Maharashtra Board Class 9 Science Chapter 11 Reflection of Light Solutions

Class 9 Science Chapter 11 Reflection Of Light Question Answer Maharashtra Board

1. Answer The Following Questions.

Question 1.a. Explain the difference between a plane mirror, a concave mirror and a convex mirror with respect to the type and size of the images produced.
Answer:

	Plane mirror	Concave mirror	Convex mirror
Type of image	Virtual and Erect	Virtual (erect) as well as Real (inverted)	Virtual and Erect
Size of image	Same size	Diminished, Same size and magnified	Diminished

In simple words: Plane mirrors produce same-sized, virtual, erect images. Concave mirrors can produce various images (virtual/real, erect/inverted, diminished/same-size/magnified) depending on object position. Convex mirrors always produce virtual, erect, and diminished images.

🎯 Exam Tip: When differentiating mirrors, focusing on image characteristics (type, size, orientation) is crucial for a complete answer.

Question 1.b. Describe the positions of the source of light with respect to a concave mirror in
1. Torch light
2. Projector lamp
3. Floodlight
Answer:
(a) Torch light: The source of light is placed at the focus.
(b) Projector lamp : The source of light is placed at the centre of curvature.
(c) Flood light : The source of light is placed just beyond the centre of curvature.

In simple words: For a torch, the light source is at the focus, producing a parallel beam. For a projector lamp, it's at the center of curvature for image projection. For a floodlight, it's just beyond the center of curvature for a broader beam.

🎯 Exam Tip: Remember these specific placements to understand how different light beams (parallel, converging, diverging) are formed for various applications.

Question 1.c. Why are concave mirrors used in solar devices?
Answer:
- Solar devices like solar cooker or solar water heater use solar energy to cook food or heat water.
- When sun rays fall on the concave mirror, they converge and come together in the focal plane.
- Due to convergence, the intensity of sun rays increases and the food or water is heated faster. Hence, concave mirrors are used in solar- devices.

In simple words: Concave mirrors concentrate parallel sun rays to a single point (focal plane), increasing heat intensity, which is ideal for cooking or heating water in solar devices.

🎯 Exam Tip: The key concept here is 'convergence' of light rays and the resulting increase in intensity, which makes concave mirrors suitable for solar applications.

Question 1.d. Why are the mirrors fitted on the outside of cars convex?
Answer:
- A convex mirror is used as rear view mirror because they form erect, virtual, and diminished images.
- This, allows the driver to view a large area in a small mirror.

In simple words: Convex mirrors are used as rear-view mirrors because they provide a wider field of view by forming small, upright images, helping drivers see more of the road behind them.

🎯 Exam Tip: The two crucial advantages of convex mirrors for rear-view applications are the 'erect and virtual image' and the 'wider field of view'.

Question 1.e. Why does obtaining the image of the sun on a paper with the help of a concave mirror burn the paper?
Answer:
- When sunrays fall on the concave mirror, they converge and come together in the focal plane.
- Due to convergence, the intensity of sunrays increases.
- Hence, image of the sun on a paper with the help of concave mirror bums the paper.

In simple words: A concave mirror focuses parallel sun rays into a tiny, intensely hot spot at its focal point, causing paper placed there to burn due to the concentrated energy.

🎯 Exam Tip: Emphasize the 'concentration' of sun's energy at the focal point, leading to increased 'intensity' and subsequently, burning.

Question 1.f. If a spherical mirror breaks, what type of mirrors are the individual pieces?
Answer:
- When a spherical mirror breaks into smaller pieces, the radius of curvature and focal length does not change.
- Hence, it will continue to behave like a spherical mirror only.

In simple words: Even if a spherical mirror breaks, each piece retains the same curvature and focal length, so they still act as spherical mirrors, just smaller ones.

🎯 Exam Tip: The key takeaway is that the fundamental optical properties (radius of curvature, focal length) are inherent to the mirror's surface, not its overall size, hence individual pieces retain these properties.

 

Question 2. What sign conventions are used for reflection from a spherical mirror?
Answer:According to the Cartesian sign convention, the pole of the mirror is taken as the origin. The principal axis is taken as the X-axis of the frame of reference. The sign conventions are as follows.
1. The object is always kept on the left of the mirror. All distances parallel to the principal axis are measured from the pole of the mirror.
2. All distances measured towards the right of the pole are taken to be positive, while those measured towards the left are taken to be negative.
3. The distance measured vertically upwards from the principal axis are taken to be positive.
4. The distance measured vertically downwards from the principal axis are taken to be negative.
5. The focal length of a concave mirror is negative while that of a convex mirror is positive.

In simple words: The Cartesian sign convention defines the pole as the origin, principal axis as the X-axis. Objects are left of the mirror. Distances to the right are positive, to the left negative. Upward measurements are positive, downward negative. Concave mirrors have negative focal length, convex mirrors have positive.

🎯 Exam Tip: Understanding and correctly applying the Cartesian sign conventions is critical for solving numerical problems related to spherical mirrors accurately.

