strong base in the subject.

The latest term-wise CBSE Syllabus 2021-2022 ensures that the chapters are suitable for the students to progress in their respective streams.  NCERT Solutions for Class 10 Science Chapter 10-Light Reflection and Refraction will help students to solve their queries and doubts if any. Solutions cover various concepts of the chapter and are considered ideal study material. . No wonder students have complete faith and trust in Extramarks. Register at Extramarks’ website for more details on NCERT Solutions Class 10 Science Chapter 10-Light Reflection and Refraction.

### Key Topics Covered In NCERT Solutions for Class 10 Science Chapter 10

Below we have given brief notes about the key topics from the Class 10 Science Light Reflection and Refraction Chapter. Students are recommended to register on Extramarks’ website to get access to our NCERT Solutions for Class 10 Science Chapter 10 and explore a repository of educational resources available at Extramarks.

Reflection of light:

Reflection occurs when a light beam hits a smooth polished surface and bounces back. The incident ray is that ray that hits the surface, whereas the reflected ray is the ray that reflects from the surface. Also, a vertical line between these two rays is known as Normal.

The two most important laws of reflection are as follows:

1. The angle of incidence(i) = The angle of reflection(r).
2. The incident, regular, and reflected ray must lie in the same plane. These laws help us determine the type of reflection that hits a plane surface, mirror, metal surface, water, etc.

The three types of reflection are:

1. Regular reflection: A plane mirror or glass is used, which is heavily coated with a uniform layer of highly reflective material like powder, making the surface reflect all the rays falling on it with minimal variation and no obscurity.
2. Diffused Reflection: The reflective surfaces used for diffused reflection are comparatively rough, with scratches, dust or dents. The material used is other than glass and thus affecting the quality and brightness of the reflection. As the incident ray falls on different points and gets reflected in a different direction, the incident angle no longer stays equal to the angle of reflection.
3. Multiple Reflection: When a reflected ray of light reflects back and forth many times on being incident on another surface, it is said to have multiple reflections. This type of reflection is possible with two mirrors where the light intensity is very low. Infinite images will generate multiple reflections based on the angle between two mirrors. The more we decrease the angle or curve, the more images are formed. Each image is the result of another image. When the angle between the two mirrors is zero, i.e., they are parallel, the number of images becomes infinite.

Calculation of the number of images for diffused light reflection using the following formula, the number of images= (360 degree/angle between mirrors)-1.

Spherical Mirror: A Spherical mirror has the shape of a piece carved out of a sphere.

The two types of Spherical mirrors are:

1. Concave mirrors: Reflecting surface of the concave mirror is curved inwards to one focal light, just as compared to the inside of the spoon. They are used as reflectors, converging lights, solar cookers, motor vehicle headlights, and telescopes.
2. Convex mirrors: Reflecting surface of the convex mirror is curved outwards, just as compared to the outside of a balloon. The bulging surface reflects the light rays outwards and makes the virtual image the center of curvature(2F) and focal point(F). These are the imaginary points in the mirror that cannot be arrived at. They are often used in stores, schools, driveways, roads, automated teller machines, and passenger side mirrors on a car.

Important definitions are given below:

• Scattering of light: It is the circumstance in which the light rays deviate from their straight path and hit an obstacle like gas molecules, dust, water droplets, etc., resulting in the scattering of light. This diffusion of light is affected by the factors like size of the particle and the wavelength of light.
• Example of Reflections of light in Daily life: One of the fascinating examples of reflection of light in our daily life is the sunset appears red, as the sky has scattered most blue light while traveling around the atmosphere during the day. As a result, the red light dominates during the sunset.
• Refraction of light: It is the most usually observed phenomenon in which the change in speed of a light ray results in a shift in its direction. Optical equipment such as magnifying glasses, lenses, and prisms are the best examples. The law of refraction describes that the incident ray, the refracted ray, and the normal interface all lie in the same plane.
• Mirror formula and magnification: The mirror formula is 1/f=1/v+1/u. It gives the relationship between the distance of the object(u), the distance of the image(v), and the focal length(f).

