Goyal Brothers Solutions for ICSE Class 9 Physics Chapter 8 Light

Unit I

Exercise 1

Question 1. (a) What do you understand by the following terms?
1. Light
2. Diffused light
(b) By giving one example and one use explain or define
1. Regular reflection
2. Irregular reflection.
Answer: (a)
(i) Light : Light is a form of energy which produces in us sensation of seeing.
(ii) Diffused light : Light obtained after reflection from rough surface is known as diffused light. It is a soft light with neither the intensity nor the glare of direct light. It is scattered and comes from all directions. It does not cause harsh shadows.
(b)
(i) Regular reflection : The phenomenon due to which a parallel beam of light travelling through a certain medium, on striking some polished surface, bounces off from it, as parallel beam, in some other direction is called regular reflection.
For example : Reflection taking place from the objects like looking glass, still water, oil, highly polished metals is regular reflection.
Regular reflection is useful in the formation of images. We can see our face in a mirror only due to regular reflection.
(ii) Irregular reflection or Diffused reflection : The phenomenon due to which a parallel beam of light, travelling through some medium, gets reflected in various possible directions, on striking some rough surface is called irregular reflection.
For example : Reflection taking place from ground, walls, trees, suspended particles in air is irregular reflection.
Use : It helps in the general illumination of places and helps us to see things around us.
In simple words: Light is what lets us see. Regular reflection is like a ball bouncing perfectly off a smooth floor (like a mirror), while irregular reflection is like a ball hitting a pile of rocks and bouncing everywhere (like light hitting a wall).

📝 Teacher's Note: Use the analogy of a polished floor versus a gravel path to explain why we can see our reflection in one but not the other.

🎯 Exam Tip: Remember that even in irregular reflection, the laws of reflection are obeyed at each individual point of contact.

Question 2. By drawing a neat diagram define the following :
1. Mirror
2. Incident ray
3. Reflected ray
4. Angle of incidence
5. Angle of reflection
6. Normal
Answer:
1. Mirror : Mirror is a highly polished and smooth surface which reflects almost the entire light falling on it. A plane mirror is made by silvering one side of a glass plate as shown in figure.
2. Incident ray : The light ray striking a reflecting surface is called the incident ray.
3. Reflected ray : The light ray obtained after reflection from the surface, in the same medium in which the incident ray is travelling, is called the reflected ray.
4. Angle of incidence : The angle which the incident ray makes with the normal at the point of incidence is called the angle of incidence. It is denoted by the letter “i”.
5. Angle of reflection : The angle which the reflected ray makes with the normal at point of incidence, is called the angle of reflection. It is denoted by the letter “r”.
6. Normal : The perpendicular drawn at the point of incidence, to the surface of mirror is called normal.
In simple words: An incident ray is the incoming light, the reflected ray is the outgoing light, and the normal is an imaginary line sticking straight up from the mirror at 90 degrees.

📝 Teacher's Note: When drawing, always use a ruler and show arrows on light rays. The normal should always be a dotted line.

🎯 Exam Tip: Remember that angles \( i \) and \( r \) are measured from the Normal, not from the mirror surface.

Question 3. State the laws of reflection.
Answer: Laws of reflection :
1. The incident ray, the reflected ray and the normal ray at the point of incidence, lie in the same plane.
2. The angle of incidence \( i \) is equal to the angle of reflection \( r \) i.e. \( \angle i = \angle r \).
In simple words: The two main rules are: (1) all parts of the bounce stay on the same flat surface, and (2) the angle going in is the exact same as the angle coming out.

📝 Teacher's Note: Demonstrate Law 2 using a laser pointer and a mirror on a protractor to show that the angles are always equal.

🎯 Exam Tip: Do not just write \( i = r \); state the first law about the "same plane" as well to get full credit.

Question 4. A ray of light strikes a plane mirror, such that angle with the mirror is \( 20^\circ \). What is value of angle of reflection? What is the angle between the incident ray and the reflected ray?
Answer:
\( \because \) Light ray makes an angle of \( 20^\circ \) with the mirror
\( \therefore \angle \text{ABM} = 20^\circ \)
Angle of incidence \( = \angle i = 90^\circ - 20^\circ \)
\( \angle i = 70^\circ \)
\( \because \angle i = \angle r \therefore \angle r = 70^\circ \)
Angle between incident ray and reflected ray \( = \angle i + \angle r = 70^\circ + 70^\circ = 140^\circ \)
In simple words: Since the ray is 20 degrees from the mirror, it is 70 degrees from the normal. Because the bounce is equal, the reflected ray is also 70 degrees away, making a total gap of 140 degrees between the two rays.

📝 Teacher's Note: This is a common trap. Students often take the angle with the mirror as the angle of incidence. Remind them: Incidence is always from the Normal!

🎯 Exam Tip: Always subtract the "glancing angle" from \( 90^\circ \) to find the angle of incidence.

Question 5. Prove experimentally that images are formed as far behind in a plane mirror as the object is in front of it.
Answer: Consider an object ‘O’ situated in front of a plane mirror MM,. A ray of light which starts from point ‘O’ perpendicularly, is reflected back along the same path (See figure). However, another ray which moves along OB is reflected along BC, obeying the laws of reflection, such that BN is the normal. Produce OA and CB backward, such that they meet at point I. Then ‘I’ is image of ‘O’.
We have to prove that OA = IA.
\( \angle i = \angle r \) [By the laws of reflection]
Also, \( \angle i = \angle 1 \) ... (i) [Pair of alternate angles]
\( \angle r = \angle 2 \) [Pair of corresponding angles] ... (ii)
Comparing (i) and (ii), \( \therefore \angle 1 = \angle 2 \)
In \( \triangle \text{BAI} \) and \( \triangle \text{BAO} \)
\( \angle 1 = \angle 2 \) [Proved]
\( \angle 3 = \angle 4 \) [Each \( = 90^\circ \)]
BA = BA [Common to both \( \triangle s \)]
\( \therefore \triangle \text{BAI} \cong \triangle \text{BAO} \)
Thus, in particular OA = IA.
In simple words: By using geometry rules like congruent triangles, we show that the distance from the mirror to the object is exactly the same as the distance to where the image seems to be inside the mirror.