 

Question 3. Draw ray diagrams for the cases of images obtained in concave mirrors as described in the table on page 122.
Answer:

No.	Position of the object	Position of the image	Nature of image	Size of the image
1	Between pole and focus	Behind the mirror	Erect, virtual	Magnified
2	At the focus	At infinity	Inverted, real	Very large
3	Between focus and centre of curvature	Beyond the centre of curvature	Inverted, real	Magnified
4	At the centre of curvature	At the centre of curvature	Inverted, real	Same as the object
5	Beyond the centre of curvature	Between the centre of curvature and focus	Inverted, real	Diminished
6	At a very large (infinite) distance	At focus	Inverted, real	Point image

ℹ️ चित्र व्याख्या (Diagram Explanation): यह आरेख एक अवतल दर्पण के लिए अनंत पर स्थित वस्तु से निकलने वाली किरणों को दर्शाता है। समांतर किरणें दर्पण से परावर्तन के बाद फोकस (F) पर अभिसरित होती हैं, जिससे एक बिंदु-आकार, वास्तविक और उलटा प्रतिबिंब बनता है।
Object at infinity for a concave mirror
ℹ️ चित्र व्याख्या (Diagram Explanation): यह आरेख एक अवतल दर्पण के सामने, वक्रता केंद्र (C) से परे रखी गई वस्तु के लिए किरण आरेख को दर्शाता है। परावर्तित किरणें वक्रता केंद्र (C) और फोकस (F) के बीच मिलती हैं, जिससे एक छोटा, वास्तविक और उलटा प्रतिबिंब बनता है। An object beyond centre of curvature for a concave mirror
ℹ️ चित्र व्याख्या (Diagram Explanation): यह आरेख एक अवतल दर्पण के सामने, वक्रता केंद्र (C) पर रखी गई वस्तु के लिए किरण आरेख को दर्शाता है। परावर्तित किरणें वक्रता केंद्र (C) पर ही मिलती हैं, जिससे वस्तु के समान आकार का, वास्तविक और उलटा प्रतिबिंब बनता है। Object at centre of Curva fu re be a concave mirror.
ℹ️ चित्र व्याख्या (Diagram Explanation): यह आरेख एक अवतल दर्पण के सामने, फोकस (F) और वक्रता केंद्र (C) के बीच रखी गई वस्तु के लिए किरण आरेख को दर्शाता है। परावर्तित किरणें वक्रता केंद्र (C) से परे मिलती हैं, जिससे एक बड़ा, वास्तविक और उलटा प्रतिबिंब बनता है। Object between F & C for a concave mirror
ℹ️ चित्र व्याख्या (Diagram Explanation): यह आरेख एक अवतल दर्पण के सामने, फोकस (F) पर रखी गई वस्तु के लिए किरण आरेख को दर्शाता है। परावर्तित किरणें समांतर हो जाती हैं और अनंत पर मिलती प्रतीत होती हैं, जिससे अत्यधिक बड़ा, वास्तविक और उलटा प्रतिबिंब बनता है। Object at focus for a concave mirror.
ℹ️ चित्र व्याख्या (Diagram Explanation): यह आरेख एक अवतल दर्पण के सामने, ध्रुव (P) और फोकस (F) के बीच रखी गई वस्तु के लिए किरण आरेख को दर्शाता है। परावर्तित किरणें दर्पण के पीछे अपसरित होती हुई प्रतीत होती हैं, जिससे एक बड़ा, आभासी और सीधा प्रतिबिंब बनता है। Object between pole and focus for a concave mirror

Image position	Nature of image
At focus	Real, inverted and point image
Between the centre of curvature and focus.	Real, inverted and diminished.
At the centre of curvature.	Real, inverted and same size
Beyond the centre of curvature.	Real, inverted and magnified.
At infinity.	Real, inverted and highly magnified.
Behind the mirror.	Virtual, erect and magnified.

In simple words: Concave mirrors produce a variety of images (real, inverted, magnified, diminished, same size, or virtual and erect) depending on where the object is placed relative to its focal point and center of curvature, as detailed in the diagrams.

🎯 Exam Tip: Practicing and memorizing these six ray diagrams and their corresponding image characteristics is essential for understanding concave mirrors and scoring well.

 

Question 4. Which type of mirrors are used in the following? Periscope, floodlights, shaving mirror, kaleidoscope, street lights, headlamps of a car.
Answer:

Objects	Type of Mirror
Periscope	Plane mirror
Floodlights	Concave mirror
Shaving mirror	Concave mirror
Kaleidoscope	Plane mirror
Street lights	Convex mirror
Head lamps of car	Concave mirror

In simple words: Plane mirrors are used in periscopes and kaleidoscopes for simple reflections. Concave mirrors are for converging light (floodlights, headlamps) or magnified images (shaving mirrors). Convex mirrors are used in street lights for diverging light.

🎯 Exam Tip: This question tests practical applications of different mirror types; remember the primary function (converging, diverging, simple reflection, magnification) for each device.

 

5. Solve The Following Examples

Question 5.a. An object of height 7 cm is kept at a distance of 25 cm in front of a concave mirror. The focal length of the mirror is 15 cm. At what distance from the mirror should a screen be kept so as to get a clear image? What will be the size and nature of the image?
Answer:Solution:
Given: Object size \(h_1\) = 7 cm
Object distance \(u\) = -25 cm
Focal length \(f\) = -15cm
To find: Image distance \(v\) = ?
Image size \(h_2\) = ?
Formula:
\( \frac{1}{v} + \frac{1}{u} = \frac{1}{f} \)
\( M = \frac{h_2}{h_1} = - \frac{v}{u} \)
Solution:
\( \frac{1}{v} + \frac{1}{u} = \frac{1}{f} \)

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

\( \implies \frac{1}{v} = \frac{1}{-15} - \frac{1}{-25} \)

\( \implies \frac{1}{v} = \frac{-1}{15} + \frac{1}{25} \)

\( \implies \frac{1}{v} = \frac{-5+3}{75} \)

\( \implies \frac{1}{v} = \frac{-2}{75} \)

\( \implies v = \frac{-75}{2} \)

\( \implies v = -37.5 \text{ cm} \)
The screen should be kept 37.5 cm in front of the mirror. The image is real.

