Selina Concise Solutions for ICSE Class 9 Physics Chapter 7 Reflection Of Light

Exercise 7(A)

Question 1S. Define reflection of light.
Answer: The return of light into the same medium after striking a surface is called reflection.
In simple words: Reflection is when light hits a surface and bounces back, just like a ball bounces off a wall.

📝 Teacher's Note: Use the analogy of a rubber ball bouncing off a smooth floor versus a carpet to explain how the surface affects reflection.

🎯 Exam Tip: The phrase "return of light into the same medium" is the most important part of this definition.

Question 2S. Which surface reflects most of the light incident on it?
Answer: Black silvered surface reflects most of the light incident on it.
In simple words: A surface coated with a special shiny silver layer is the best at reflecting light.

📝 Teacher's Note: Though the source says "Black silvered", in practice, mirrors use a highly polished silver or aluminum coating protected by paint.

🎯 Exam Tip: Silver is considered the best reflector of light among metals.

Question 3S. Define the following terms related to reflection:
(a) Plane mirror
(b) Incident ray
(c) Reflected ray
(d) Angle of incidence
(e) Angle of reflection
Answer:
(a) Plane mirror: Plane mirror is a highly polished and smooth reflecting surface made from a clear plane glass sheet, usually thin and silvered with suitable reflecting abrasive (for example, mercury) on one side. Once this pasting is done, then the glass becomes opaque but due to the reflecting property of the abrasive, the plane glass sheet becomes a plane glass reflector or a plane glass mirror.
(b) Incident ray: The light ray striking a reflecting surface is called the incident ray.
(c) 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.
(d) 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 \( i \).
(e) Angle of reflection: The angle which the reflected ray makes with the normal at the point of incidence is called the angle of reflection. It is denoted by the letter \( r \).
In simple words: A mirror has a front reflecting side and a back coating. The light coming in is the incident ray, and the light bouncing out is the reflected ray. We measure the angles of these rays from a dotted line in the middle called the "Normal."

📝 Teacher's Note: Always emphasize that angles in optics are measured from the Normal, not the mirror surface. This is a very common mistake.

🎯 Exam Tip: Draw clear diagrams with arrows on rays to show the direction of light. Label the point 'O' as the point of incidence.

Question 4S. Differentiate between regular and irregular reflection.
Answer: Regular reflection occurs when a beam of light falls on a smooth and polished surface and irregular reflection occurs when a beam of light falls on a rough surface. Since the surface is uneven, from different points light rays get reflected in different directions and give rise to irregular reflection.
In simple words: Regular reflection is like light bouncing off a calm lake—you get a clear image. Irregular reflection is like light hitting bumpy ground—it scatters everywhere, so you can see the surface but not a reflection.

📝 Teacher's Note: Explain that irregular reflection is the reason why we can see non-luminous objects around us, like a wall or a book, from all directions.

🎯 Exam Tip: Mention that in irregular reflection, the rays are parallel when they hit the surface but are not parallel after reflection.

Question 5S. Give examples of regular and irregular reflection.
Answer: Reflection of light from a plane mirror is regular reflection and reflection of light from plane sheet of paper is irregular reflection of light.
In simple words: A shiny mirror gives a perfect bounce (regular), while a piece of paper scatters light in many ways (irregular).

📝 Teacher's Note: Ask students if they can see their face in a piece of paper. Since they can't, it proves the reflection is irregular/diffuse.

🎯 Exam Tip: Polished metal is another good example of regular reflection.

Question 6S. State the laws of reflection.
Answer:
Laws of reflection:
1. The angle of incidence is equal to the angle of reflection.
2. The incident ray, the reflected ray and the normal at the point of incidence, lie in the same plane.
In simple words: First, light bounces off at the same angle it came in. Second, the light and the middle line all stay on the same flat surface, like a drawing on a piece of paper.

📝 Teacher's Note: Demonstrate the "same plane" law by using a laser and a sheet of paper. If you tilt the paper, the reflected spot disappears from that plane.

🎯 Exam Tip: These laws are universal and apply to both regular and irregular reflection.

Question 7S. Describe an experiment to verify the laws of reflection.
Answer: Fix a white sheet of paper on a drawing board and draw a line \( MM_1 \) as shown in figure. On this line, take a point O nearly at the middle of it and draw a line OA such that \( \angle MOA \) is less than \( 90^\circ \). Then draw a normal ON on line \( MM_1 \) at the point O, and place a small plane mirror vertical by means of a stand with its silvered surface along \( MM_1 \).
Next fix two pins P and Q at some distance (\(\approx 5 \text{ cm}\)) apart vertically on line OA, on the board. Keeping eye on the other side of normal (but on the same side of mirror), see clearly images P’ and Q’ of the pins P and Q. Next fix a pin R such that it is in line with the images of pins P and Q as observed in the mirror. Next, fix one more pin S such that the pin S is in line with the pin R as well as images P’ and Q’ of pins P and Q.
Draw small circles on paper around the positions of pins as shown in figure. Remove the pins and draw a line OB joining the pin points S and R, which meets the surface of mirror at O. The angles AON and BON are measured and recorded.
The experiment is then repeated for the angle of incidence \( \angle AON \) equal to \( 40^\circ, 50^\circ, 60^\circ \). From results, it is observed that angle of incidence is equal to the angle of reflection. This verifies the first law of reflection.
The experiment has been performed on a flat drawing board, with mirror normal to the plane of board on which white sheet of paper is being fixed. Since the lower tips of all the pins also lie on the same plane (i.e., the plane of paper), it proves the second law of reflection.
In simple words: We place pins on a line hitting a mirror and then place more pins where we see the reflections. When we measure the angles between these pins and the center line, they are always identical.

📝 Teacher's Note: Emphasize that the pins must be placed vertically and at least 5 cm apart to ensure the lines are drawn accurately.

🎯 Exam Tip: When describing the experiment, clearly state the two separate conclusions that prove Law 1 (\( \angle i = \angle r \)) and Law 2 (same plane).

Question 8S. What is the angle of incidence and reflection for a light ray falling normally on a plane mirror?
Answer:
(a) \( 0^\circ \)
(b) Same as the incident ray (The ray retraces its path).
In simple words: If you shine a light straight down into a mirror, it bounces straight back up. Both the incoming and outgoing angles are zero.

📝 Teacher's Note: This is a special case. Since the incident ray lies on the normal, the angle between them is zero.

🎯 Exam Tip: "Normal incidence" means \( \angle i = 0^\circ \). Be careful not to say \( 90^\circ \), which is the angle with the mirror surface, not the normal.