Magnification Is derived from the size of the image formed by the mirror. It can be represented by:

In two ways of ratio, i.e., m=h/h, where h is the height of the image and h is the object’s height,  m=-v/u, where v is image distance, and u is object distance.

It can be defined as the ability of the lens to bend the light. The measuring unit for the power of the lens is dioptres, and the formula is P=1/f.

NCERT Solutions for Class 10 Science Chapter 10-Light Reflection and Refraction provided by Extramarks gives a thorough explanation of the above phenomenons, in simple and easy language. We recommend students visit the Extramarks’ website and access this study resource. This will undoubtedly assist them in achieving excellent results in the examinations.

### NCERT Solutions for Class 10 Science Chapter 10: Exercises &  Solutions

NCERT Solutions for Class 10 Science Chapter 10 provided by Extramarks, covers all crucial topics commonly asked in the board examinations. Our NCERT Solutions also contains a variety of questions such as MCQs, short and long answer questions, and much more. Students may refer to and practice different types of questions consolidated in our NCERT Solutions for Light Reflection and Refraction by clicking on the link given below.

 NCERT SOLUTIONS FOR CLASS 10 SCIENCE Chapter 1 Chemical Reactions and Equations Chapter 2 Acids, Bases and Salts Chapter 3 Metals and Non-metals Chapter 4 Carbon and Its Compounds Chapter 5 Periodic Classification of Elements Chapter 6 Life Processes Chapter 7 Control and Coordination Chapter 8 How do Organisms Reproduce? Chapter 9 Heredity and Evolution Chapter 10 Light Reflection and Refraction Chapter 11 Human Eye and Colourful World Chapter 12 Electricity Chapter 13 Magnetic Effects of Electric Current Chapter 14 Sources of Energy Chapter 15 Our Environment Chapter 16

Class 10 Science Chapter 10 NCERT Solutions – Exercise and Answers

NCERT Solutions – MCQ’s

Furthermore, in addition to NCERT Solutions for Class 10 Science Chapter 10, students may access NCERT Solutions for all other chapters. Students may click on the respective links to refer to NCERT Solutions for other classes. All these solutions are available for Classes 1 to 12. Students need to make use of these solutions to excel in academics and stay ahead of the pack.

### NCERT Exemplar Class 10 Science

NCERT Exemplar Class 10 Science will allow students to revise difficult questions on Light Reflection and Refraction in the final examination. By practicing these questions thoroughly, students will understand all the concepts and will be able to handle not-so-easy questions with ease.

We recommend students refer to the exemplar problem and solutions for Class 10 Science on the Extramarks’ website.

NCERT Solutions Class 10 Science -Exemplar Solutions

### Key Features of NCERT Solutions for Class 10 Science Chapter 10

• Here are some important features of the chapter-Light Reflection and Refraction
• NCERT solutions provide detailed answers to the chapter questions that help students secure better grades in the examinations.
• The notes are prepared by highly qualified and experienced faculty who meticulously follow the NCERT textbooks and CBSE guidelines to provide authentic and reliable study material.
• NCERT Solutions for Class 10 Science Chapter 10 provides well-explained solutions to all the questions in the textbook and provides sample question papers to be thorough in their respective subjects.
• The solutions provided by the Extramarks help students be confident and improve their learning abilities to understand those complex topics with ease and prepare well for the examinations to get excellent results.
• The content on NCERT solutions is regularly updated as per the CBSE board.

We recommend students sign up for  NCERT Solutions Class 10 Chapter 10 on Extramarks’ website for a  detailed explanation of the Light Reflection and Refraction chapter.

Q.1 Which one of the following materials cannot be used to make a lens?