\( M = \frac{h_2}{h_1} = - \frac{v}{u} \)

\( \implies h_2 = - \frac{vh_1}{u} \)

\( \implies h_2 = - \left( \frac{-75}{2} \times \frac{7}{-25} \right) \)

\( \implies h_2 = \frac{-21}{2} \)

\( \implies h_2 = -10.5 \text{ cm} \)
The height of the image is 10.5 cm, it is an inverted and enlarged image.

In simple words: The screen must be placed 37.5 cm in front of the concave mirror to get a real, inverted, and enlarged image of 10.5 cm height.

🎯 Exam Tip: Remember to use the correct sign conventions for object distance (u), focal length (f), and image distance (v) when applying the mirror formula and magnification formula. Always state the nature and size of the image clearly.

 

Question 5.b. A convex mirror has a focal length of 18 cm. The image of an object kept in front of the mirror is half the height of the object. What is the distance of the object from the mirror?
Answer:Solution:
Given: Image size \(h_2\) = \( \frac{1}{2} h_1 \)
Focal length \(f\) = 18 cm
To find: Object distance \(u\) = ?
Formula:
(i) \( M = - \frac{v}{u} \)
(ii) \( \frac{1}{v} + \frac{1}{u} = \frac{1}{f} \)
Solution: \( M = \frac{h_2}{h_1} \)
\( M = \frac{1/2 h_1}{h_1} \)
\( M = \frac{1}{2} \)
Now \( M = - \frac{v}{u} \)

\( \implies \frac{v}{u} = - \frac{1}{2} \)

\( \implies v = - \frac{u}{2} \)
\( \frac{1}{v} + \frac{1}{u} = \frac{1}{f} \)

\( \implies \frac{1}{-u/2} + \frac{1}{u} = \frac{1}{18} \)

\( \implies \frac{-2}{u} + \frac{1}{u} = \frac{1}{18} \)

\( \implies \frac{(-2+1)}{u} = \frac{1}{18} \)

\( \implies \frac{-1}{u} = \frac{1}{18} \)

\( \implies u = -18 \text{ cm} \)
The object is placed in front of the convex mirror at a distance of 18 cm.

In simple words: Given a convex mirror with a focal length of 18 cm, if the image height is half the object height, the object is located 18 cm in front of the mirror.

🎯 Exam Tip: For convex mirrors, the focal length is always positive. Ensure you use the magnification formula correctly along with the mirror formula to find the unknown distance.

 

Question 5.c. A 10 cm long stick is kept in front of a concave mirror having focal length of 10 cm in such a way that the end of the stick closest to the pole is at a distance of 20 cm. What will be the length of the image?
Answer:Solution:
Given: Object size \(h_1\) = 10 cm
Object distance \(u\) = -20 cm
Focal length \(f\) = -10 cm
To find : Image size \(h_2\) = ?
Formula:
\( \frac{1}{v} + \frac{1}{u} = \frac{1}{f} \)
\( M = \frac{h_2}{h_1} = - \frac{v}{u} \)
Solution:
\( \frac{1}{v} + \frac{1}{u} = \frac{1}{f} \)

\( \implies \frac{1}{v} + \frac{1}{-20} = \frac{1}{-10} \)

\( \implies \frac{1}{v} = \frac{1}{-10} + \frac{1}{20} \)

\( \implies \frac{1}{v} = \frac{-2+1}{20} \)

\( \implies \frac{1}{v} = \frac{-1}{20} \)

\( \implies v = -20 \text{ cm} \)
\( M = \frac{h_2}{h_1} = - \frac{v}{u} \)

\( \implies h_2 = - \frac{v \times h_1}{u} \)

\( \implies h_2 = - \frac{(-20 \times 10)}{-20} \)

\( \implies h_2 = -10 \text{ cm} \)
The height of the image is 10 cm and it is a real and inverted image.

In simple words: For a 10 cm stick placed 20 cm from a concave mirror with a 10 cm focal length, the image formed is 10 cm tall, real, and inverted.

🎯 Exam Tip: For problems involving extended objects, determine the image position and size for both ends of the object if it extends along the principal axis. Here, since the stick is 10cm long and the end closest to the pole is at 20cm, it's simpler to calculate the image for the end at 20cm first as `u` in the formula, then we will calculate the length of image. Pay attention to negative signs for real images and inverted images.

 

Question 6. Three mirrors are created from a single sphere. Which of the following: pole, centre of curvature, radius of curvature, principal axis - will be common to them and which will not be common?
Answer:
ℹ️ चित्र व्याख्या (Diagram Explanation): यह आरेख एक एकल गोले को तीन खंडों में विभाजित करके तीन दर्पणों के निर्माण को दर्शाता है। प्रत्येक खंड अपने मूल गोले के वक्रता केंद्र और वक्रता त्रिज्या को साझा करेगा, लेकिन प्रत्येक दर्पण का अपना ध्रुव और प्रमुख अक्ष होगा।
Three mirror created from single sphere
- Centre of curvature and Radius of curvature will be common for all three pieces.
- Pole and Principal axis will not be common.

In simple words: If a single sphere is cut into multiple mirrors, they will all share the same center of curvature and radius of curvature, but each individual mirror piece will have its own unique pole and principal axis.

🎯 Exam Tip: The center of curvature and radius of curvature are properties of the original sphere from which the mirror is cut, making them common. The pole and principal axis are specific to the individual mirror's surface and orientation.