Question 10S. A light ray strikes a mirror such that it makes an angle of \( 30^\circ \) with the mirror surface. Calculate: (a) Angle of incidence (b) Angle between incident and reflected ray.
Answer:
(a) Angle of incidence = \( 90^\circ - 30^\circ = 60^\circ \)
(b) Angle between the incident ray and reflected ray = Angle of incidence + Angle of reflection
Angle of reflection = Angle of incidence = \( 60^\circ \)

\( \implies \) Angle between incident and reflected ray = \( 60^\circ + 60^\circ = 120^\circ \)
In simple words: The total corner between the mirror and the center line is 90 degrees. If the ray is 30 degrees from the mirror, it must be 60 degrees from the center line. The bounce is the same, so the total gap between the rays is 60 + 60 = 120.

📝 Teacher's Note: The angle between the mirror surface and the incident ray is called the "Glancing Angle". The Angle of Incidence is \( 90^\circ - \text{Glancing Angle} \).

🎯 Exam Tip: Always subtract the given angle from \( 90^\circ \) if the angle is measured from the *surface* rather than the *normal*.

Question 11S. Draw a ray diagram to show the formation of a point image by a plane mirror. Why is the image called virtual?
Answer:
(a) and (b) formation of image diagram:
(c) The image formed is virtual because the reflected rays do not actually meet; they only appear to meet when produced backwards.
In simple words: The light doesn't really go behind the mirror to make a picture. It just bounces into your eyes in a way that tricks your brain into thinking there is an object inside the mirror.

📝 Teacher's Note: A "Virtual Image" cannot be caught on a screen. Use this to differentiate it from images in a cinema or projector (Real Images).

🎯 Exam Tip: Use dashed lines for all rays behind the mirror to indicate they are "imaginary" or "produced" rays.

Exercise 7(A)

Question 13S. (a) Three characteristics of image formed by plane mirror:
1. Image formed in erect (upright)
2. Image formed is virtual
3. Image formed is of the same size as the object
(b) The image is situated at the same perpendicular distance behind the mirror as the object in front of it.
Answer: (a) Three characteristics of image formed by plane mirror:
1. Image formed is erect (upright)
2. Image formed is virtual
3. Image formed is of the same size as the object
(b) The image is situated at the same perpendicular distance behind the mirror as the object in front of it.
In simple words: When you look in a flat mirror, your reflection is right-side up, the exact same size as you, and looks like it is just as far "inside" the mirror as you are standing outside.

📝 Teacher's Note: Use a simple mirror in class and have students move closer and farther away to observe that their image does the same. This reinforces the concept of equal distance.

🎯 Exam Tip: "Virtual," "Erect," and "Same size" are the three mandatory keywords for this question. Practice drawing the ray diagram for these points as well.

Question 14S. Differentiate between a real image and a virtual image.
Answer:

In simple words: A real image can be projected onto a wall (like in a movie theater) and is usually upside down. A virtual image is like the one you see in a bathroom mirror; it looks like it is there, but you cannot catch it on a piece of paper.

📝 Teacher's Note: Explain that "produced backwards" means the light rays don't actually go behind the mirror; our brain just follows the straight lines to imagine an intersection point.

🎯 Exam Tip: The ability to be "obtained on a screen" is the most fundamental difference often tested in one-mark questions.

Question 15S. Define lateral inversion and show the image formation of a letter P in a plane mirror.
Answer: The interchange of the left and right sides in the image of an object in a plane mirror is called lateral inversion.
The letter P appears in the plane mirror as q.
In simple words: Lateral inversion is when a mirror makes your left side look like your right side. It flips the image sideways, which is why a 'P' looks like a 'q' in the mirror.

📝 Teacher's Note: Ask students to write their name on a piece of paper and hold it up to a mirror. This is the easiest way for them to experience lateral inversion firsthand.

🎯 Exam Tip: When defining lateral inversion, always use the terms "interchange" of "left and right sides."

Question 16S. Why are the letters on the front of an ambulance written laterally inverted?
Answer: The letters on the front of a ambulance are written laterally inverted, so that the driver of the vehicle moving ahead of the ambulance reads these words laterally inverted as AMBULANCE, in his rear view mirror, and gices side to pass the ambulance first.
In simple words: Since mirrors flip things sideways, writing "Ambulance" backward on the van makes it look perfectly normal in a driver's rear-view mirror so they can get out of the way quickly.

📝 Teacher's Note: This is a real-world safety application of physics. Point out that the word needs to be readable at a glance for drivers in stressful traffic.

🎯 Exam Tip: Explain that the driver in front sees a "laterally inverted image of a laterally inverted text," which results in the text appearing normal.

Question 17S. Why is it difficult to read a page of text in a plane mirror?
Answer: Due to lateral inversion, , it becomes difficult to read the image of the text of a page formed due to reflection by a plane mirror.
In simple words: Mirrors swap the left and right sides of everything, so letters like 'b' become 'd' and sentences look like they are written in a secret backward code.

📝 Teacher's Note: Explain that symmetrical letters like 'A', 'H', and 'O' don't seem to change because their left and right sides are identical.

🎯 Exam Tip: The scientific term to use for this phenomenon is always "Lateral Inversion."

Question 1M. For a plane mirror, the relationship between the angle of incidence (i) and the angle of reflection (r) is
(a) \( i > r \)
(b) \( i = r \)
(c) \( i < r \)
(d) \( i \neq r \)
Answer: (b) \( i = r \)
In simple words: The angle light hits a mirror is exactly the same as the angle it bounces off — it is a perfect match.

📝 Teacher's Note: This law holds true for both smooth mirrors and rough surfaces at the microscopic level.

🎯 Exam Tip: Angles of incidence and reflection are always measured from the "Normal" (the perpendicular line to the mirror surface).

Question 2M. The image formed by a plane mirror is always
(a) Real and of same size
(b) Inverted and smaller
(c) Erect and of same size
(d) Erect and larger
Answer: (c) Erect and of same size
In simple words: A flat mirror doesn't make you look upside down or change your size; it shows you exactly as you are.

📝 Teacher's Note: "Erect" is the formal scientific term for an image that is right-side up.

🎯 Exam Tip: Plane mirrors never magnify or diminish an object; the magnification is always 1.

Question 3M. A plane mirror forms a _____ image.
(a) real with lateral inversion
(b) virtual with lateral inversion
(c) real without lateral inversion
(d) virtual without lateral inversion
Answer: (b) virtual with lateral inversion
In simple words: The image looks real but isn't, and your left side appears on the right side.

📝 Teacher's Note: This summaries the two most important technical features of plane mirror reflections.

🎯 Exam Tip: Always look for the combination of "Virtual" and "Lateral Inversion" for plane mirrors.

Question 1N. If the sum of the angle of incidence (i) and angle of reflection (r) is \( 90^\circ \), find the value of i.
Answer: Angle of incidence (i) + Angle of reflection(r) = \( 90^\circ \)
But, as per the laws of reflection, \( i = r \)
Therefore, \( 2 i = 90^\circ \)
Or, \( i = r = 45^\circ \)
In simple words: Since the incoming and outgoing angles are always equal, if they add up to 90, each one must be half of that, which is 45.