📝 Teacher's Note: Use a transparent glass sheet and two candles to show that the reflection of the lit candle perfectly overlaps a second unlit candle placed at the same distance behind the glass.

🎯 Exam Tip: Mentioning congruent triangles (\( \triangle \text{BAI} \cong \triangle \text{BAO} \)) is the key mathematical step in this proof.

Question 6. Prove geometrically that when plane mirror turns through a certain angle, the reflected ray turns through twice the angle.
Answer: Consider a ray of light AB, incident on plane mirror in position MM’, such that BC is the reflected ray and BN is the normal.
\( \angle \text{ABM} = \angle \text{CBN} = \angle i \)
\( \angle \text{ABC} = 2 \angle i \) …(i)
Let the mirror be rotated through an angle \( \theta \) about point B, such that \( \text{M}_1\text{M}_1 \) is the new position of the mirror and \( \text{BN}_1 \) is the new position of normal. As the position of the incident ray remains same, therefore new angle of the incidence is \( \angle \text{ABN}_1 \) whose magnitude is \( (i + \theta) \). Let BD be the reflected ray, such that \( \angle \text{DNB}_1 \) is the new angle of reflection.
\( \angle \text{ABD} = \angle \text{ABN}_1 + \angle \text{DBN}_1 \)
\( = (i + \theta) + (i + \theta) \)
\( = 2 \angle i + 2 \angle \theta \) ... (ii)
Subtracting (i) from (ii),
\( \angle \text{ABD} - \angle \text{ABC} = 2 \angle i + 2 \angle \theta - 2 \angle i \)
\( \therefore \angle \text{CBD} = 2 \angle \theta \)
Thus, for a given incident ray, if the plane mirror is rotated through a certain angle, then the reflected ray rotates through twice the angle.
In simple words: If you tilt your mirror by 5 degrees, the light beam reflecting off it will shift by 10 degrees. The reflection moves double the amount you move the mirror.

📝 Teacher's Note: This is why even a small shake of a mirror can make a reflected light spot dance wildly on a distant wall.

🎯 Exam Tip: The final derivation \( \angle \text{CBD} = 2 \theta \) is the most important part of the answer.

Question 7. (a) What do you understand by the term lateral inversion?
(b) A printed card has letters PHYSICS. By drawing the diagram show the appearance of the letters. (No ray diagram is required).
Answer: (a) Lateral Inversion: The phenomenon due to which the image of an object turns through an angle of \( 180^\circ \) through verticle axis rather the horizontal axis, such that right side of image appears as left or vice-versa is called Lateral Inversion. During lateral inversion the left side of object appears as right side of image and vice-versa. In a way the image turns through the angle of \( 180^\circ \) about vertical axis.
(b)
PHYSICS \( \implies \) (S C I S Y H P in mirror writing)
In simple words: Lateral inversion is when the left and right sides swap in a mirror. That's why the word AMBULANCE is written backwards on the front of the vehicle—so drivers can read it correctly in their rearview mirrors.

📝 Teacher's Note: Have students write their names on paper and hold them up to a mirror to see lateral inversion in action.

🎯 Exam Tip: Be careful! Mirror images swap left-to-right, but not top-to-bottom.

Question 8. (a) State the mirror formula for the formation of total number of images formed in two plane mirrors, held at an angle.
(b) Calculate the number of images formed in two plane mirrors, when they are held at the angle of (i) \( 72^\circ \) (ii) \( 36^\circ \).
Answer: (a) If \( \theta \) = Angle of inclination between two mirrors
\( n \) = number of images formed
Then \( n = \frac{360^\circ}{\theta} - 1 \)
Numeral one is subtracted because of the loss of one image due to overlapping of the images.
(b) (i) When \( \theta = 72^\circ \)
\( n = \frac{360^\circ}{72^\circ} - 1 \)
\( n = 5 - 1 = 4 \)
(ii) When \( \theta = 36^\circ \)
\( n = \frac{360^\circ}{36^\circ} - 1 \)
\( n = 10 - 1 = 9 \)
In simple words: The smaller the angle between the mirrors, the more times the light bounces and the more images you see. You subtract 1 because two of the reflections sit right on top of each other.

📝 Teacher's Note: Use two hinged mirrors and a coin to show how the number of coins increases as you close the "book" of mirrors.

🎯 Exam Tip: This formula only works when \( 360/\theta \) is an even number. If it is odd and the object is asymmetrical, different rules may apply, but for this grade, stick to this formula.

Question 9. Draw a neat two ray diagram for the formation of images in two plane mirrors, when mirrors are
1. at right angles to each other
2. facing each other.
Answer: (a) When two mirrors are inclined at right angles
‘O’ is an object placed in between two mirrors XY and XZ, inclined at an angle of \( 90^\circ \). Taking normal incidence, \( \text{I}_1 \) and \( \text{I}_2 \) are the images formed in the plane mirror XY and XZ respectively as far behind the mirrors, as point ‘O’ is in front of them. However, image \( \text{I}_1 \) acts as a virtual object for image mirror \( \text{XZ}_1 \) and forms an image \( \text{I}_3 \). Similarly, image \( \text{I}_2 \) acts as a virtual object for the image mirror \( \text{XY}_1 \) and forms the image \( \text{I}_4 \). The images \( \text{I}_3 \) and \( \text{I}_4 \) overlap to form a very bright image. Thus, on the whole three images are seen.
(b) When two mirrors are parallel to each other (facing each other), they form infinite number of images.
In simple words: At 90 degrees, you see three images. When mirrors face each other perfectly, it's like standing in a barber shop with mirrors on both walls—you see a never-ending tunnel of yourself!

📝 Teacher's Note: This is a great way to introduce the concept of "Virtual Objects." The reflection in the first mirror becomes the "thing" that the second mirror reflects.

🎯 Exam Tip: For mirrors at \( 90^\circ \), remember there are 3 images total. The 4th image is hidden behind the 3rd.

Question 10. Why are infinite images not seen when two plane mirrors are facing each other?
Answer: Infinite images are not seen when two plane mirrors are facing each other because :
1. After every successive reflection, some amount of light energy is absorbed. Thus luminosity of image goes on decreasing, till they are no longer visible.
2. As the distance of images from the eye goes on increasing, it is unable to resolve far off images.
In simple words: Every time light bounces, it gets a little bit weaker. Eventually, the reflections become too dim and too small for our eyes to see.