(a) Water
(b) Glass
(c) Plastic
(d) Clay

Ans-

The correct option is (d).
Explanation: The light can pass through a lens. As clay does not allow light to pass through it hence, it cannot be used to make a lens.

Q.2 The image formed by a concave mirror is observed to be virtual, erect and larger than the object. Where should be the position of the object?

(a) Between the principle focus and the centre of curvature
(b) At the centre of curvature
(c) Beyond the centre of curvature
(d) Between the pole of the mirror and its principle focus.

Ans-

The correct option is (d).
Explanation: Concave mirror forms a virtual, erect and enlarged image of an object when the object is placed between the pole and focus of the mirror.

Q.3 Where should an object be placed in front of a convex lens to get a real image of the size of the object?

(a) At the principle focus of the lens
(b) At twice the focal length
(c) At infinity
(d) Between the optical centre of the lens and its principle focus.

Ans-

The correct option is (b).

Explanation: On placing the object at a distance twice of the focal length (at centre of curvature) in front of a convex lens, the image is formed at the centre of curvature on the other side of the lens. This image is real, inverted, and of the same size as the object.

Q.4 A spherical mirror and a thin spherical lens have each a focal length of –15 cm. The mirror and the lens are likely to be

(a) both concave.
(b) both convex.
(c) the mirror is concave and the lens is convex.
(d) the mirror is convex and the lens is concave.

Ans-

The correct option is (a).
Explanation: As per sign convention, the focal length of a concave mirror and a concave lens are considered as negative.

Q.5 No matter how far you stand from a mirror, your image appears erect. The mirror is likely to be-

(a) only plane.
(b) only concave.
(c) only convex.
(d) either plane or convex.

Ans-

The correct option is (d).
Explanation: A plane mirror always forms a virtual and erect image of same size as that of the object. In the same way, a convex mirror forms a virtual and erect image of smaller size of the object placed in front of it.

Q.6 Which of the following lenses would you prefer to use while reading small letters found in a dictionary?

(a) A convex lens of focal length 50 cm
(b) A concave lens of focal length of 50 cm
(c) A convex lens of focal length of 5 cm
(d) A concave lens of focal length 5 cm

Ans-

The correct option is (c).
Explanation: When a convex lens is placed between the radius of curvature and focal length, it forms magnified image of the given object. Moreover, magnification is more for convex lenses having shorter focal length.

Q.7 We wish to obtain an erect image of an object, using a concave mirror of focal length 15 cm. What should be the range of distance of the object from the mirror? What is the nature of the image? Is the image larger or smaller than the object? Draw a ray diagram to show the image formation in this case.

Ans- A concave mirror always forms an erect image when object is placed between pole (P) and the principal focus (F). Therefore, the object should be placed anywhere between the pole and the focus of the concave mirror. The image formed in this case will be virtual, erect, and magnified in nature.

Q.8 Name the type of mirror used in the following situations.

(b) Side/rear-view mirror of a vehicle.
(c) Solar furnace

Ans-

(a) Concave mirror
(b) Convex mirror
(c) Concave mirror.

Reasons:
(a) In the headlights of cars, concave mirror is used as it produces parallel beam of light when the light source (bulb) is at the principal focus of the concave mirror.
(b) A convex mirror forms a virtual, erect, and diminished image of the objects placed in front of. Hence, a driver can see most of the traffic behind him.
(c) A concave mirror can converge the light incident on it at a single point known as principal focus. Therefore, it can be used to produce a large amount of heat at that point.

Q.9 One-half of a convex lens is covered with a black paper. Will this lens produce a complete image of the object? Verify your answer experimentally. Explain your observations.

Ans-

The convex lens will produce a compete image of the object kept in front of it even half of the lens is covered with black paper.

i. When the upper half of the lens is covered: The rays of light coming from the object will be refracted by the lower half of the lens. These rays meet at the other side of the lens to form the image of the given object.

ii. When the lower half of the lens is covered: The rays of light coming from the object are refracted by the upper half of the lens. These rays meet at the other side of the lens to form the image of the given object.