📝 Teacher's Note: This is a standard geometric application of the First Law of Reflection. Ensure students state \( i=r \) before solving.

🎯 Exam Tip: Showing the step \( 2i = 90^\circ \) helps secure partial marks even if you make a calculation error.

Question 2N. The distance between a man and his image in a plane mirror is 6 m. How far is the man from the mirror?
Answer: Distance between man and his image = 6m
Distance between man and mirror + distance between mirror and image = 6m
But, Distance between man and mirror (object distance) = distance between mirror and image (image distance)
Therefore, distance of man from mirror = \( 6/2 = 3\text{m} \)
In simple words: The total gap is from you to the mirror and then from the mirror to the image. Since those two parts are equal, just divide the total gap by two.

📝 Teacher's Note: Remind students that the mirror is always the exact midpoint between the object and its image.

🎯 Exam Tip: Object distance (\( u \)) is always equal to image distance (\( v \)) for plane mirrors.

Question 3N. An insect is sitting 1 m in front of a mirror. (a) Where is its image formed? (b) Find the distance between the insect and its image.
Answer: (a) Image of the insect is formed 1m behind the mirror.
(b) Distance between the insect and his image = \( 1 + 1 = 2 \text{ m} \)
In simple words: The image forms just as deep inside the mirror as the insect is standing outside. So, the twin is 1m in and the insect is 1m out, making a 2m total gap.

📝 Teacher's Note: Use a ruler to show that if the object moves, the image moves the same amount on the other side.

🎯 Exam Tip: Pay attention to whether the question asks for distance from the "mirror" or from the "image."

Question 4N. An object is placed 60 cm from a plane mirror. If the mirror is moved 25 cm away from the object, find the shift in the position of the image.
Answer: Initially, distance of the object from the mirror = 60 cm.
Therefore, image is formed at a distance 60 cm from the mirror, behind it.
Thus, initial distance between the object and image = \( 60 + 60 = 120 \text{ cm} \)
If the mirror is moved 25 cm away from the object,
The new distance of the object from the mirror = \( 60 + 25 = 85 \text{ cm} \)
The new image is now at a distance 85 cm from the mirror behind it.
Thus, new distance of the image from the object = \( 85 + 85 = 170 \text{ cm} \)
Taking the position of the object as reference point, the distance between the two positions of the image = new distance of image from the object – initial distance of the image from the object
= \( (170 – 120) \text{ cm} = 50 \text{ cm} \)
Thus, the image shifts 50 cm away.
In simple words: If you move a mirror 25 cm away from yourself, your reflection seems to jump 50 cm away. This is because both the distance to the mirror and the reflection depth increase at once.

📝 Teacher's Note: A useful shortcut for students: Image shift = \( 2 \times \text{Mirror shift} \) for a stationary object.

🎯 Exam Tip: Calculating the initial and final total distances (\( 120 \text{ cm} \) and \( 170 \text{ cm} \)) is the safest way to avoid mistakes in this problem.

Question 5N. A patient is sitting 3 m from a chart. A plane mirror is fixed 2 m in front of the patient. Find the distance of the chart as seen by the patient.
Answer: Distance between man and chart = 3m
Distance between man and mirror = 2m
Therefore, distance between chart and mirror = 5 m
Now, final image is formed on the mirror, which is at a distance of 2 m from the man, therefore, the chart as seen by patient is \( (5\text{m} + 2\text{m} =) 7\text{m} \) away.
In simple words: The patient looks at the mirror 2m away. The reflection of the chart (which is 5m from the mirror) is 5m deep inside. So the total look distance is 2m plus the 5m depth, equaling 7m.

📝 Teacher's Note: This is a classic eye-doctor setup used to double the effective testing distance in a small room.

🎯 Exam Tip: Draw a simple line diagram with the patient, mirror, and chart marked to help visualize the additions.

Exercise 7(B)

Question 1S. How is the number of images formed by two inclined mirrors calculated?
Answer: If two mirrors make an angle \( \theta \) with each other and object is placed in between the two mirrors, the number of images formed is \( n \) or \( (n – 1) \) depending upon \( n = 360^\circ / \theta^\circ \) is odd or even.
(a) If \( n = 360^\circ / \theta^\circ \) is odd,
(i) The number of images formed is \( n \), when the object is placed asymmetrically between the mirrors.
(ii) The number of images formed is \( n-1 \), when the object is placed symmetrically between the mirrors.
(b) If \( n = 360^\circ / \theta^\circ \) is even, the number of images is always \( n-1 \).
In simple words: The narrower the angle between mirrors, the more "clones" you see. We divide 360 by the angle to find the number, then sometimes subtract one if the images overlap.

📝 Teacher's Note: Use the example of 90 degrees: \( 360/90 = 4 \). Since 4 is even, we see \( 4-1 = 3 \) images.

🎯 Exam Tip: Memorize the formula \( n = (360/\theta) - 1 \) for the even case, as it is the most frequently tested part of this law.

Question 2S. What happens to the number of images if the angle between the mirrors is gradually increased?
Answer: The number of images formed is given as \( n = \frac{360}{\theta} \). So, if \( \theta \) is gradually increased, \( n \) decreases.
In simple words: If you open the mirrors wider, the reflections have less space to bounce around, so you see fewer and fewer images.

📝 Teacher's Note: This is an inverse relationship. Bigger denominator (\( \theta \)) equals smaller result (\( n \)).

🎯 Exam Tip: State that the number of images is "inversely proportional" to the angle between the mirrors.

Question 3S. How many images are formed when two mirrors are kept perpendicular to each other? Show with a diagram.
Answer: For two mirrors kept perpendicular to each other, three images are formed for an object kept in between them.
In simple words: When mirrors are at a right angle (like a corner), you see three images. One in each mirror and one tucked right in the corner where they meet.

📝 Teacher's Note: The fourth image overlap occurs because the image of the first reflection falls on the second mirror, and vice-versa, creating a coincident image.

🎯 Exam Tip: Note that for perpendicular mirrors, \( 360/90 = 4 \), which is even, so images \( = 4-1 = 3 \).

Question 4S. How many images are formed when two mirrors are kept parallel to each other?
Answer: For two mirrors kept parallel to each other, an infinite number of images are formed for an object kept in between them.
In simple words: This is like an "infinite tunnel" effect. The light bounces back and forth forever between the mirrors, making a never-ending line of reflections.

📝 Teacher's Note: Use the example of a barber shop or a dress fitting room where mirrors on opposite walls create this effect.

🎯 Exam Tip: When \( \theta = 0^\circ \), then \( n = \infty \). Mention "infinite" or "very large number" as the answer.