📝 Teacher's Note: Mention that no mirror is \( 100\% \) efficient; they always soak up a tiny bit of light with each bounce.

🎯 Exam Tip: The two key reasons are "absorption of light" and "distance from the eye."

Question 11. (a) State four characteristics of the image formed in a plane mirror.
(b) State three ways in which the image formed in a plane mirror differs from the image formed in a pin hole camera.
Answer: (a) Characteristics of Image formed by a Plane Mirror:
1. Image is of the same size as that of object.
2. Image is laterally inverted.
3. It is upright (erect).
4. It is virtual (cannot be caught on screen).
5. Image formed is as far behind the mirror as the object in front of the mirror.
(b) Comparison :

In simple words: A mirror shows you "you" but swaps left and right. A pinhole camera shows a tiny version of the world that is flipped upside down.

📝 Teacher's Note: This is a very common comparison question. Use the acronym V-E-S-D (Virtual, Erect, Same size, Distance) for mirror images.

🎯 Exam Tip: "Virtual" and "Real" are the most important technical terms to use when comparing these two images.

Question 12. (a) What should be the minimum size of a plane mirror, so that a person \( 182 \text{ cm} \) high can see himself completely?
(b) A boy stands \( 4 \text{ m} \) away from plane mirror. If the boy moves \( 1/2 \text{ m} \) towards mirror, what is now the distance between the boy and his image? Give a reason for your answer.
Answer: (a) Size (height) of a person \( = 182 \text{ cm} \)
We know to see the full length image of a person, we need a plane mirror of size as half of the size of the person.
So, size of plane mirror required \( = \text{Height of person}/2 = 182/2 = 91 \text{ cm} \).
(b) Distance of boy from the plane mirror \( = u = 4 \text{ m} \). When boy moves \( 1/2 \text{ m} \) towards the mirror:
Then, distance between boy and mirror \( = 4 - 0.5 = 3.5 \text{ m or } 7/2 \text{ m} \).
We know, in a plane mirror, the image formed is as behind the mirror as the object is in front of it.
So, distance between the mirror and image of boy \( = 3.5 \text{ m} \).
Distance between the boy and his image \( = \text{Distance of boy from mirror} + \text{Distance of image from mirror} \)
\( = 3.5 + 3.5 = 7 \text{ m} \).
In simple words: To see your whole body, you only need a mirror half as tall as you. If you walk closer to the mirror, your reflection walks closer too. The total distance is always double your distance to the glass.

📝 Teacher's Note: This proves that the minimum mirror size does not depend on how far away you stand.

🎯 Exam Tip: For part (b), remember the question asks for distance between the boy and the image, not the boy and the mirror.

Question 13. State four uses of a plane mirror.
Answer: Four uses of plane mirror are :
1. Plane mirrors are used in construction of reflecting periscope.
2. They are used as looking glass for dressing.
3. They are used in solar cookers for reflecting the rays of the sun into the cooker.
4. They are used for signalling purpose.
In simple words: We use mirrors to see ourselves, to peek over crowds with periscopes, to catch sun rays for cooking, and even to send light signals to others.

📝 Teacher's Note: Ask students to think of other places they see plane mirrors, like in a kaleidoscope or as decorative wall panels.

🎯 Exam Tip: "Rear-view mirror" is usually a convex mirror, not a plane mirror. Don't mix them up in the exam!

Question 14. (a) Draw a neat diagram of reflecting periscope.
(b) State two advantages and two disadvantages of the reflecting periscope.
Answer: (a)
(b) Advantages of Reflecting Periscope :
1. It is used to see above the head of crowds.
2. It is used by soldiers in trench warfare to see the enemy without exposing themselves.
Disadvantages of Reflecting Periscope :
1. The final images is not brightly illuminated as light energy is absorbed due to two successive reflections.
2. Any deposition of moisture of dust on the mirror reduces the reflection almost to nil.
In simple words: A periscope lets you see "around corners" using two tilted mirrors. It's useful for submarines or soldiers, but the view can get dusty or dim.

📝 Teacher's Note: The mirrors must be parallel to each other and tilted at exactly \( 45^\circ \) to the tube axis for the periscope to work correctly.

🎯 Exam Tip: Always label the angle of mirrors as \( 45^\circ \) in your diagram.

Question 15. What must be the minimum length of a plane mirror in which a person can see himself full length? Draw a diagram to justify your answer. Does the distance of person from the mirror affect the above answer?
Answer: Consider a person AB, such that A represents the highest point on his head, and B the lowest point on the foot such that E is the fixed eye level. The person will be able to see every part of his body if he can see points A and B. Let MN be the minimum length of mirror fixed on the wall, such that the rays AM and BN, after reflection, reach the eye of person, thereby forming an image \( \text{A}_1\text{B}_1 \) when produced backward.
In \( \triangle \text{AEA}_1 \), CM is parallel to AE and C is the mid-point of \( \text{AA}_1 \). M is the mid-point of \( \text{A}_1\text{E} \).
Similarly, in \( \triangle \text{BEB}_1 \), ND is parallel to BE and D is the mid-point of \( \text{BB}_1 \).
∴ N is the mid-point of \( \text{B}_1\text{E} \).
Now in \( \triangle \text{A}_1\text{B}_1\text{E} \), M is mid-point of \( \text{A}_1\text{E} \) and N is mid-point of \( \text{B}_1\text{E} \).
\( \therefore \) MN is parallel and half of \( \text{A}_1\text{B}_1 \).
But, \( \text{A}_1\text{B}_1 \) = AB (Height of person).
So, MN = 1/2 AB.
Thus, in order to see full length, a person requires a plane mirror which is half his own height. This relation is true for any distance of object from plane mirror.
In simple words: No matter how far back you stand, you still need the same half-size mirror to see your whole body. Moving back doesn't help you "fit" more of yourself in a tiny mirror!

📝 Teacher's Note: This is a counter-intuitive fact for many students. Use the geometry of similar triangles to show that as you move back, the "cone" of light from your body shrinks at the same rate the mirror area appears to shrink.

🎯 Exam Tip: The final conclusion is: "Mirror length = \( 1/2 \times \text{Object height} \)." This is the most common result to remember.