 

Class 9 Science Chapter 11 Reflection Of Light Intext Questions And Answers

Question 1. What is light
Answer:Light is a form of electromagnetic radiation that produces the sensation of vision.

In simple words: Light is energy that travels as waves and allows us to see.

🎯 Exam Tip: A concise definition that includes 'electromagnetic radiation' and 'sensation of vision' is key for a complete answer.

 

Question 2. What is a mirror?
Answer:A mirror is a reflecting surface which reflects light and creates clear images.

In simple words: A mirror is a smooth, shiny surface that bounces back light, forming images.

🎯 Exam Tip: The essential components of a mirror's definition are its 'reflecting surface' and its ability to 'create clear images'.

 

Question 3. Principal Focus of Concave and Convex Mirror.
Answer:

Principal Focus of the Concave Mirror	Principal Focus of the Convex Mirror
(i) Incident rays which are parallel to the principal axis of a concave mirror, after reflection from the mirror, meet at a particular point in front of the mirror on the principal axis. This point (F) is called the principal focus of the concave mirror.	(i) Incident rays parallel to the principal axis, after reflection, appear to come from a particular point behind the mirror lying along the principal axis. This point is called the principal focus of the convex mirror.
(ii) It is formed in front of the mirror.	(ii) It is formed behind the mirror.
(iii) Focus of concave mirror is real.	(iii) Focus of convex mirror is virtual.

In simple words: For a concave mirror, the principal focus is where parallel rays actually meet in front of the mirror (real focus). For a convex mirror, it's where parallel rays appear to diverge from behind the mirror (virtual focus).

🎯 Exam Tip: Clearly differentiate between the real focus of a concave mirror (where rays actually converge) and the virtual focus of a convex mirror (where rays appear to diverge from).

 

Question 4. If we hold a page of a book in front of a mirror, we see laterally inverted letters in the mirror. Why does it happen?
Answer:
- When we hold a page of a book in front of the mirror, the image of the words appear laterally inverted.
- The image of every point of the word is formed behind the mirror at the same distance from the mirror
- Because of this the left and right side of the image is interchanged.
- Hence, if we hold a page of a book in front of a mirror, we see laterally inverted letters in the mirror.

In simple words: Lateral inversion occurs in a plane mirror because the mirror reverses objects along a horizontal axis, swapping left and right parts of the image, while maintaining vertical orientation.

🎯 Exam Tip: Explain that lateral inversion is the interchange of left and right, not top and bottom, due to the way light reflects off a plane mirror surface.

 

Question 5. Which letters of the English alphabet form images that look the same as the original letters?
Answer:A, H, I, M, O, T, U, V, W, X, Y

In simple words: Letters that appear symmetrical when reflected horizontally (like A or H) look unchanged in a mirror.

🎯 Exam Tip: These letters possess horizontal symmetry, meaning their left half is a mirror image of their right half, causing them to appear identical upon lateral inversion.

 

Question 6. When a person stands in front of a plane mirror, how is the image formed? What is the nature of the image?
Answer:
- The image of a person is formed from every point of the source, thereby forming an extended image of the whole source.
- The image formed would be virtual, upright and left-right reversed.

In simple words: A plane mirror forms a virtual, upright image that is laterally inverted (left-right reversed), appearing as far behind the mirror as the person is in front.

🎯 Exam Tip: Remember the four key characteristics of an image formed by a plane mirror: virtual, erect, laterally inverted, and same size/distance as the object.

 

Answer The Following Questions:

Question 1. Place two plane mirrors at an angle of 90a to each other. Place a small object between them. Images will be formed in both mirrors. How many images do you see? Now change the angle between the mirrors as given in the following table and count the number of images each time. How is this number related to the measure of the angle?
Answer:The Relation between the angle between the mirrors and the number of images formed is given by
\( n = \frac{360^\circ}{A} - 1 \)

In simple words: The number of images formed by two plane mirrors inclined at an angle 'A' is found by dividing 360 degrees by 'A' and then subtracting one.

🎯 Exam Tip: This formula applies when \( \frac{360^\circ}{A} \) is an integer. For non-integer or even values, specific rounding or adjustments may be needed, but for most school-level problems, this formula is sufficient.

 

n = number of images,
A = angle between the mirrors

Angle	Number of images
\(120^\circ\)	2
\(90^\circ\)	3
\(60^\circ\)	5
\(45^\circ\)	7
\(30^\circ\)	11

(1) \(A = 120^\circ\)
\(n = \frac{360}{120} - 1 = 3-1 = 2\)
(2) \(A = 90^\circ\)
\(n = \frac{360}{90} - 1 = 4-1 = 3\)
(3) \(A = 60^\circ\)
\(n = \frac{360}{60} - 1 = 6-1 = 5\)
(4) \(A = 45^\circ\)
\(n = \frac{360}{45} - 1 = 8-1 = 7\)
(5) \(A = 30^\circ\)
\(n = \frac{360}{30} - 1 = 12-1 = 11\)

In simple words: The calculations demonstrate how the number of images increases as the angle between the two plane mirrors decreases, following the formula \( n = \frac{360^\circ}{A} - 1 \).

🎯 Exam Tip: When using the formula \( n = \frac{360^\circ}{A} - 1 \), ensure the angle 'A' is in degrees. This formula is accurate for symmetrical placements and generally provides the visible number of images.

 

Class 9 Science Chapter 11 Reflection Of Light Additional Important Questions And Answers

Can You Recall?

Question 1. What is meant by reflection of light and what are the types of reflection?
Answer:The bouncing back of light when it hits an opaque surface is called reflection of light. The two types of reflection are regular and irregular reflection.