Question 5S. State two uses of a plane mirror.
Answer: Two uses of plane mirror:
1. In barber’s shop for seeing the hairs at the back of head, two mirrors facing each other are fixed on opposite walls at the front and back of the viewer.
2. In solar heating devices such as a solar cooker, solar water heater, etc., a plane mirror is used to reflect the incident light rays from sun on the substance to be heated.
In simple words: We use mirrors to see parts of ourselves we can't normally see (like the back of our heads) and to catch sunlight to heat up food.

📝 Teacher's Note: Point out that in a solar cooker, the mirror acts as a light concentrator.

🎯 Exam Tip: Using "solar heating devices" is a high-mark scientific example compared to just saying "looking at ourselves."

Question 1M. How many images are formed for two mirrors at \( 60^\circ \)?
(a) 3
(b) 4
(c) 5
(d) 6
Answer: (c) 5
In simple words: We calculate \( 360/60 = 6 \). Since 6 is an even number, we take one away to get 5 total images.

📝 Teacher's Note: This is the most common MCQ on this topic. Ensure students always check if the division result is even.

🎯 Exam Tip: Remember: for even quotients, the answer is always \( n-1 \).

Question 2M. In a barber’s shop, two plane mirrors are placed
(a) Perpendicular to each other
(b) Parallel to each other
(c) At 45 degrees
(d) At 60 degrees
Answer: (b) Parallel to each other
In simple words: Placing them parallel lets you see the reflection of your back on the mirror in front of you.

📝 Teacher's Note: This creates the "infinite reflection" tunnel needed to see 360 degrees around a person's head.

🎯 Exam Tip: This is a standard application-based question for parallel mirrors.

Question 1N. Calculate the number of images formed when the angle between the mirrors is (a) \( 90^\circ \) and (b) \( 60^\circ \).
Answer: (a) Angle between the mirrors, \( \theta = 90^\circ \)
Now, \( n = 360^\circ / \theta^\circ = 360^\circ / 90^\circ = 4 \), which is even.
Hence number of images formed will be \( (n-1) \); i.e., \( 4-1 = 3 \) images

(b) Angle between the mirrors, \( \theta = 60^\circ \)
Now, \( n = 360^\circ / \theta^\circ = 360^\circ / 60^\circ = 6 \), which is even.
Hence number of images formed will be \( (n-1) \); i.e., \( 6-1 = 5 \) images
In simple words: Just divide 360 by the mirror angle. If the result is an even number, subtract one to get the final count.

📝 Teacher's Note: These are the two most fundamental calculations for this sub-chapter. Practice them until they are second nature.

🎯 Exam Tip: Always state the formula and whether the quotient is even or odd for full credit.

Question 2N. Find the number of images for mirrors at \( 50^\circ \) for (i) asymmetric and (ii) symmetric object placement.
Answer: Angle between the mirrors, \( \theta = 50^\circ \)
Now, \( n = 360^\circ / \theta^\circ = 360^\circ / 50^\circ = 7.2 \), which we round to the odd integer 7.
(i) When placed asymmetrically, number of images formed will be n, i.e. 7.
(ii) When placed symmetrically, number of images formed will be \( (n-1) \); i.e. \( 7-1 = 6 \) images
In simple words: When the division isn't perfect, we look at the whole number. If the object is right in the middle, two images merge together, so you see one less.

📝 Teacher's Note: This is an advanced case. Symmetric means the object is on the angle bisector. Asymmetric means it is off-center.

🎯 Exam Tip: For odd quotients, the number of images depends on the position of the object.

Exercise 7(C)

Question 1S. What is a spherical mirror?
Answer: A reflecting surface which is a part of a sphere is called a spherical mirror.
In simple words: Imagine cutting a piece out of a glass ball and making it shiny; that curved piece of glass is a spherical mirror.

📝 Teacher's Note: Use a shiny spoon as a real-life example of a spherical mirror — it has a concave side and a convex side.

🎯 Exam Tip: Mention that the mirror is a "part of a hollow sphere" for a complete definition.

Question 2S. Name the two types of spherical mirrors and state the difference between them.
Answer: Two kinds of spherical mirrors are concave and convex.
Distinction between concave and convex mirror: A concave mirror’s bulging surface is silvered and reflection takes place from the hollow surface but a convex mirror’s inner surface is silvered and reflection takes place from the bulging surface.
In simple words: A concave mirror is like a "cave"—you go into the hollow shiny part. A convex mirror is like a ball—it bulges out towards you.

📝 Teacher's Note: Mnemonic: "Concave" has a "Cave" in it. The reflecting surface is "caved in."

🎯 Exam Tip: When drawing, use short dashes to show the silvered/non-reflecting side.

Question 3S. Define the following terms: (a) Pole, (b) Principal axis, (c) Centre of curvature.
Answer:
Pole: The geometric centre of the spherical surface of mirror is called the pole of mirror.
Principal axis: It is the straight line joining the pole of the mirror to its centre of curvature.
Centre of curvature: The centre of curvature of a mirror is the centre of the sphere of which the mirror is a part.
In simple words: The Pole is the dead center of the mirror's face. The Centre of Curvature is where the middle of the original glass ball would have been. The Principal Axis is a long line connecting both points.

📝 Teacher's Note: Use a diagram to show these points. The Pole (\( P \)) is on the surface, while the Centre (\( C \)) is in space in front of or behind the mirror.

🎯 Exam Tip: These definitions are frequent 1-mark theory questions. Ensure they are learned accurately.

Question 5S. Explain how concave and convex mirrors affect a beam of light parallel to the principal axis.
Answer:
(i) Convex mirror diverges a beam of light falling on it.
(ii) Concave mirror converges a beam of light falling on it.
In simple words: A concave mirror acts like a "collector," bringing parallel light rays together at one spot. A convex mirror acts like a "scatterer," spreading the rays out as if they are coming from a point behind the mirror.

📝 Teacher's Note: Use the analogy of a spoon. The front side (caved in) is a concave mirror that focuses light, while the back side (bulging out) is a convex mirror that spreads light out.

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

Question 6S. Define the focus and focal length of a concave mirror.
Answer:
Focus of a concave mirror: The focus of a concave mirror is a point on the principal axis through which the light rays incident parallel to principal axis, pass after reflection from the mirror.
Focal length of a concave mirror: The distance of the focus from the pole of the concave mirror is called its focal length.
In simple words: The Focus is the "meeting point" where all parallel light rays gather after bouncing off the mirror. The Focal Length is just the distance from the center of the mirror to that meeting point.

📝 Teacher's Note: Explain that the focus of a concave mirror is a "real" focus because the light rays actually cross each other at that point.

🎯 Exam Tip: Remember the relationship \( f = R/2 \). The focal length is always exactly half of the radius of curvature.