Question 16. An insect is sitting in front of a plane mirror at a distance of one metre from it.
1. Where is the image of insect formed?
2. What is the distance between insect and its image?
3. State any two characteristics of image formed in a plane mirror.
Answer:
(i) Distance of insect from plane mirror \( = u = 1 \text{ m} \).
We know in case of plane mirror, distance of object from mirror is equal to distance of image from the mirror.
\( \therefore \) Distance of the image of insect from plane mirror \( = v = 1 \text{ m} \). The image is formed behind the mirror.
(ii) Distance between insect and its image \( = u + v = 1 + 1 = 2 \text{ m} \).
(iii) Characteristics of the image formed by a plane mirror :
(a) Image formed is virtual.
(b) Image formed is erect.
(c) Image formed is of same size as that of the object.
In simple words: The image is 1 meter "inside" the mirror. The total distance from the actual insect to its reflection is 2 meters.

📝 Teacher's Note: Use the term "Virtual" to describe an image that light doesn't actually reach, but only appears to come from.

🎯 Exam Tip: When a question asks "Where is the image formed?", always specify "behind the mirror."

Question 17. (i) Draw a diagram to show reflection of a ray of light using plane mirror. In the diagram label the incident ray, the reflected ray, the normal, the angle of incidence and angle of reflection. (ii) State the laws of reflection.
Answer:
(i)
(ii) Laws of reflection :
1. See Q. No. (2) of Exercise (1).
2. See Q. No. (3) of Exercise (1).
In simple words: When light hits a mirror, it bounces off like a ball. The angle it comes in at (i) is always equal to the angle it bounces out at (r).

📝 Teacher's Note: Remind students that the Normal is an imaginary line perpendicular to the mirror. All angles must be measured from this Normal, not the mirror surface.

🎯 Exam Tip: Always draw the rays with arrowheads to show the direction of light. A ray diagram without arrows is technically incorrect.

Question 18. (i) Parallel rays are incident :
1. on regular surface and
2. on irregular surface. In what respect do reflected rays in (1) differ from those of (2)?
(ii) Write down four characteristics of image formed in a plane mirror.
Answer:
(i) Parallel rays of light after reflection from a regular surface goes in a particular direction while after reflection from irregular surface go in different directions.
(ii) Characteristics of Image formed by a Plane Mirror.
(a) Image is of the same size as that of object.
(b) Image is laterally inverted.
(c) It is upright.
(d) It is virtual.
(e) Image formed is as far behind the mirror as the object in front of the mirror.
In simple words: A smooth surface keeps light rays organized (regular), while a rough surface scatters them everywhere (irregular). A mirror image is just like you, but swapped left-to-right and trapped "inside" the mirror.

📝 Teacher's Note: Use the "organized soldiers" vs. "dispersing crowd" analogy to explain the difference between regular and irregular reflection.

🎯 Exam Tip: "Laterally inverted" is a technical keyword. It means the left side of the object appears as the right side of the image.

Question 19. How many images will be formed when an object is placed between two parallel plane mirrors with their reflecting surfaces facing each other? Why do more distant images appear fainter?
Answer: Angle between two mirrors facing each other \( = \theta = 0^\circ \)
No. of images formed \( = n = \frac{360^\circ}{\theta} = \frac{360^\circ}{0} = \infty \)
So, infinite number of images are formed when an object is placed between two parallel plane mirrors with their reflecting surfaces facing each other.
More distant images appear fainter when two plane mirror with their reflecting surfaces facing each other because,
After every successive reflection, some amount of light energy is absorbed. Thus luminosity of images goes on decreasing and hence they appear fainter.
In simple words: When mirrors are parallel, light bounces back and forth forever, creating endless images. However, each bounce "steals" a little bit of light, so the far-off images look dimmer.

📝 Teacher's Note: This is commonly seen in barbershops or trial rooms. Explain that "infinite" is the theoretical answer, but practically we only see a few dozen before they fade away.

🎯 Exam Tip: For parallel mirrors, always mention that the angle \( \theta \) is \( 0^\circ \) and the result is "infinite" or \( \infty \).

Question 20. (a) Write down the letters of the word ‘POLEX’ as seen in a plane mirror, held parallel to the plane of this paper. (b) Distinguish between real and virtual image.
Answer:
(a)
(b)
Real image :
1. Can be taken on the screen.
2. It is inverted.
3. It is formed when light rays after reflection (or refraction) actually meet.
Virtual image :
1. Cannot be taken on screen.
2. It is erect.
3. It is formed when rays of light after reflection (or refraction) appear to meet.
In simple words: A mirror swaps letters around (lateral inversion). A real image is like a movie on a screen, while a virtual image is like a ghost that only looks like it's behind the mirror.

📝 Teacher's Note: To demonstrate a real image, use a concave mirror and a candle to project an image onto a piece of white paper.

🎯 Exam Tip: The word 'POLEX' is a classic test because 'O' and 'X' look the same, but 'P', 'L', and 'E' change significantly.

Question 21. (a) Describe the principle of simple periscope through an outline ray diagram. Give one of its uses. (b) Draw diagrams to show difference between regular and irregular reflection.
Answer:
(a) Simple periscope is based upon the principle of reflection. It consists of a cardboard or wooden tube, bent twice at right angles and is provided with two openings as shown in figure. Two plane mirrors are fixed at the bends of the tube at an angle \( 45^\circ \) to the framework, such that the mirrors face each other. The tube is completely blackened from inside to avoid any reflection from its sides.
The parallel rays coming from an object at a higher plane, strike the plane mirror at an angle of \( 45^\circ \) and hence are reflected through an angle of \( 45^\circ \).
These reflected rays strike the second mirror at an angle of \( 45^\circ \) and hence are further reflected through an angle of \( 45^\circ \). These reflected rays on reaching the eye form the image on retina.
Use of periscope : It is used by soldiers in trench warfare.
(b)
In simple words: A periscope uses two mirrors at \( 45^\circ \) to let you see over things. Regular reflection is tidy like a mirror, and irregular is messy like a bumpy road.

📝 Teacher's Note: The mirrors in a periscope must be parallel to each other. If they aren't, the image will be tilted or distorted.

🎯 Exam Tip: In the periscope diagram, make sure to show the light rays bouncing at right angles (\( 90^\circ \) turns).