Q.10 An object 5 cm in length is held 25 cm away from a converging lens of focal length 10 cm. Draw the ray diagram and find the position, size and the nature of the image formed.

Ans- $\begin{array}{l}\text{Given},\text{object distance (u) = -25 cm}\\ {\text{object height (h}}_{\text{o}}\text{)=5 cm}\\ \text{focal length (f) = +10 cm}\\ \text{As per lens formula,}\\ \text{}\frac{\text{1}}{\text{v}}\text{\hspace{0.17em}}-\text{\hspace{0.17em}}\frac{\text{1}}{\text{u}}\text{\hspace{0.17em}}=\text{\hspace{0.17em}}\frac{\text{1}}{\text{f}}\\ ⇒\text{}\frac{\text{1}}{\text{v}}\text{\hspace{0.17em}}\text{=}\text{\hspace{0.17em}}\frac{\text{1}}{\text{f}}\text{\hspace{0.17em}}\text{+}\text{\hspace{0.17em}}\frac{\text{1}}{\text{u}}\text{\hspace{0.17em}}\text{=}\text{\hspace{0.17em}}\frac{\text{1}}{\text{10}}\text{\hspace{0.17em}}-\text{\hspace{0.17em}}\frac{\text{1}}{\text{25}}\text{\hspace{0.17em}}\text{=}\text{\hspace{0.17em}}\frac{\text{15}}{\text{250}}\\ ⇒\text{v}\text{\hspace{0.17em}}\text{=}\text{\hspace{0.17em}}\frac{\text{250}}{\text{15}}\text{\hspace{0.17em}}\text{=}\text{\hspace{0.17em}}\text{16}\text{.66 cm}\\ \text{As image distance is positive hence, the image is}\\ \text{formed at the other side of the lens}\text{.}\\ \text{Magnification (m) =}\frac{image\text{distance}}{object\text{distance}}\text{\hspace{0.17em}}\text{=}\text{\hspace{0.17em}}-\frac{\text{v}}{\text{u}}\text{\hspace{0.17em}}\text{=}\text{\hspace{0.17em}}-\frac{\text{16}\text{.66}}{\text{25}}\\ \text{=}\text{\hspace{0.17em}}-\text{0}\text{.66}\\ \text{Here, negative sign shows that the image is real and}\\ \text{formed behind the lens}\text{.}\\ \text{Again, Magnification (m)}\text{\hspace{0.17em}}\text{=}\text{\hspace{0.17em}}\frac{\text{image height}}{\text{object height}}\text{\hspace{0.17em}}\text{=}\text{\hspace{0.17em}}\frac{{\text{h}}_{\text{i}}}{{\text{h}}_{\text{o}}}\text{\hspace{0.17em}}\text{=}\text{\hspace{0.17em}}\frac{{\text{h}}_{\text{i}}}{\text{5}}\\ {\text{or, h}}_{\text{i}}\text{\hspace{0.17em}}\text{=}\text{\hspace{0.17em}}\text{m}\text{\hspace{0.17em}}\text{×}\text{\hspace{0.17em}}{\text{h}}_{\text{o}}\text{\hspace{0.17em}}\text{=}\text{\hspace{0.17em}}-\text{0}\text{.66}\text{\hspace{0.17em}}\text{×}\text{\hspace{0.17em}}\text{5}\text{\hspace{0.17em}}\text{=}\text{\hspace{0.17em}}-\text{3}\text{.3 cm}\\ \text{Hence, the image formed is inverted}\text{.}\end{array}$

Q.11 A concave lens of focal length 15 cm forms an image 10 cm from the lens. How far is the object placed from the lens? Draw the ray diagram.