Question 7S. Define the focus and focal length of a convex mirror.
Answer:
Focus of a convex mirror: The focus of a convex mirror is a point on the principal axis from which, the light rays incident parallel to principal axis, appear to come, after reflection from the mirror.
Focal length of a convex mirror: The distance of the focus from the pole of the convex mirror is called its focal length.
In simple words: For a convex mirror, light rays scatter away. If you follow those scattered lines backward into the mirror, they seem to meet at one point called the Focus. The distance to this "imaginary" point is the focal length.

📝 Teacher's Note: Contrast this with the concave mirror. The focus here is "virtual" because the light rays don't actually pass through it; they only appear to originate from it.

🎯 Exam Tip: When defining the focus of a convex mirror, the phrase "appear to come from" is crucial for a technically correct answer.

Question 8S. Why does a ray of light directed towards the centre of curvature of a spherical mirror retrace its path?
Answer: Incident ray is directed towards the centre of curvature because the ray is normal to the spherical mirror, so \( \angle i = \angle r = 0 \).
In simple words: Any line from the center of a circle to its edge hits the edge at a perfect 90-degree angle. Since the ray hits the mirror straight-on, it bounces straight back the way it came.

📝 Teacher's Note: Use a geometry analogy: a radius of a circle is always perpendicular to the tangent at that point. In optics, hitting the mirror perpendicularly means the angle of incidence is zero.

🎯 Exam Tip: Be careful! Hitting the mirror "normally" means the angle of incidence is \( 0^\circ \), not \( 90^\circ \).

Question 12S. Name two convenient rays that are chosen to construct the image by a spherical mirror for a given object.
Answer:
Two convenient rays that are chosen to construct the image by a spherical mirror for a given object:

(i) A ray passing through the centre of curvature: A ray of light passing through the centre of curvature of a concave mirror or a ray directed in the direction of centre of curvature of a convex mirror is reflected back along the same path after reflection.
(ii) A ray parallel to the principal axis: A ray of light parallel to the principal axis, after reflection pass through the principal focus in case of a concave mirror or appears to diverge from it in case of convex mirror.
In simple words: To draw where an image forms, we use two easy rules: one ray that goes through the center (it just bounces straight back) and one ray that goes straight across (it bounces through the focus point).

📝 Teacher's Note: We only need two rays to find the intersection point, which determines the image position. Using these two standard rays makes drawing diagrams much simpler and more accurate.

🎯 Exam Tip: When drawing Ray (ii) for a convex mirror, ensure you use a dashed line behind the mirror to show where the ray "appears" to go toward the focus.

Question 16S. Describe the nature and position of the image formed by a concave mirror when the object is placed between the pole and the focus.
Answer: The image formed is virtual, erect and magnified.
In simple words: When you put something very close to a concave mirror, you see a giant, right-side-up reflection that looks like it's inside the mirror. This is how makeup or shaving mirrors work.

📝 Teacher's Note: This is the only position for a concave mirror that produces a virtual and erect image. In all other positions (beyond F), the image is real and inverted.

🎯 Exam Tip: "Magnified" means the image is larger than the object. Be prepared to state the use of the mirror in this case (e.g., dentist's mirror).

Question 17S. Describe the nature and position of the image formed by a concave mirror when the object is placed between the centre of curvature and the focus.
Answer: The image formed is real, inverted and magnified.
In simple words: If the object is a bit further away, the mirror projects a large, upside-down image that is further back than the object itself.

📝 Teacher's Note: Explain that since the image is "real," it can be caught on a screen placed at the image position.

🎯 Exam Tip: Note the trend: as the object moves from F towards C, the real image moves from infinity towards C and decreases in size (though it remains magnified until it reaches C).

Question 19S. What type of images are always produced by a convex mirror?
Answer: Convex mirror always produces erect and virtual images. The images formed are diminished, i.e. the size of the image is shorter than the size of the object.
In simple words: A convex mirror always makes things look smaller and right-side up. This allows you to see a much wider area in a small mirror.

📝 Teacher's Note: This is why convex mirrors are used as rear-view mirrors in cars—they give a "wide-angle" view of the traffic behind.

🎯 Exam Tip: If the question asks "Which mirror always forms a virtual image?", the answer is Convex mirror (and Plane mirror). If it asks "Which mirror always forms a diminished image?", the answer is Convex mirror.

Question 20S. (a) Describe the image formed when an object is placed between the pole and focus of a concave mirror. (b) What is the nature of this image?
Answer:
(a) If the object is placed between the pole and focus of a concave mirror, the image formed is magnified and erect.
(b) The image is virtual.
In simple words: Putting an object very close to a concave mirror creates a zoomed-in, right-side-up image that you can't touch.

📝 Teacher's Note: This specific case is used in dentistry to see small cavities in teeth more clearly.

🎯 Exam Tip: This is the only case where a concave mirror produces a virtual image. It's a very frequent exam topic.

Question 21S. (a) Describe the image formed when an object is placed at the centre of curvature of a concave mirror. (b) What is its nature?
Answer:
(a) If the object is placed at the centre of curvature of a concave mirror, the image formed is of same size.
(b) The image formed is real and inverted.
In simple words: At this one special spot (the center), the mirror reflects an image that is exactly the same height as the object, but completely upside down.

📝 Teacher's Note: This is a "unity" case where the magnification is exactly -1. The object and image are at the same distance from the mirror.

🎯 Exam Tip: Mention that the image is formed exactly at the "centre of curvature" (\( C \)) itself.

Question 22S. (a) Define a real image. (b) Which spherical mirror can produce it? (c) Does it form a real image for all object locations?
Answer:
(a) An image which can be obtained on a screen is called a real image.
(b) A concave mirror can be used to obtain a real image of an object.
(c) No, it does not form real image for all locations of the object.
In simple words: A real image is like a movie projected on a wall. Concave mirrors can do this, but not if the object is too close (inside the focus point).

📝 Teacher's Note: Real images are formed by the actual meeting of reflected rays. Virtual images are formed when rays only *appear* to meet.

🎯 Exam Tip: For part (c), clarify that when the object is between the pole and focus, the image becomes virtual.

Question 23S. How does the image in a concave mirror change as an object is moved from infinity towards the pole?
Answer: When an object is moved from infinity towards the pole of mirror, the image formed moves away from the mirror. The image formed is real and inverted.
In simple words: As you bring an object closer to the mirror, its upside-down reflection starts small at the focus and grows larger as it moves further away from the mirror.

📝 Teacher's Note: Note that once the object crosses the focus point (\( F \)), the image behavior completely flips—it jumps to the back of the mirror and becomes virtual and erect.

🎯 Exam Tip: The image "moves from focus to infinity" as the object "moves from infinity to focus."

Question 24S. Describe the general characteristics of images formed by a convex mirror.
Answer: In a convex mirror, the image formed is always virtual, upright and diminished. It is always situated between its pole and focus, irrespective of the distance of object in front of the mirror.
In simple words: No matter where you put an object in front of a convex mirror, its reflection will always be small, right-side up, and stuck in the space behind the mirror.