Question 22. An object is placed \( 2 \text{ cm} \) from a plane mirror. If the object is moved by \( 1 \text{ cm} \) towards the mirror, what will be the distance between the object and its new image?
Answer:
Distance of the object from plane mirror \( = 2 \text{ cm} \)
If object is moved by \( 1 \text{ cm} \) towards the mirror, then distance between the object and plane mirror \( = u = 1 \text{ cm} \)
We know in case of plane mirror,
Distance of object from plane mirror \( = \) Distance of the image of the object from plane mirror
\( \implies v = u = 1 \text{ cm} \)
\( \therefore \) Distance between its object and new image \( = u + v = 1 + 1 = 2 \text{ cm} \)
In simple words: If you are 1 cm from the mirror, your reflection is also 1 cm deep "inside." So, the total gap between you and your reflection is 2 cm.

📝 Teacher's Note: This is a logic test. Always update the object distance first, then double it to get the object-to-image distance.

🎯 Exam Tip: Many students accidentally answer "1 cm." Read carefully: "distance between object and image."

Unit II

Practice Problem 1

Question 1. An object \( 3 \text{ cm} \) high produces a real image \( 4.5 \text{ cm} \) high, when placed at a distance of \( 20 \text{ cm} \) from a concave mirror. Calculate :
1. the position of image
2. focal length of the concave mirror.
Answer:
Size of object \( = h_o = 3 \text{ cm} \)
Size of image \( = h_i = -4.5 \text{ cm} \) [Image formed is real, hence inverted]
Distance of the object from the concave mirror \( = u = -20 \text{ cm} \)
Distance of the image from the concave mirror \( = v = ? \)
Focal length of the concave mirror \( = f = ? \)
Linear magnification \( = m = \frac{h_i}{h_o} = -\frac{v}{u} \)

\( \frac{-4.5}{3} = -\frac{v}{-20} \)

\( v = \frac{-20 \times 4.5}{3} = -30 \text{ cm} \)
Now, using mirror formula:

\( \frac{1}{u} + \frac{1}{v} = \frac{1}{f} \)

\( \frac{1}{-20} + \frac{1}{-30} = \frac{1}{f} \)

\( \frac{1}{f} = \frac{-3-2}{60} = -\frac{5}{60} = -\frac{1}{12} \)

\( \implies f = -12 \text{ cm} \)
In simple words: Because the image is real, it's upside down. We used the magnification formula to find that the image is 30 cm from the mirror, and then calculated that the mirror's focus point is 12 cm away.

📝 Teacher's Note: Explain the New Cartesian Sign Convention. Distances in front of the mirror (real space) are negative, and heights below the axis (inverted) are negative.

🎯 Exam Tip: Always include the negative sign in your final answer for focal length of a concave mirror to show you understand sign conventions.

Practice Problem 2

Question 1. An object \( 1.5 \text{ cm} \) high when placed in front of a concave mirror, produces a virtual image \( 3 \text{ cm} \) high. If the object is placed at a distance of \( 6 \text{ cm} \) from the pole of the mirror, calculate :
1. the position of the image
2. the focal length of the mirror.
Answer:
Size of object \( = h_o = 1.5 \text{ cm} \)
Size of image \( = h_i = +3 \text{ cm} \) [Image formed by a concave mirror is virtual, hence erect]
Distance of the object from the pole of the mirror \( = u = -6 \text{ cm} \)
Distance of the image from the pole of mirror \( = v = ? \)
(i) Distance of the image from the pole of mirror \( = v = ? \)
Linear magnification \( = m = \frac{h_i}{h_o} = -\frac{v}{u} \)

\( \frac{+3}{1.5} = -\frac{v}{-6} \)

\( v = \frac{6 \times 3}{1.5} \)

\( v = +12 \text{ cm} \)
So, image is formed at a distance of \( 12 \text{ cm} \) behind the concave mirror.
(ii) Focal length of the concave mirror \( = f = ? \)
Using mirror formula :

\( \frac{1}{u} + \frac{1}{v} = \frac{1}{f} \)

\( \frac{1}{-6} + \frac{1}{12} = \frac{1}{f} \)

\( \frac{1}{f} = \frac{-2+1}{12} = -\frac{1}{12} \)

\( \implies f = -12 \text{ cm} \)
In simple words: A virtual image in a concave mirror is like a magnifying glass. The image is "inside" the mirror 12 cm deep and looks twice as big as the object.

📝 Teacher's Note: This specific case happens only when the object is between the Focus and the Pole (\( u < f \)).

🎯 Exam Tip: When \( v \) is positive, explicitly state that the image is formed "behind the mirror."

Question 2. A converging mirror, forms a three times magnified virtual image when an object is placed at a distance of \( 8 \text{ cm} \) from it. Calculate :
1. the position of the image
2. the focal length of the mirror.
Answer:
Let size of the object \( = h_o = x \)
Size of image \( = h_i = +3 \times x \) (Virtual image)
Distance of the object from the mirror \( = u = -8 \text{ cm} \)
Distance of the image from the mirror \( = v = ? \)
Focal length of the mirror \( = f = ? \)
(i) Magnification \( = m = \frac{h_i}{h_o} = -\frac{v}{u} \)

\( \frac{3x}{x} = -\frac{v}{-8} \)

\( v = 8 \times 3 = 24 \text{ cm} \)
(ii) Using mirror formula :

\( \frac{1}{u} + \frac{1}{v} = \frac{1}{f} \)

\( \frac{1}{-8} + \frac{1}{24} = \frac{1}{f} \)

\( \frac{1}{f} = -\frac{1}{8} + \frac{1}{24} \)

\( \frac{1}{f} = \frac{-3+1}{24} = \frac{-2}{24} = -\frac{1}{12} \)

\( \implies f = -12 \text{ cm} \)
In simple words: To get a 3x magnification, the object had to be very close to the mirror. The image forms 24 cm behind the mirror surface.

📝 Teacher's Note: "Converging mirror" is a synonym for concave mirror. Using different names helps test if students recognize the mirror types.

🎯 Exam Tip: "Magnified virtual image" for a single mirror always indicates a concave mirror.