Ans-

$Given, focal length of concave lens ( f )= −15 cm image distance ( v )= −10 cm On using lens formula, 1 v – 1 u = 1 f ⇒ 1 v – 1 u = 1 f =− 1 10 cm − 1 ( -15 cm ) =− 5 150 cm ⇒ u =−30 cm Here, negative object distance shows that the object is placed 30 cm in front of the lens. MathType@MTEF@5@5@+=feaaguart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLnhiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr4rNCHbGeaGqiVv0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq=Jc9vqaqpepm0xbba9pwe9Q8fs0=yqaqpepae9pg0FirpepeKkFr0xfr=xfr=xb9adbaqaaeaacaGaaiaabaqaamaabaabaaGceaqabeaacaqGhbGaaeyAaiaabAhacaqGLbGaaeOBaiaabYcacaqGGaGaaeOzaiaab+gacaqGJbGaaeyyaiaabYgacaqGGaGaaeiBaiaabwgacaqGUbGaae4zaiaabshacaqGObGaaeiiaiaab+gacaqGMbGaaeiiaiaabogacaqGVbGaaeOBaiaabogacaqGHbGaaeODaiaabwgacaqGGaGaaeiBaiaabwgacaqGUbGaae4CaiaabccadaqadaqaaiaabAgaaiaawIcacaGLPaaacaqG9aGaaeiiaiabgkHiTiaabgdacaqG1aGaaeiiaiaabogacaqGTbaabaGaaeiiaiaabccacaqGGaGaaeiiaiaabccacaqGGaGaaeiiaiaabccacaqGGaGaaeiiaiaabccacaqGGaGaaeiiaiaabccacaqGGaGaaeiiaiaabccacaqGGaGaaeiiaiaabccacaqGGaGaaeiiaiaabccacaqGGaGaaeiiaiaabccacaqGGaGaaeiiaiaabccacaqGGaGaaeiiaiaabccacaqGGaGaaeyAaiaab2gacaqGHbGaae4zaiaabwgacaqGGaGaaeizaiaabMgacaqGZbGaaeiDaiaabggacaqGUbGaae4yaiaabwgacaqGGaWaaeWaaeaacaqG2baacaGLOaGaayzkaaGaaeypaiaabccacqGHsislcaqGXaGaaeimaiaabccacaqGJbGaaeyBaaqaaiaab+eacaqGUbGaaeiiaiaabwhacaqGZbGaaeyAaiaab6gacaqGNbGaaeiiaiaabYgacaqGLbGaaeOBaiaabohacaqGGaGaaeOzaiaab+gacaqGYbGaaeyBaiaabwhacaqGSbGaaeyyaiaabYcaaeaacaqGGaGaaeiiaiaabccacaqGGaGaaeiiaiaabccacaqGGaGaaeiiaiaabccacaqGGaGaaeiiaiaabccacaqGGaGaaeiiaiaabccacaqGGaWaaSaaaeaacaqGXaaabaGaaeODaaaacaqGTaWaaSaaaeaacaqGXaaabaGaaeyDaaaacaqG9aWaaSaaaeaacaqGXaaabaGaaeOzaaaaaeaacqGHshI3caqGGaGaaeiiaiaabccacaqGGaGaaeiiaiaabccacaqGGaGaaeiiaiaabccacaqGGaGaaeiiaiaabccadaWcaaqaaiaabgdaaeaacaqG2baaaiaab2cadaWcaaqaaiaabgdaaeaacaqG1baaaiaab2dadaWcaaqaaiaabgdaaeaacaqGMbaaaiaab2dacqGHsisldaWcaaqaaiaabgdaaeaacaqGXaGaaeimaaaacqGHsisldaWcaaqaaiaabgdaaeaadaqadaqaaiaab2cacaqGXaGaaeynaaGaayjkaiaawMcaaaaacaqG9aGaeyOeI0YaaSaaaeaacaqG1aaabaGaaeymaiaabwdacaqGWaaaaaqaaiabgkDiElaabccacaqGGaGaaeiiaiaabccacaqGGaGaaeiiaiaabccacaqGGaGaaeiiaiaabccacaqGGaGaaeiiaiaabccacaqG1bGaaeiiaiaab2dacqGHsislcaqGZaGaaeimaiaabccacaqGJbGaaeyBaaqaaiaabIeacaqGLbGaaeOCaiaabwgacaqGSaGaaeiiaiaab6gacaqGLbGaae4zaiaabggacaqG0bGaaeyAaiaabAhacaqGLbGaaeiiaiaab+gacaqGIbGaaeOAaiaabwgacaqGJbGaaeiDaiaabccacaqGKbGaaeyAaiaabohacaqG0bGaaeyyaiaab6gacaqGJbGaaeyzaiaabccacaqGZbGaaeiAaiaab+gacaqG3bGaae4CaiaabccacaqG0bGaaeiAaiaabggacaqG0bGaaeiiaiaabshacaqGObGaaeyzaiaabccacaqGVbGaaeOyaiaabQgacaqGLbGaae4yaiaabshacaqGGaGaaeyAaiaabohacaqGGaGaaeiCaiaabYgacaqGHbGaae4yaiaabwgacaqGKbGaaeiiaiaabodacaqGWaGaaeiiaiaabogacaqGTbGaaeiiaiaabMgacaqGUbGaaeiiaiaabAgacaqGYbGaae4Baiaab6gacaqG0bGaaeiiaiaab+gacaqGMbGaaeiiaiaabshacaqGObGaaeyzaiaabccacaqGSbGaaeyzaiaab6gacaqGZbGaaeOlaiaabccaaeaacaqGGaGaaeiiaiaabccacaqGGaGaaeiiaiaabccacaqGGaGaaeiiaiaabccacaqGGaGaaeiiaiaabccacaqGGaGaaeiiaiaabccaaaaa@3563@$ Q.12 An object is placed at a distance of 10 cm from a convex mirror of focal length 15 cm. Find the position and nature of the image.