In simple words: Reflection is when light bounces off a surface, and it can be either regular (like from a mirror) or irregular (like from a rough surface).

🎯 Exam Tip: Defining reflection clearly and naming its two types (regular and irregular/diffused) with brief descriptions is crucial for scoring well.

 

Question 2. What are the laws of reflection.
Answer:
- The incident ray, reflected ray and normal all lie in the same plane at the point of incidence.
- The angle of incidence is equal to the angle of reflection.
- The incident ray and the reflected ray lie on opposite sides of the normal.

In simple words: The laws of reflection state that the incoming light ray, the bounced-back ray, and the line perpendicular to the surface all exist in the same flat area, and the angle at which light hits the surface is equal to the angle at which it leaves.

🎯 Exam Tip: Clearly state all three laws. The first law deals with the coplanarity of rays and normal, and the second law (angle of incidence = angle of reflection) is fundamental.

 

Choose And The Correct Option:

Question 1. If the reflected rays do not actually meet, such an image is called as image.
(a) real
(b) virtual
(c) magnified
(d) inverted
Answer: (b) virtual

In simple words: An image formed when reflected light rays only appear to meet, but don't actually converge, is called a virtual image.

🎯 Exam Tip: Virtual images cannot be projected onto a screen, unlike real images where light rays actually intersect.

 

Question 2. In a plane mirror, the perpendicular distance of the image from the mirror is equal to
(a) the perpendicular distance of the source from the object.
(b) the perpendicular distance of the source from the mirror.
(c) the parallel distance of the source from the object.
(d) the parallel distance of the source from the mirror.
Answer: (b) the perpendicular distance of the source from the mirror

In simple words: For a plane mirror, the image is formed exactly as far behind the mirror as the object is in front of it, along the perpendicular line to the mirror.

🎯 Exam Tip: This property (object distance = image distance) is a defining characteristic of images formed by plane mirrors.

 

Question 3. The image formed in a convex mirror is always
(a) virtual, smaller and behind the mirror
(b) virtual, smaller and in front of the mirror
(c) real, smaller and behind the mirror
(d) real, smaller and in front of the mirror
Answer: (a) virtual, smaller and behind the mirror

In simple words: A convex mirror always creates an image that appears behind the mirror, is smaller than the actual object, and cannot be projected onto a screen (virtual).

🎯 Exam Tip: Remember that convex mirrors always produce virtual, erect, and diminished images, regardless of the object's position.

Question 4. images can be displayed on a screen.
(a) Virtual
(b) Real
(c) Virtual and erect
(d) Virtual and inverted
Answer: (b) Real
In simple words: Real images can be projected onto a screen because they are formed where light rays actually converge. Virtual images cannot be projected.

🎯 Exam Tip: Understanding the difference between real and virtual images is crucial for optics questions.

Question 5. A concave mirror is also called as a mirror.
(a) converging
(b) diverging
(c) plane
(d) outward curved
Answer: (a) converging
In simple words: A concave mirror is called a converging mirror because it brings parallel rays of light together to a single point (its focus) after reflection.

🎯 Exam Tip: Memorize the alternative names for concave (converging) and convex (diverging) mirrors.

Question 6. The centre of the mirror surface is called its
(a) pole
(b) centre of curvature
(c) principal axis
(d) focus
Answer: (a) pole
In simple words: The pole is the geometric center of the spherical mirror's reflecting surface. It lies on the principal axis.

🎯 Exam Tip: Clearly distinguish between the pole (center of the reflecting surface) and the center of curvature (center of the sphere from which the mirror is cut).

Question 7. According to the new sign convention, the of the mirror is taken as origin.
(a) focus
(b) pole
(c) optical centre
(d) centre of curvature
Answer: (b) pole
In simple words: In the Cartesian sign convention, the pole of the mirror is used as the origin (0,0) for all distance measurements.

🎯 Exam Tip: Always remember that the pole is the reference point for all measurements in spherical mirror sign conventions.

Question 8. A convex mirror is also called as a mirror.
(a) converging
(b) plane
(c) diverging
(d) inward curved
Answer: (c) diverging
In simple words: A convex mirror is called a diverging mirror because it spreads out parallel rays of light after reflection, making them appear to come from a focal point behind the mirror.

🎯 Exam Tip: Knowing the converging/diverging nature of mirrors is essential for understanding image formation.

Question 9. In order to see the full image of a person standing in front of a mirror, the minimum height of the mirror must be
(a) same height as that of the person
(b) double the height of the person
(c) half the height of the person
(d) quarter the height of the person
Answer: (c) half the height of the person
In simple words: To see your full image in a plane mirror, the mirror's height must be at least half your own height.

🎯 Exam Tip: This is a common application-based question related to plane mirrors; remember the "half height" rule.

Question 10. If the inner surface of the spherical mirror is reflecting, then it is a mirror, and if the outer surface is reflecting then it is mirror.
(a) convex, concave
(b) convex, plane
(c) concave, plane
(d) concave, convex
Answer: (d) concave, convex
In simple words: An inward curving reflecting surface makes a concave mirror, while an outward bulging reflecting surface makes a convex mirror.

🎯 Exam Tip: Clearly differentiate between concave (reflecting from inside) and convex (reflecting from outside) mirrors.

Question 11. The image formed by a concave mirror is
(a) always virtual and erect
(b) always virtual and inverted
(c) virtual if the object is placed between the pole and the focus
(d) virtual if the object is beyond the focus
Answer: (c) virtual if the object is placed between the pole and the focus
In simple words: A concave mirror usually forms real and inverted images, but when the object is very close (between the pole and focus), it forms a virtual, erect, and magnified image.