Practice Problems 3

Question 1. An object \( 5 \text{ cm} \) high forms a virtual image of \( 1.25 \text{ cm} \) high, when placed in front of a convex mirror at a distance of \( 24 \text{ cm} \). Calculate :
1. the position of the image
2. the focal length of the convex mirror.
Answer:
Size of the object \( = h_0 = 5 \text{ cm} \)
Size of the image \( = h_1 = +1.25 \text{ cm} \)
Distance of the object from mirror \( = u = -24 \text{ cm} \)
Distance of the image from the mirror \( = v = ? \)
Focal length of the mirror \( = f = ? \)
(i) Magnification \( = m = \frac{h_i}{h_o} = -\frac{v}{u} \)

\( \frac{1.25}{5} = -\frac{v}{-24} \)

\( v = \frac{24 \times 0.25}{5} \) [Wait, \( 1.25/5 = 0.25 \)]

\( v = 24 \times 0.25 = 6 \text{ cm} \)
(ii) Using mirror formula :

\( \frac{1}{u} + \frac{1}{v} = \frac{1}{f} \)

We have,

\( \frac{1}{-24} + \frac{1}{6} = \frac{1}{f} \)

\( \frac{1}{f} = \frac{-1+4}{24} = \frac{3}{24} = \frac{1}{8} \)

\( \implies f = 8 \text{ cm} \)
In simple words: This is a convex mirror, which always makes things look smaller. The image is just 6 cm inside the mirror, and the mirror's focus is at 8 cm.

📝 Teacher's Note: For convex mirrors, the focal length (\( f \)) and the image distance for real objects (\( v \)) are always positive because they are behind the mirror.

🎯 Exam Tip: Convex mirrors only ever produce virtual, erect, and diminished images.

Question 2. An object forms a virtual image which is 1/8th of the size of the object. If the object is placed at a distance of \( 40 \text{ cm} \) from the convex mirror, calculate :
1. the position of the image
2. the focal length of the convex mirror.
Answer:
Let size of the object \( = h_o = x \)
Size of the image \( = h_i = \frac{x}{8} = +0.125 \text{ cm} \) (if x=1)
Distance of the object from the convex mirror \( = u = -40 \text{ cm} \)
Distance of the image from the convex mirror \( = v = ? \)
(i) Magnification \( = m = \frac{h_i}{h_o} = -\frac{v}{u} \)

\( \frac{x/8}{x} = -\frac{v}{-40} \)

\( v = \frac{x}{8x} \times 40 = 5 \text{ cm} \)
(ii) Using mirror formula :

\( \frac{1}{u} + \frac{1}{v} = \frac{1}{f} \)

We have,

\( \frac{1}{-40} + \frac{1}{5} = \frac{1}{f} \)

\( \frac{1}{f} = -\frac{1}{40} + \frac{1}{5} \)

\( \frac{1}{f} = \frac{-1+8}{40} = \frac{7}{40} \)

\( \implies f = \frac{40}{7} = 5.71 \text{ cm} \)
In simple words: The object is far away (40 cm), so the convex mirror makes it look very tiny (1/8th size). The image is formed only 5 cm deep in the mirror.

📝 Teacher's Note: This problem shows that even with a high reduction ratio, the image in a convex mirror still stays between the focus and the pole.

🎯 Exam Tip: Fraction problems are common. Set \( h_o = 1 \) and \( h_i = 1/8 \) to make the math easier for yourself.

Exercise 2

(A) Objective Questions

Question 1. A concave mirror is made by cutting a portion of a hollow glass sphere of radius 30 cm. The focal length of the concave mirror is :
(a) 24 cm
(b) 12 cm
(c) 15 cm
(d) 60 cm
Answer: (c) 15 cm
Explanation :
Radius of curvature \( = R = 30 \text{ cm} \)
Focal length of concave mirror \( f = \frac{R}{2} \)
\( f = \frac{30}{2} = 15 \text{ cm} \)
In simple words: The focus point of a curved mirror is always halfway between the mirror and the center of the ball it was cut from.

📝 Teacher's Note: This relationship \( R = 2f \) applies to all spherical mirrors regardless of their size.

Question 2. A mirror forms a virtual image (diminished) of an object, whatever be the position of object :
(a) it must be a concave mirror
(b) it must be a convex mirror
(c) it must be a plane mirror
(d) it may be (b) or (c) or both
Answer: (b) it must be a convex mirror
In simple words: A convex mirror is the only one that *always* makes things look smaller and trapped inside. A plane mirror doesn't change size, and a concave mirror can make things look bigger.

📝 Teacher's Note: This property makes convex mirrors excellent for security mirrors in shops and rearview mirrors in cars.

Question 3. A ray of light is incident on a concave mirror. If it is parallel to principal axis, the reflected ray will :
(a) pass through its principal focus
(b) pass through its centre of curvature
(c) pass through its pole
(d) retraces its path
Answer: (a) pass through its principal focus
In simple words: Rays that come in straight and parallel to the center line always bounce and head straight for the Focus point.

📝 Teacher's Note: This is the definition of "Focus." It's where parallel light rays "focus" their energy after hitting the mirror.

Question 4. If an incident ray passes through the centre of curvature of a spherical mirror, the reflected ray will :
(a) pass through its pole
(b) retraces its path
(c) pass through its focus
(d) be parallel to principal axis
Answer: (b) retraces its path
In simple words: Any ray going through the center hits the mirror perfectly straight (at \( 90^\circ \)), so it just bounces straight back where it came from.

📝 Teacher's Note: This ray hits the mirror surface "normally." The angle of incidence is \( 0^\circ \), so the angle of reflection is also \( 0^\circ \).

Question 5. In case of concave mirror, the minimum distance between an object and its real image is :
(a) f
(b) 2f
(c) 4f
(d) zero
Answer: (d) zero
In simple words: If you put the object exactly at the Center of Curvature (\( 2f \)), the image forms right on top of the object. Since they are at the same spot, the distance between them is zero.

📝 Teacher's Note: This is a special case where the image is real, inverted, and exactly the same size as the object.

Question 6. Looking into a mirror one finds her image diminished, the mirror is :
(a) concave
(b) convex
(c) cylindrical
(d) parabolic
Answer: (b) convex
In simple words: Diminished means smaller. Only a convex mirror makes you look smaller at every distance.

📝 Teacher's Note: While concave mirrors can diminish images, they only do so when you are far away. Convex mirrors do it all the time.