Ans-

$Given, focal length ( f ) = +15 cm object distance ( u ) = −10 cm As per mirror formula, 1 v + 1 u = 1 f ⇒ 1 v = 1 f − 1 u = 1 15 cm − 1 ( -10 cm ) = 25 150 cm v = 6 cm As ( v ) is positive hence, image is formed behind the mirror. Magnification ( m )= image distance object distance =− v 4 =− 6 ( −10 ) =+0.6 Here, positive sign shows that the image formed is virtual and erect. MathType@MTEF@5@5@+=feaaguart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLnhiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr4rNCHbGeaGqiVv0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq=Jc9vqaqpepm0xbba9pwe9Q8fs0=yqaqpepae9pg0FirpepeKkFr0xfr=xfr=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@6A3C@$

Q.13 The magnification produced by a plane mirror is +1. What does this mean?

Ans-

$\begin{array}{l}For\text{a mirror, magnification (m) is given as,}\\ \text{m =}\frac{{h}_{i}}{{h}_{o}}\\ Where,{\text{h}}_{\text{i}}={\text{image height , h}}_{\text{o}}=\text{object height}\\ \text{When magnification produced by a plane mirror is +1,}\\ \text{the image formed would be of the same size as that}\\ \text{of object}\text{. Also, the positive sign indicates that the}\\ \text{image formed is virtual and erect}\text{.}\end{array}$

Q.4 An object 5.0 cm in length is placed at a distance of 20 cm in front of a convex mirror of radius of curvature 30 cm. Find the position of the image, its nature and size.