📝 Teacher's Note: This "always diminished" property is what gives the mirror its wide field of view, making it essential for security and driving.

🎯 Exam Tip: The phrase "irrespective of the distance" is a key technical detail to include.

Question 25S. Identify the type of spherical mirror used in the following cases:
(a) Shaving mirror
(b) Doctor's head mirror
(c) Rear-view mirror in vehicles
(d) Reflector in searchlights
Answer:
(a) Concave, (b) Concave, (c) Convex and (d) Concave
In simple words: We use concave mirrors when we need to zoom in or create a strong beam of light. We use convex mirrors when we need to see a wide, zoomed-out view of the road.

📝 Teacher's Note: Shaving mirrors are concave to show a magnified image. Rear-view mirrors are convex to cover more area behind the car.

🎯 Exam Tip: This is a classic "application-based" question. Memorize this list as it appears almost every year in exams.

Question 26S. Focal length is half the radius of curvature of a spherical mirror.
Answer: \( f = \frac{R}{2} \)
In simple words: The focal point of a curved mirror is always exactly halfway between the mirror's surface and its center point.

📝 Teacher's Note: This relationship is a fundamental property of paraxial rays. It shows that the curvature of the mirror directly determines where it focuses light.

🎯 Exam Tip: This formula is often used as a starting step in numerical problems. Always check if the question gives you the radius (\( R \)) or the focal length (\( f \)).

Question 27S. State the spherical mirror's formula.
Answer: The spherical mirror's formula is
\( \frac{1}{u} + \frac{1}{v} = \frac{1}{f} \)
Here, \( u \) is the object distance, \( v \) is the image distance and \( f \) is the focal length of the mirror.
In simple words: This equation allows us to calculate exactly where an image will appear if we know how far the object is and how curved the mirror is.

📝 Teacher's Note: Remind students to always use the standard sign convention (Cartesian sign convention) when substituting values into this formula.

🎯 Exam Tip: When writing the formula, clearly define what \( u \), \( v \), and \( f \) stand for to get full marks for the theory part.

Question 28S. Define magnification for a spherical mirror.
Answer: Magnification is the ratio of the length of image to the length of the object.
\( m = \frac{I}{O} \)
It is also given as
\( m = -\frac{v}{u} \)
Where, \( u \) and \( v \) is the object and image distance, respectively.
Hence, we have
a. For real image: \( u \) and \( v \) are negative. So, \( m \) is negative.
b. For virtual image: \( u \) is negative and \( v \) is positive. So, \( m \) is positive.
In simple words: Magnification tells us how many times bigger or smaller the reflection is compared to the real thing. A negative answer means the image is upside down.

📝 Teacher's Note: A magnification of 1 means the image is the same size. If \( |m| > 1 \), the image is enlarged; if \( |m| < 1 \), it is diminished.

🎯 Exam Tip: The sign of \( m \) is a quick way to identify the nature of the image: negative for real/inverted and positive for virtual/erect.

Question 29S. What is the maximum image distance for a convex mirror?
Answer: The image formed by a convex mirror is always between pole and focus. Hence, the maximum distance that can be obtained in convex mirror is the focal length. For this case the object has to be at infinity.
In simple words: No matter how far away an object is, a convex mirror will always squeeze its reflection into the small space between the mirror and its focal point.

📝 Teacher's Note: This property is why convex mirrors are useful for surveillance; they show a very wide area in a compact reflection.

🎯 Exam Tip: For a convex mirror, the image distance \( v \) will always be less than or equal to \( f \).

Question 30S. At what position of the object does a concave mirror form an image at the maximum distance?
Answer: The maximum distance that can be obtained in concave mirror is infinity. For this case the object has to be at focus.
In simple words: If you put an object exactly at the focal point of a concave mirror, the reflected light rays go perfectly straight and never meet, meaning the image forms "at infinity."

📝 Teacher's Note: This is the principle used in searchlights and car headlights to create a powerful, straight beam of light.

🎯 Exam Tip: Remember: Object at \( F \implies \) Image at \( \infty \). Conversely, Object at \( \infty \implies \) Image at \( F \).

Question 31S. How can you distinguish between a plane mirror, concave mirror and convex mirror without touching them?
Answer: To distinguish between a plane mirror, concave mirror and convex mirror, the given mirror is held near the face and image is seen. There can be following three cases:
Case (i): If the image is upright, of same size and it does not change in size by moving the mirror towards or away from the face, the mirror is plane.
(ii) If the image is upright and magnified, and increases in size on moving the mirror away, the mirror is concave.
(iii) If the image is upright and diminished and decreases in size on moving the mirror away, the mirror is convex.
In simple words: Look closely at the mirror. If you look normal, it's a plane mirror. If you look like a giant, it's concave. If you look tiny, it's convex.

📝 Teacher's Note: This test only works when the object (the face) is placed close to the mirror (within the focal length for the curved mirrors).

🎯 Exam Tip: Use the terms "Erect," "Magnified," and "Diminished" to distinguish the three mirrors clearly.

Question 32S. State two uses of a concave mirror.
Answer: Two uses of concave mirror:
1. It is used as a shaving mirror.
2. It is used as reflector in torch, head light of automobiles etc.
In simple words: Concave mirrors are used to zoom in on your face while shaving and to focus the light from a tiny bulb into a strong beam for car headlights.

📝 Teacher's Note: In use (1), the face is placed between \( P \) and \( F \). In use (2), the bulb is placed at \( F \).

🎯 Exam Tip: For the headlight example, mention that the light source is placed at the "focus" to produce parallel rays.

Question 33S. Identify the mirror used in the following: (a) To see a magnified image of the face. (b) To see a magnified image of teeth.
Answer: (a) Concave mirror
(b) Concave mirror
In simple words: Both cases need a "zoom" effect, which only a concave mirror can provide when the object is close.

📝 Teacher's Note: Shaving mirrors and dental mirrors both rely on the "Object between \( P \) and \( F \)" case of concave reflection.

🎯 Exam Tip: Concave mirrors are the only mirrors that can produce a magnified virtual image.

Question 34S. When a person uses a concave mirror as a shaving mirror, (a) where is his face situated and (b) what is the nature of the image?
Answer: (a) The person’s face is between the pole and focus of the mirror.
(b) The image formed is erect, virtual and magnified.
In simple words: The person stands very close to the mirror. This creates a reflection that is right-side up and much larger than real life, making it easy to see every hair.

📝 Teacher's Note: This is a specific application of Case (ii) in Solution 31S.

🎯 Exam Tip: "Virtual, erect, and magnified" is the standard trio of words to describe the image in this position.

Question 35S. Why is a convex mirror preferred as a rear view mirror?
Answer: A convex mirror is preferred as a rear view mirror because it has a wider field of view as compared to a plane mirror of same size.
In simple words: Because a convex mirror makes everything look smaller, it can fit much more of the road behind you into the mirror's frame.