Question 7. Which mirror is used in periscope?
(a) Convex mirror
(b) Concave mirror
(c) Plane mirror
(d) Parabolic mirror
Answer: (c) Plane mirror
In simple words: A periscope needs to show you exactly what is there without changing its size, so it uses two flat plane mirrors.

📝 Teacher's Note: Prisms are used in modern military periscopes for better clarity, but simple periscopes use plane mirrors.

(B) Subjective Questions

Question 1. Define the following terms :
1. spherical mirror
2. convex mirror
3. concave mirror
Answer:
1. Spherical mirror : “A mirror which is made from a part of a hollow sphere is called Spherical Mirror.
2. Convex mirror : “A mirror made by silvering the inner surface such that reflection takes place from the bulging surface” is called Convex Mirror. Centre of curvature is towards the silvered surface.
3. Concave mirror : “A mirror made by silvering the outer or the bulging surface such that the reflection takes place from the concave surface.” Centre of curvature is towards the reflecting surface.
In simple words: A spherical mirror is a curved mirror cut from a glass ball. A concave mirror curves inward like a cave, and a convex mirror bulges outward like the back of a spoon.

📝 Teacher's Note: Use a stainless steel spoon to demonstrate this in class. The side you eat from is concave, and the back side is convex.

🎯 Exam Tip: When defining these, always specify which surface is silvered (polished) and which surface reflects the light.

Question 2. Define the following terms in relation to concave mirror.
1. Pole
2. Centre of curvature
3. Principal axis
4. Principal focus
5. Focal length
6. Radius of curvature
7. Aperture
Answer:
1. Pole : Pole “is the mid-point of the mirror”.
2. Centre of curvature : The centre of hollow sphere of which the mirror forms a part, is called centre of curvature.
3. Principal axis : An imaginary line passing through the pole and the centre of curvature of a spherical mirror is called principal axis
4. Principal focus : It is a point on the principal axis, where a beam of light, parallel to principal axis, after reflection actually meet.
5. Focal length : The linear distance between the pole and the principal focus is called focal length.
6. Radius of curvature : The linear distance between the pole and the centre of curvature is called radius of curvature.
7. Aperture : The diameter of a spherical mirror is called its aperture.
In simple words: The pole is the center of the mirror face. The center of curvature is the center of the original glass ball. The focus is where light rays gather, and focal length is the distance from the mirror to that focus.

📝 Teacher's Note: Draw a large circle on the board first, then erase most of it to show the "mirror" part. This helps students visualize that C is the center of that original circle.

🎯 Exam Tip: Remember that for a concave mirror, the focus is "real" (light actually meets there), while for a convex mirror, it is "virtual."

Question 3. (a) Define the term principal focus in case of convex mirror. Draw a convex mirror and show its principal focus and focal length clearly.
(b) What is the relation between focal length and radius of curvature of a concave mirror?
Answer: (a) Focal length is the distance between pole (P) and focus (F). (F) Focus is a point on principal axis where rays of light appear to meet.
(b) Focal length of a spherical mirror is equal to half of the radius of curvature of spherical mirror.
\( \text{Focal length} = \frac{\text{Radius of curvature}}{2} \)
\( f = \frac{R}{2} \)
In simple words: For a convex mirror, light rays bounce off and spread out, but if you trace them backward, they look like they are coming from a single spot called the Focus. The focus is always halfway between the mirror and its center.

📝 Teacher's Note: Emphasize that the focus of a convex mirror is *behind* the mirror. This is why we call it a virtual focus.

🎯 Exam Tip: When drawing the focal length, ensure it is the distance between 'P' and 'F'. The formula \( f = R/2 \) is a must-know for numericals.

Question 4. (a) What do you understand by the term real image?
(b) What type of mirror is used to obtain a real image?
(c) Does the mirror named by your form real image for all locations? Give reason for your answer.
(d) Is real image always inverted?
Answer:
(a) Real image : When rays of light after reflection or refraction actually meet at some other point” the image is real.
(b) Concave mirror.
(c) No, this mirror does not give real image of the object that lies between principal focus and pole.
(d) Yes. Real image is always ihverted.
In simple words: A real image is formed when light rays actually hit a spot. You can catch a real image on a piece of paper or a screen. These images are always upside down.

📝 Teacher's Note: Use the analogy of a movie projector. The image on the screen is a real image because light actually travels to the screen. The image in your bathroom mirror is virtual because light doesn't actually go "inside" the wall.

🎯 Exam Tip: Keywords for real images: "Actually meet" and "Inverted."

Question 5. Copy the figure. By taking two rays from point A, show the formation of image. State four characteristics of the image.
Answer:
Characteristics of the image :
1. Image formed is real.
2. Image formed is inverted.
3. Image formed is enlarged.
4. Image is formed beyond centre of curvature in front of concave mirror.
In simple words: When an object is far from a concave mirror, the image is upside down, real, and changes size depending on exactly how far it is.

📝 Teacher's Note: The OCR says "enlarged" and "beyond C" for this specific setup, but usually, if the object is far beyond C, the image is diminished and between C and F. Check the object position in the original diagram carefully.

🎯 Exam Tip: Always show the intersection of at least two rays to locate the image.

Question 6. Draw a neat two ray diagram to illustrate how a concave mirror is used as a shaving mirror.
Answer: We know that when an object is placed between P and F of a concave mirror, it forms a virtual and enlarged image. Thus, by using concave mirror we can have a proper shave, as the tiny hairs are clearly visible.
In simple words: If you get very close to a concave mirror (closer than its focus point), your face looks right-side up and much bigger. This makes it easy to see tiny details while shaving.

📝 Teacher's Note: This is the *only* case where a concave mirror forms a virtual, erect, and magnified image. It's a very high-yield exam topic.

🎯 Exam Tip: The object must be between 'P' and 'F' for the mirror to act as a magnifier.

Question 7. Copy the figure. By taking two rays from point A, show the formation of image. State four characteristics of the image.
Answer: Characteristics of image :
1. Image formed is virutal.
2. Image formed is erect.
3. Image formed is diminished.
4. Image is always formed between pole and principal focus, behind the convex mirror.
In simple words: A convex mirror always makes things look smaller and upright. The image is "virtual," meaning it looks like it's inside the mirror.

📝 Teacher's Note: Even though the object moves, the image in a convex mirror always stays upright and small. This is why they are so reliable for safety.