🎯 Exam Tip: Remember the special case of image formation by a concave mirror when the object is within its focal length.

Question 12. No matter how far you stand from a spherical mirror, your image appears erect. The mirror may be
(a) plane
(b) concave
(c) convex
(d) either plane or convex
Answer: (d) either plane or convex
In simple words: Both plane mirrors and convex mirrors always produce erect images, regardless of the object's distance. Concave mirrors produce erect images only when the object is very close.

🎯 Exam Tip: Understand the characteristics of images formed by different types of mirrors, especially concerning erectness and virtual nature.

Question 13. In case of a concave mirror, an erect image is
(a) real and enlarged
(b) real and diminished
(c) virtual and diminished
(d) virtual and enlarged
Answer: (d) virtual and enlarged
In simple words: For a concave mirror, an erect image is only formed when the object is between the pole and the principal focus, and this image is always virtual and magnified.

🎯 Exam Tip: Link the properties of the image (erect/inverted, real/virtual, magnified/diminished) to the object's position for concave mirrors.

Question 14. A rear view mirror of a car is
(a) plane mirror
(b) concave mirror
(c) convex mirror
(d) cylindrical mirror
Answer: (c) convex mirror
In simple words: Convex mirrors are used as rear view mirrors because they always form virtual, erect, and diminished images, providing a wider field of view.

🎯 Exam Tip: Know the practical applications of each type of mirror, such as convex mirrors for rear-view mirrors due to their wide field of view.

Question 15. An image of an object placed at infinite distance from a concave mirror is formed at
(a) the focus of the mirror
(b) behind the mirror
(c) centre of curvature
(d) infinity
Answer: (a) the focus of the mirror
In simple words: When an object is placed at a very large (infinite) distance from a concave mirror, parallel rays of light from the object converge at the principal focus to form a real, inverted, and highly diminished image.

🎯 Exam Tip: This is a fundamental concept in ray optics; parallel rays from infinity always converge or appear to diverge from the focal point.

Question 16. A ray of light parallel to principal axis after reflection from concave mirror passes through
(a) centre of curvature
(b) focus
(c) pole
(d) optical centre
Answer: (b) focus
In simple words: By definition, rays parallel to the principal axis of a concave mirror converge at the principal focus after reflection.

🎯 Exam Tip: This is one of the key rules for drawing ray diagrams for spherical mirrors.

Question 17. The image made by a plane mirror is a image.
(a) real
(b) virtual
(c) inverted
(d) diminished
Answer: (b) virtual
In simple words: A plane mirror always forms a virtual image, which means the light rays only appear to originate from the image, but do not actually converge there.

🎯 Exam Tip: Remember the fixed characteristics of images formed by plane mirrors: virtual, erect, same size, and laterally inverted.

Question 18. The size of the image of an object placed at the focus of a concave mirror is
(a) erect
(b) very large
(c) same size
(d) diminished
Answer: (b) very large
In simple words: When an object is placed at the focus of a concave mirror, the reflected rays become parallel, forming a real, inverted, and highly magnified image at infinity.

🎯 Exam Tip: The image formed at infinity implies it is very large; this is the inverse of an object at infinity forming an image at the focus.

Question 19. For virtual images, the height is while for real images, it is
(a) positive, positive
(b) negative, positive
(c) negative, negative
(d) positive, negative
Answer: (d) positive, negative
In simple words: By convention, the height of an erect (virtual) image is positive, and the height of an inverted (real) image is negative.

🎯 Exam Tip: Sign conventions for image height are important for numerical problems. Erect images are positive, inverted are negative.

Find The Odd Man Out:

Question 1. Torches, flood lights, head lamps of vehicles, rear view mirror.
Answer: Rear view mirror - In rear view mirrors, convex mirror is used. Concave mirrors are used in the rest.
In simple words: The rear view mirror uses a convex mirror for a wide field of view, unlike the other devices which use concave mirrors to focus light.

🎯 Exam Tip: Pay attention to the specific mirror type used for each application.

Question 2. Side mirrors of cars, parking mirrors, flood lights, mirror fitted in shops.
Answer: Flood lights - In flood lights concave mirror is used. Convex mirrors are used in the rest.
In simple words: Floodlights use concave mirrors to project a strong, focused beam, while the other items use convex mirrors for a wider, diminished view.

🎯 Exam Tip: Understand why certain mirror types are chosen for specific uses based on their image formation properties.

Question 3. Virtual and enlarged, virtual and diminished, real and inverted, real and magnified
Answer: Virtual and diminished type of image is not formed by a concave mirror. All the other types of images are formed by a concave mirror.
In simple words: A concave mirror can't produce a virtual and diminished image. It can form virtual and enlarged, real and inverted (diminished, same size, or magnified).

🎯 Exam Tip: Thoroughly know all possible image characteristics formed by concave mirrors for different object positions.

Question 4. Image is laterally inverted, image is of same size, image is at same distance, image is diminished.
Answer: Image is diminished is not a characteristic of image formed in a plane mirror. Rest of them are characteristics of plane mirror.
In simple words: Plane mirrors always form images that are the same size as the object; they never diminish them.

🎯 Exam Tip: Recall the four key characteristics of images formed by plane mirrors: virtual, erect, same size, same distance, and laterally inverted.