Ans-

$\begin{array}{l}\text{Given,}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{u}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{=}\text{\hspace{0.17em}}\text{-20}\text{\hspace{0.17em}}\text{cm,}\text{\hspace{0.17em}}{\text{h}}_{\text{o}}\text{\hspace{0.17em}}\text{=}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{5}\text{\hspace{0.17em}}\text{cm,}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{R}\text{\hspace{0.17em}}\text{=}\text{\hspace{0.17em}}\text{30}\text{\hspace{0.17em}}\text{cm}\\ \text{focal}\text{\hspace{0.17em}}\text{length}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{(f)}\text{\hspace{0.17em}}\text{=}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\frac{\text{R}}{\text{2}}\text{\hspace{0.17em}}\text{=}\text{\hspace{0.17em}}\frac{\text{30 cm}}{\text{2 cm}}\text{\hspace{0.17em}}\text{=}\text{\hspace{0.17em}}\text{15}\text{\hspace{0.17em}}\text{cm}\\ \text{As}\text{\hspace{0.17em}}\text{per}\text{\hspace{0.17em}}\text{mirror}\text{\hspace{0.17em}}\text{formula,}\\ \text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\frac{\text{1}}{\text{v}}\text{\hspace{0.17em}}\text{+}\text{\hspace{0.17em}}\frac{\text{1}}{\text{u}}\text{\hspace{0.17em}}\text{=}\text{\hspace{0.17em}}\frac{\text{1}}{\text{f}}\\ ⇒\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\frac{\text{1}}{\text{v}}\text{\hspace{0.17em}}\text{=}\text{\hspace{0.17em}}\frac{\text{1}}{\text{f}}\text{\hspace{0.17em}}\text{–}\text{\hspace{0.17em}}\frac{\text{1}}{\text{u}}\text{\hspace{0.17em}}\text{=}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\frac{\text{1}}{\text{15 cm}}\text{\hspace{0.17em}}\text{+}\text{\hspace{0.17em}}\frac{\text{1}}{\text{20 cm}}\text{\hspace{0.17em}}\text{=}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\frac{\text{7}}{\text{60 cm}}\\ \text{}\text{}\text{}\text{}\text{}\text{}\text{}\text{}\text{}\text{}\text{}\text{}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{v =}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{8}\text{.57}\text{\hspace{0.17em}}\text{cm}\\ \text{As}\text{\hspace{0.17em}}\text{v}\text{\hspace{0.17em}}\text{is}\text{\hspace{0.17em}}\text{positive}\text{\hspace{0.17em}}\text{hence,}\text{\hspace{0.17em}}\text{image}\text{\hspace{0.17em}}\text{is}\text{\hspace{0.17em}}\text{formed}\text{\hspace{0.17em}}\text{behind}\text{\hspace{0.17em}}\text{the}\text{\hspace{0.17em}}\text{mirror}\text{.}\\ \text{Magnification}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{(m)}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{=}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{–}\frac{\text{v}}{\text{u}}\text{\hspace{0.17em}}\text{=}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{–}\text{\hspace{0.17em}}\frac{\text{8}\text{.57 cm}}{\text{(-20 cm)}}\text{\hspace{0.17em}}\text{=}\text{\hspace{0.17em}}\text{0}\text{.428}\\ \text{The}\text{\hspace{0.17em}}\text{positive}\text{\hspace{0.17em}}\text{value}\text{\hspace{0.17em}}\text{of}\text{\hspace{0.17em}}\text{magnification}\text{\hspace{0.17em}}\text{shows}\text{\hspace{0.17em}}\text{that}\text{\hspace{0.17em}}\text{the}\text{\hspace{0.17em}}\text{image}\text{\hspace{0.17em}}\text{formed}\text{\hspace{0.17em}}\text{is}\text{\hspace{0.17em}}\text{virtual}\text{.}\\ \text{Again,}\text{\hspace{0.17em}}\text{Magnification}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{(m)}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{=}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\frac{{\text{h}}_{\text{i}}}{{\text{h}}_{\text{o}}}\\ \text{or,}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}{\text{h}}_{\text{i}}\text{\hspace{0.17em}}\text{=}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}{\text{m × h}}_{\text{o}}\text{\hspace{0.17em}}\text{=}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{0}\text{.428}\text{\hspace{0.17em}}\text{×5}\text{cm}\text{=}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{2}\text{.14}\text{\hspace{0.17em}}\text{cm}\\ \text{The}\text{\hspace{0.17em}}\text{positive}\text{\hspace{0.17em}}\text{value}\text{\hspace{0.17em}}\text{of}\text{\hspace{0.17em}}\text{image}\text{\hspace{0.17em}}\text{height}\text{\hspace{0.17em}}\text{indicates}\text{\hspace{0.17em}}\text{that}\text{\hspace{0.17em}}\text{the image}\text{\hspace{0.17em}}\text{formed}\text{\hspace{0.17em}}\text{is}\text{\hspace{0.17em}}\text{erect}\text{.}\\ \text{Therefore,}\text{\hspace{0.17em}}\text{the}\text{\hspace{0.17em}}\text{image}\text{\hspace{0.17em}}\text{formed}\text{\hspace{0.17em}}\text{is}\text{\hspace{0.17em}}\text{virtual,}\text{\hspace{0.17em}}\text{erect}\text{\hspace{0.17em}}\text{and}\text{\hspace{0.17em}}\text{smaller in}\text{\hspace{0.17em}}\text{size}\text{.}\end{array}$