🎯 Exam Tip: "Virutal" is a common spelling mistake in students' work—ensure you use the correct spelling "Virtual" in your answers.

Question 8. Why do automobile drivers prefer convex mirror as a rear view mirror? Illustrate your answer.
Answer: A convex mirror always forms a small and upright image between pole and focus. That means in small area of mirror driver can see all the traffic coming from behind.
In simple words: Drivers use these mirrors because they show a "wide view" of the road. It makes cars behind you look small, but it lets you see many lanes of traffic at once.

📝 Teacher's Note: The main advantage is the "wide field of view." A plane mirror of the same size would show much less of the road.

🎯 Exam Tip: Always mention both "erect image" and "wider field of view" to get full marks.

Question 9. Give two uses of
1. convex mirror
2. concave mirror.
Answer:
(i) Two uses of convex mirror are :
(a) It is used as a rear view mirror in automobile to see the traffic behind.
(b) It is used as reflector for street light bulbs to spread light over a large area.
(ii) Two uses of concave mirror are :
(a) Concave mirror is used as A Reflector in head lights of cars and in search light. The source of light (bulb) is placed at the principal focus and the reflector forms parallel beam of light.
(b) For doctors to examine throat, ear, nose and eyes, light is focused with the help of concave mirror.
In simple words: Convex mirrors help us see more of the road and spread street lights. Concave mirrors help doctors focus light into small spaces and help car headlights shine a powerful beam forward.

📝 Teacher's Note: Explain the difference: Convex *spreads* light (diverges), while Concave *collects* light (converges).

🎯 Exam Tip: Don't forget that "Shaving mirror" or "Makeup mirror" are also excellent examples for concave mirrors.

Question 10. You are provided a convex mirror, a concave mirror and a plane mirror. How will you distinguish between them, without touching or using any other apparatus?
Answer: We can identify convex mirror, concave mirror and a plane mirror by looking at them one by one.
1. If the size of the image of an object is of the same size as that of the object, then that mirror is a plane mirror.
2. If the size of image of an object increases as the object is brought closer to the mirror and size of the image of the object decreases when the object is taken away from the mirror, then that mirror is a concave mirror.
3. If the size of the iamge of an object remains diminished, either the object is moved away from the mirror or moved towards the mirror, then that mirror is a convex mirror.
In simple words: Just look at your face! If it looks normal, it's a plane mirror. If it looks tiny, it's convex. If it looks huge when you get close, it's concave.

📝 Teacher's Note: This is a very common practical-based question. It tests the understanding of image characteristics for all three mirrors.

🎯 Exam Tip: Use the "object brought closer" condition to specifically identify the concave mirror.

Question 11. Compare the characteristics of an image formed by a convex mirror and a concave mirror, when object is beyond centre of curvature, but not at infinity.
Answer: Characteristics of the image formed by concave mirror when object is beyond centre of curvature, but not at infinity are :
1. Image formed is real.
2. Image formed is inverted.
3. Image formed is diminished.
4. Image is formed between centre of curvature (C) and principal focus (F), in front of the concave mirror.
Characteristics of the image formed by convex mirror when object is beyond centre of curvature, but not at infinity are :
1. Image formed is virutal.
2. Image formed is erect.
3. Image formed is diminished.
4. Image is always formed between pole and principal focus and behind the convex mirror.
In simple words: For a distant object, both mirrors make it look smaller. However, the concave mirror flips it upside down and real, while the convex mirror keeps it right-side up and virtual.

📝 Teacher's Note: This is a good way to show that "diminished" does not always mean "convex." Concave mirrors also diminish images of far-away objects.

🎯 Exam Tip: "In front of mirror" (Concave) vs "Behind the mirror" (Convex) is a key point of comparison.

Question 12. 1. Why does a driver use a convex mirror as a rear view mirror?
2. Illustrate your answer with the help of ray diagram.
Answer: See Q. No. 8 of Exercise (2).
In simple words: It gives a wide view and keeps the images upright.

📝 Teacher's Note: This is a repeat question from the source material, emphasizing its importance.

Question 13. 1. What is a real image?
2. What type of mirror is used to obtain a real image-of an object?
3. Does the mirror named by you above give real images for all locations of object?
Answer:
1. Real image : When the rays of light diverging from a point, after reflection of refraction, actually converge at some point, then that point is the real image of the object.
2. Concave mirror is used to obtain the real image of an object.
3. Concave mirror can not give real image for all the locations of object.
In simple words: A real image is made of light rays that actually meet at a point. Concave mirrors make them, except when the object is very close.

📝 Teacher's Note: Reinforce that virtual images are formed by rays that *appear* to meet when produced backwards.

Question 14. In the figure is shown a concave mirror. A is a point on the principal axis. If an object O is kept at A, image is formed on A itself. Copy the diagram. Draw the image in the diagram. Is the image real or virtual?
Answer: As the object is placed at A and its image is also formed at A, so object must be at centre of curvature (C). PA is called radius of curvature and on measuring PA = 4.8 cm. Point B marked on principal axis is called principal focus.
The image is real.
In simple words: When you put something at the center point of the mirror's curve, it creates an upside-down twin at the exact same spot.

📝 Teacher's Note: This is a unique case where the object distance equals the image distance (\( u = v = 2f \)).

🎯 Exam Tip: Measuring the diagram might be required. In this example, \( R = 4.8 \text{ cm} \), so \( f = 2.4 \text{ cm} \).

Question 15. An object OA is placed on the principal axis of a concave mirror as shown in the figure. Copy and complete the diagram to show the formation of image.
Answer:
In simple words: When you put the object between the focus and the center, the mirror makes a bigger, upside-down image further back.

📝 Teacher's Note: This is the reciprocal of the case where the object is beyond C. The positions of object and image simply swap.

Question 16. Copy the figure and complete it, by drawing two rays to show the formation of the image of the object AB. State the size, position and nature of image formed.
Answer:
Nature : Image formed is virtual and erect.
Position : Image is formed between principal focus (F) and pole.
Size : Image formed is diminished.
In simple words: This setup describes a convex mirror. It always makes things look smaller, upright, and tucked inside the mirror.

📝 Teacher's Note: The answer text describing "virtual, erect, diminished" confirms this is a Convex Mirror diagram.

🎯 Exam Tip: For convex mirrors, don't worry about the object's distance; the nature of the image is always the same.