Answer The Following In One Sentence:

Question 1. What kind of mirror will a doctor use to concentrate on teeth, eyes, ears etc.?
Answer: The doctor will use a concave mirror to concentrate on teeth, eyes, ears etc.
In simple words: Doctors use concave mirrors because they can produce magnified images of small areas, which helps in examination.

🎯 Exam Tip: Concave mirrors are often used in medical examinations for magnification.

Question 2. What do the nature, position and size of the image depend on?
Answer: The nature, position and size of the image depend upon the distance of the object from the reflecting surface.
In simple words: The characteristics of an image depend directly on how far the object is from the mirror's surface.

🎯 Exam Tip: Understanding the relationship between object distance and image characteristics is fundamental to geometrical optics.

Question 3. Give the expression for mirror formula.
Answer: \( \frac{1}{v} + \frac{1}{u} = \frac{1}{f} \)
In simple words: The mirror formula relates the image distance (v), object distance (u), and focal length (f) of a spherical mirror.

🎯 Exam Tip: This formula is crucial for solving all numerical problems related to spherical mirrors. Remember the correct signs for variables.

Question 4. State any four uses of concave mirror.
Answer: Concave mirrors are used in torches, headlights, shaving mirrors, dentists' mirrors, solar devices etc.
In simple words: Concave mirrors are used where focusing light or magnifying images is required, like in torches, car headlights, and dental examinations.

🎯 Exam Tip: Be prepared to list applications of both concave and convex mirrors, explaining the reason behind each use.

Question 5. What are the two types of spherical mirror?
Answer: Convex mirror and concave mirror are the two types of spherical mirror.
In simple words: Spherical mirrors are categorized into two main types: those that curve inwards (concave) and those that curve outwards (convex).

🎯 Exam Tip: These are the basic classifications; understanding their distinct shapes is key.

Match The Columns:

Question 1.

Column 'A'	Column 'B'
(1) Plane mirror	(a) Rear view mirror
(2) Concave mirror	(b) At laughing gallery
(3) Convex mirror	(c) At a hair dresser
(4) Irregular curved mirror	(d) At a dentist

Answer:
(1 - c),
(2 - d),
(3 - a),
(4 - b)
In simple words: Plane mirrors are used by hairdressers, concave mirrors by dentists, convex mirrors as rear-view mirrors, and irregular curved mirrors are found in laughing galleries.

🎯 Exam Tip: Practice matching mirror types with their common applications to reinforce understanding.

Question 2.

Column 'A'	Column 'B'
(1) Plane mirror	(a) Virtual and diminished image
(2) Concave mirror	(b) Virtual and same size image
(3) Convex mirror	(c) Real and inverted image

Answer:
(1 - b),
(2 - c),
(3 - a)
In simple words: Plane mirrors give virtual, same-size images; concave mirrors can give real, inverted images; and convex mirrors always give virtual, diminished images.

🎯 Exam Tip: This question tests your knowledge of image characteristics for different mirror types. Ensure you know all possibilities.

State Whether The Following Statements Are True Or False. Correct The False Statements:

(1) If the mirrors are kept at right angle to each other, then the number of images formed will be 4.
Answer: False. if the mirrors are kept at right angle to each other then the number of images formed will be 3.
In simple words: When two mirrors are placed at 90 degrees, the formula (360/angle) - 1 gives 3 images, not 4.

🎯 Exam Tip: Remember the formula \( n = \frac{360^\circ}{\theta} - 1 \) for calculating the number of images formed by two inclined plane mirrors.

(2) A convex mirror is used in flood lights.
Answer: False. a concave mirror is used in flood lights.
In simple words: Floodlights require a focused beam, which is provided by a concave mirror, not a convex mirror that spreads light.

🎯 Exam Tip: Distinguish between converging (concave) and diverging (convex) properties and their respective applications.

(3) False. a concave mirror can sometimes form a diminished image as well.
Answer: False. a concave mirror can sometimes form a diminished image as well.
In simple words: Concave mirrors can form diminished images when the object is placed beyond the center of curvature.

🎯 Exam Tip: Review the different image sizes formed by concave mirrors based on object position.

(4) Images formed by convex mirrors are always virtual.
Answer: True
In simple words: Convex mirrors always produce virtual images, meaning light rays appear to diverge from them.

🎯 Exam Tip: This is a defining characteristic of convex mirrors; they never form real images.

(5) The distance between the focus and the pole is called the radius of curvature.
Answer: False. the distance between the focus and the pole is called the focal length.
In simple words: The distance from the pole to the focus is the focal length, while the distance from the pole to the center of curvature is the radius of curvature.

🎯 Exam Tip: Be precise with definitions of focal length (f) and radius of curvature (R); remember \( R = 2f \).

(6) Reflection from a spherical mirror obeys laws of reflection.
Answer: True
In simple words: The fundamental laws of reflection (angle of incidence equals angle of reflection) apply to all reflecting surfaces, including spherical mirrors.

🎯 Exam Tip: The laws of reflection are universal and apply to all types of mirrors (plane, spherical).

(7) The reflecting surface of a concave mirror is curved.
Answer: True
In simple words: A concave mirror has a reflecting surface that is curved inwards, like the inner surface of a spoon.

🎯 Exam Tip: Understand the physical shape of concave and convex mirrors.

(8) Distances measured in the direction of the incident light are taken as positive.
Answer: True
In simple words: According to the Cartesian sign convention, distances measured in the same direction as the incident light are considered positive.

🎯 Exam Tip: Always apply the Cartesian sign convention consistently for numerical problems.

(9) If the image is erect, the height of the image is negative.
Answer: False. if the image is erect, the height of the image is positive.
In simple words: Erect images are always measured as having a positive height, while inverted images have a negative height.