📝 Teacher's Note: This "wider field of view" is due to the outward curvature which allows rays from a larger angular range to reach the eye.

🎯 Exam Tip: The keyword "wider field of view" is essential for scoring full marks on this reasoning question.

Question 36S. Explain why a driver can see all the traffic approaching from behind using a convex mirror.
Answer: A convex mirror diverges the incident beam and always forms a virtual, small and erect image between its pole and focus. Thus, a driver can see all the traffic approaching from behind. This fact enables the driver to use it as a rear view in vehicles to see all the traffic approaching from behind.
In simple words: The mirror's bulge catches light from the sides and bends it toward your eyes, letting you see a broad view of the entire road behind you in one glance.

📝 Teacher's Note: Point out that images in a convex mirror are closer than they appear—this is because our brain associates smaller size with greater distance.

🎯 Exam Tip: In your diagram, show that the rays from a wide area are reflected into a narrow cone toward the eye.

Question 1M. When a ray of light falls normally on a spherical mirror, the reflected ray
(a) Passes through focus
(b) Passes through center of curvature
(c) Retraces its path
(d) None of the options
Answer: (c) Retraces its path
In simple words: "Normally" means hitting the mirror straight-on. Just like a ball thrown straight at a wall, the light bounces directly back the way it came.

📝 Teacher's Note: A normal to a spherical surface is always a radius of the sphere. Therefore, it always passes through the center of curvature.

🎯 Exam Tip: Normal incidence means the angle of incidence \( i = 0^\circ \).

Question 2M. The image formed by a convex mirror is
(a) Real and magnified
(b) Virtual and magnified
(c) Erect and diminished
(d) Inverted and diminished
Answer: (c) Erect and diminished
In simple words: A convex mirror always shows you a right-side up but shrunken version of the object.

📝 Teacher's Note: Unlike concave mirrors, a convex mirror's nature never changes regardless of the object's position.

🎯 Exam Tip: Remember: Convex = Always Diminished. This is its defining characteristic for common use cases.

Question 3M. A mirror which is used by a dentist is a
(a) Plane mirror
(b) Convex mirror
(c) Concave mirror
(d) None of the options
Answer: (c) Concave mirror
In simple words: Dentists need to see teeth much larger than they really are, and only a concave mirror can act like a magnifying glass.

📝 Teacher's Note: The tooth must be held closer than the focal length of the mirror to see the enlarged upright image.

🎯 Exam Tip: If the question asks "why," mention the ability of concave mirrors to form magnified virtual images.

Question 1N. If the radius of curvature of a spherical mirror is 40 cm, what is its focal length?
Answer: Focal length = \( \frac{1}{2} \) (Radius of curvature)
Or, \( f = 40/2 = 20 \text{ cm} \)
In simple words: The focal point is halfway to the center. So, half of 40 is 20.

📝 Teacher's Note: This is the simplest calculation in optics. Ensure students include the unit "cm" in the final answer.

🎯 Exam Tip: Always state the formula \( f = R/2 \) before doing the division.

Question 2N. If the focal length of a spherical mirror is 10 cm, what is its radius of curvature?
Answer: Radius of curvature = \( 2 \times \text{focal length} \)
Or, \( R = 2f = 2 \times 10 = 20 \text{ cm} \)
In simple words: The center of the circle is twice as far as the focus. Double 10 is 20.

📝 Teacher's Note: This is the inverse of 1N. Multiplying by 2 instead of dividing.

🎯 Exam Tip: Be careful! If it's a convex mirror, the radius is considered positive; if concave, it's negative (though the magnitude is the same).

Question 3N. An object is placed at a distance of 12 cm in front of a concave mirror of focal length 20 cm. Find the nature and position of the image.
Answer: The image is 30 cm in front of the mirror, 3 cm high, real, inverted and magnified.
In simple words: Since the object is closer to the mirror than its focal point, the reflection jumps to the other side and looks big and right-side up.

📝 Teacher's Note: Using the mirror formula: \( 1/v = 1/(-20) - 1/(-12) = -1/20 + 1/12 = (-3+5)/60 = 2/60 = 1/30 \). So \( v = +30 \text{ cm} \). Positive \( v \) means virtual and behind the mirror.

🎯 Exam Tip: When \( u < f \) for a concave mirror, the image is ALWAYS virtual, erect, and magnified.

Question 4N. Where will the image be formed if an object is placed 4 cm from a concave mirror of focal length 12 cm?
Answer: The image is 6 cm behind the mirror. Yes the image is magnified.
In simple words: Because the object is very close, the reflection appears 6 cm deep inside the mirror and looks larger than the object.

📝 Teacher's Note: This is a classic numerical testing the Cartesian sign convention. \( u = -4, f = -12 \).

🎯 Exam Tip: If the result \( v \) is positive, the image is virtual and forms "behind the mirror."

Question 5N. At what position of the object is the image size equal to the object size for a concave mirror?
Answer: The size of the image is equal to the size of the object if the object is placed at the centre of curvature of a concave mirror.
In simple words: There is one special "magic" spot at the center of the mirror's curve. If you put something there, its reflection is exactly the same height, but upside down.

📝 Teacher's Note: At this point, \( u = v = R = 2f \). The magnification \( m = -1 \).

🎯 Exam Tip: This is the most common conceptual position question for concave mirrors. Remember: Object at \( C \implies \) Image at \( C \).

Question 6N. An object is placed 50 cm from a concave mirror. Use a ray diagram to find its position.
Answer: Hence, the object should be placed at 50 cm. (Assuming this is the center of curvature case).
In simple words: This diagram shows that the mirror reflects the image exactly at the same spot but flipped.

📝 Teacher's Note: The source diagram shows the object and image at the same horizontal coordinate, implying the object is at \( C = 50 \text{ cm} \) and \( f = 25 \text{ cm} \).

🎯 Exam Tip: On a ray diagram, use a ruler for straight lines and arrows to show the direction of light.

Question 8N. A parallel beam of light hits a concave mirror. Where does it focus?
Answer: A ray passing parallel to the principal axis passes through the focal point after reflection. Hence, the focal length is 12 cm. (Assuming values from the diagram).
In simple words: All light rays coming in straight will bounce off and meet at a single spot called the Focus.

📝 Teacher's Note: This defines the fundamental optical power of a concave mirror—it converges parallel light.

🎯 Exam Tip: This ray diagram is a very frequent question. Mark the 'F' point clearly at the intersection.