Q.15 An object of size 7.0 cm is placed at 27 cm in front of a concave mirror of focal length 18 cm. At what distance from the mirror should a screen be placed, so that a sharp focussed image can be obtained? Find the size and the nature of the image.

Ans-

$\begin{array}{l}\mathrm{Given},\text{u = -27 cm}\\ {\text{h}}_{\text{o}}=\text{7 cm}\\ \text{f = -18 cm}\\ \text{As per mirror formula,}\\ \text{}\frac{1}{\mathrm{v}}+\frac{1}{\mathrm{u}}=\frac{1}{\mathrm{f}}\\ ⇒\text{}\frac{1}{\mathrm{v}}=\frac{1}{\mathrm{f}}-\frac{1}{\mathrm{u}}=\frac{1}{-18}-\frac{1}{\left(-27\right)}=-\frac{1}{54}\\ \text{v= -54 cm}\\ \mathrm{Hence},\text{the screen should be placed at a distance of}\\ \text{54 cm in front of the given mirror.}\\ \text{Magnification (m) =}-\frac{\mathrm{v}}{\mathrm{u}}=-\frac{\left(-54\right)}{\left(-27\right)}=-2\\ \mathrm{Here},\text{negative sign shows that the image formed is real.}\\ \mathrm{Again},\text{m =}\frac{{\mathrm{h}}_{\mathrm{i}}}{{\mathrm{h}}_{\mathrm{o}}}\\ \mathrm{or},{\text{h}}_{\text{i}}=\text{m}×{\mathrm{h}}_{\mathrm{o}}=7×\left(-2\right)=-14\text{}\mathrm{cm}\\ \mathrm{The}\text{negative value of image height shows that the image}\\ \text{formed is inverted.}\end{array}$

Q.16 Find the focal length of a lens of power – 2.0 D. What type of lens is this?

Ans-

$\begin{array}{l}\text{Given, power of lens, P = -2 D}\\ \text{P =}\frac{1}{\text{f}\left(\text{metres}\right)}\\ \therefore \text{f =}\frac{1}{2D}=–0.5\text{m}\\ \text{As a concave lens has a negative focal length.}\\ \text{Hence, this lens is a concave.}\end{array}$

Q.17 A doctor has prescribed a corrective lens of power +1.5 D. Find the focal length of the lens. Is the prescribed lens diverging or converging?

Ans-

$\begin{array}{l}\text{Given, power of lens P = 1}\text{.5 D}\\ \text{On using the formula for power}\\ \text{P =}\frac{1}{\text{f}\left(\text{metres}\right)}\text{}\\ \therefore \text{f =}\frac{1}{1.5 D}=0.66\text{m}\\ \text{As a convex lens has a positive focal length}\text{. Hence it is convex lens}\text{.}\end{array}$