Question 9. 
Answer: 
\( O = 4 \text{ cm} \)
\( u = -30 \text{ cm} \)
\( f = -15 \text{ cm} \)

From mirror formula,
\( \frac{1}{f} = \frac{1}{v} + \frac{1}{u} \)
\( \therefore \frac{1}{v} = \frac{1}{f} - \frac{1}{u} \)

\( \therefore \frac{1}{v} = \frac{1}{-15} - \frac{1}{-30} = \frac{1}{-15} + \frac{1}{30} = \frac{-2 + 1}{30} = -\frac{1}{30} \)
\( \therefore v = -30 \text{ cm} \)
Hence, the image is formed at a distance of 30 cm in front of the mirror.

\( m = -\frac{v}{u} = \frac{I}{O} \)
\( \therefore I = \frac{-Ov}{u} = \frac{-4 \times -30}{-30} = -4 \text{ cm} \)

Negative sign indicates inverted image.
So, \( I = 4 \text{ cm} \)
Hence, the length of the image is 4 cm.

In simple words: Using math, we found that the image appears exactly where the object is (30 cm away) but it is flipped upside down. It stays the same size (4 cm).

📝 Teacher's Note: This is the special case where the object is placed at the Center of Curvature (\( u = 2f \)). In such cases, the image is always formed at the same location, is real, inverted, and of the same size.

🎯 Exam Tip: When \( u = 2f \), skip long calculations if it's a multiple-choice question; the image distance will always equal the object distance.

Question 10N.
\( O = 4 \text{ cm} \)
\( u = -30 \text{ cm} \)
\( f = -15 \text{ cm} \)

Answer: From mirror formula,
\( \frac{1}{f} = \frac{1}{v} + \frac{1}{u} \)

\( \implies \frac{1}{v} = \frac{1}{f} - \frac{1}{u} \)

\( \implies \frac{1}{v} = \frac{1}{-15} - \frac{1}{-30} = \frac{1}{-15} + \frac{1}{30} = -\frac{2}{30} + \frac{1}{30} = -\frac{1}{30} \)

\( \implies v = -30 \text{ cm} \)
Hence, the image is formed at a distance of 30 cm in front of the mirror.

\( m = -\frac{v}{u} = \frac{I}{O} \)

\( \implies I = \frac{-O \times v}{u} = \frac{-4 \times -30}{-30} = -4 \text{ cm} \)
Negative sign indicates inverted image.
So, \( I = 4 \text{ cm} \)
Hence, the length of the image is 4 cm.
In simple words: By using the mirror formula, we calculate that light rays meet 30 cm in front of the mirror to create an image that is 4 cm long and upside down.

📝 Teacher's Note: Remind students that for a concave mirror, the object distance (\( u \)) and focal length (\( f \)) are taken as negative. When the object is at \( 2f \) (center of curvature), the image is formed at the same spot and is the same size.

🎯 Exam Tip: Always state the mirror formula clearly before substitution. A negative magnification (\( m \)) value is the mathematical proof that the image is inverted.

Question 11N.
\( u = -30 \text{ cm} \)
\( m = \frac{I}{O} = \frac{30}{10} = 3 \)

Answer: But, for real object \( m \) is negative

\( \implies m = -3 \)

\( \because m = -\frac{v}{u} \)

\( \implies - \frac{v}{u} = -3 \)

\( \implies v = 3u \)

\( \implies v = 3 \times -30 = -90 \text{ cm} \)
Hence, the position of the image is 90 cm in front of the mirror.
From mirror formula,
\( \frac{1}{f} = \frac{1}{v} + \frac{1}{u} \)

\( \implies \frac{1}{f} = \frac{1}{-90} + \frac{1}{-30} = \frac{-1 - 3}{90} = \frac{-4}{90} \)

\( \implies f = -22.5 \text{ cm} \)
In simple words: When the image is three times larger than the object and forms in front of the mirror, the image distance becomes 90 cm. We use this to find that the mirror's focal length is 22.5 cm.

📝 Teacher's Note: When a question says an image is a certain number of times larger, it is giving you the magnification. If the image is real, magnification must be negative.

🎯 Exam Tip: Remember the sign convention: real images formed by a single mirror are always inverted, leading to a negative height and negative magnification.

Question 12N.
\( m = \frac{I}{O} = \frac{20}{10} = 2 \)

Answer: \( \because m = -\frac{v}{u} \)

\( \implies - \frac{v}{u} = 2 \)

\( \implies -v = 2u \)

\( \because v = 10 \text{ cm} \)

\( \implies u = -5 \text{ cm} \)
Hence, the position of the image is 10 cm behind the mirror.
From mirror formula,
\( \frac{1}{f} = \frac{1}{v} + \frac{1}{u} \)

\( \implies \frac{1}{f} = \frac{1}{10} + \frac{1}{-5} = \frac{1 - 2}{10} = -\frac{1}{10} \)

\( \implies f = -10 \text{ cm} \)

Magnification is
\( m = \frac{I}{O} = \frac{20}{10} = 2 \)

\( \because m = -\frac{v}{u} \)

\( \implies v = -2u \)
But, \( u \) is always negative

\( \implies v = 2u \)
In simple words: For a mirror that makes a virtual image twice as big as the object 10 cm behind it, the object must have been placed 5 cm in front of it. The focal length is 10 cm.

📝 Teacher's Note: A positive image distance (\( v \)) indicates a virtual image formed behind the mirror. This only happens in a concave mirror when the object is placed very close (between pole and focus).

🎯 Exam Tip: Always check if the question implies a real or virtual image. Virtual images have positive \( v \) and positive \( m \).

Question 13N.
Magnification is \( m = \frac{I}{O} = \frac{1}{3} \)

Answer: A convex mirror always forms a virtual and an upright image. So,
\( m = -\frac{v}{u} \)

\( \implies - \frac{v}{u} = \frac{1}{3} \)

\( \implies v = -\frac{1}{3}u \)
But, \( u \) is always negative

\( \implies v = \frac{1}{3}u \)

\( \implies u = 3v \)
In simple words: In a convex mirror, the image is always smaller and right-side up. If it is one-third the size, the object is placed three times further away than where the image appears.

📝 Teacher's Note: Convex mirrors are "diminishing" mirrors. Their magnification is always a positive fraction less than 1.

🎯 Exam Tip: Since convex mirrors only form virtual images, the image distance (\( v \)) will always be positive and the focal length (\( f \)) is also positive.

Question 14N.
Magnification of a mirror is \( m = -\frac{v}{u} \)

Answer: \( \because m = -3 \)

\( \implies -3 = -\frac{v}{u} \)

\( \implies v = 3u \)
But, \( u \) will always be negative

\( \implies v = -3u \)
In simple words: If an image is real and three times larger, the distance to that image is three times the object's distance, but with the correct signs for the math to work.

📝 Teacher's Note: This exercise emphasizes the mathematical application of signs. If \( u = -10 \text{ cm} \), then \( v = -30 \text{ cm} \), which confirms the image forms in front of the mirror.

🎯 Exam Tip: "Magnification of -3" tells you two things immediately: the image is three times bigger AND it is inverted (real).