🎯 Exam Tip: Positive height for erect images, negative height for inverted images – this is a standard sign convention.

(10) A real image can be displayed on a screen.
Answer: True
In simple words: Real images are formed by the actual intersection of reflected light rays and can therefore be projected onto a screen.

🎯 Exam Tip: The ability to be projected onto a screen is the hallmark of a real image.

(11) A concave mirror always forms a real and inverted image.
Answer: False. a concave mirror can also form a virtual and erect image.
In simple words: While concave mirrors mostly form real and inverted images, they produce a virtual and erect image when the object is placed between the pole and the focal point.

🎯 Exam Tip: Remember the special case where a concave mirror acts like a magnifying mirror.

(12) Doctors use diverging beam of light to study teeth, ears and eyes.
Answer: False. doctors use a converging beam of light to study teeth, ears and eyes.
In simple words: Doctors use concave (converging) mirrors to focus light and magnify small areas for examination.

🎯 Exam Tip: Concave mirrors are converging mirrors, used for focused examination and magnification.

Give Scientific Reasons:

Question 1. A concave mirror is called a converging mirror.
Answer:
• When rays of light parallel to the principal axis are incident on concave mirror, they converge.
• After convergence, they meet at one point on the principal axis, hence concave mirror is called converging mirror.
In simple words: Concave mirrors are called converging mirrors because they collect parallel light rays and bring them together at a single focal point after reflection.

🎯 Exam Tip: The ability to converge parallel rays is the defining characteristic that gives concave mirrors their "converging" name.

Question 2. Concave mirrors are used in torches and in car headlights.
Answer:
1. Concave mirrors are used in torches and car headlights because when a source of light is placed at the focus of a concave mirror, a parallel beam of light rays is obtained.
2. This helps us to see things upto a considerable distance in the darkness.
In simple words: Concave mirrors are used in torches and headlights because they produce a powerful, parallel beam of light when the bulb is placed at their focus, illuminating distant objects effectively.

🎯 Exam Tip: This is a classic application of concave mirrors, demonstrating their ability to generate a parallel beam from a source at the focus.

Question 3. A dentist uses a concave mirror while examining teeth.
Answer:
• A concave mirror produces an erect, virtual and magnified image of an object placed between its pole and focus.
• A dentist uses this principle to get a clear and distinct image of teeth, hence, a dentist uses a concave mirror.
In simple words: Dentists use concave mirrors to get a magnified, upright view of teeth by placing the mirror very close to the teeth (within the focal length), helping them see details clearly.

🎯 Exam Tip: This application highlights the concave mirror's property of producing a magnified virtual image, which is useful for close-up inspections.

Solve The Following Numericals.

Tips for solving numerical:
• Object distance (u) is always -ve
• If Image distance (u) is +ve then image is behind the mirror and virtual. if u is -ve then image is in front of the mirror and real.
• Object height (h₁) is always +ve since it is erect.
• Image height (h2) can be +ve for virtual and -ve for real.

Type - A

Question 1. A bird is sitting in front of two plane mirrors inclined at an angle of 600 to each other. How many images does the bird see in the mirror?
Solution:
Given : Angle between mirror A = \( 60^\circ \)
To find: Number of images formed n = ?
Formula: \( n = \frac{360^\circ}{A} - 1 \)
Solution:
\( n = \frac{360^\circ}{A} - 1 \)

\( n = \frac{360^\circ}{60^\circ} - 1 \)

\( = 6 - 1 \)

\( n = 5 \)
The bird sees 5 images in the mirror.
In simple words: Using the formula for images formed by inclined mirrors, a 60-degree angle results in 5 images.

🎯 Exam Tip: Always use the formula \( n = \frac{360^\circ}{\theta} - 1 \) for plane mirrors inclined at an angle \( \theta \).

Question 2. A coin is kept in front of two plane mirrors inclined to each other. If 3 images of the coin are seen then what is the angle A between the mirrors?
Solution:
Given: no. of images formed n =3
To find: Angle between mirror A =?
Formula:
\( n = \frac{360^\circ}{A} - 1 \)
Solution:
\( n = \frac{360^\circ}{A} - 1 \)

\( n+1 = \frac{360^\circ}{A} \)

\( A = \frac{360^\circ}{n+1} \)

\( A = \frac{360^\circ}{3+1} \)

\( A = \frac{360^\circ}{4} \)

\( A = 90^\circ \)
The mirrors are inclined at an angle of \( 90^\circ \) to each other.
In simple words: By rearranging the image formation formula, if 3 images are seen, the angle between the mirrors must be 90 degrees.

🎯 Exam Tip: Be comfortable with rearranging the formula to solve for the angle when the number of images is given.

Question 3. An image is formed 5 cm behind a convex mirror of focal length 10 cm. At what distance is the object placed from the mirror?
Solution:
Given: Image distance (u) = 5 cm
Focal length (f) = 10 cm
To find: Object distance (u) = ?
Formula: \( \frac{1}{v} + \frac{1}{u} = \frac{1}{f} \)
Solution:
\( \frac{1}{v} + \frac{1}{u} = \frac{1}{f} \)

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

\( \implies \frac{1}{u} = \frac{1}{10} - \frac{1}{5} \)

\( \implies \frac{1}{u} = \frac{1-2}{10} \)

\( \implies \frac{1}{u} = \frac{-1}{10} \)

\( u = -10 \text{ cm} \)
The object is placed at a distance of 10 cm in front of the mirror.
In simple words: Using the mirror formula and given image and focal length, the object distance is calculated to be 10 cm in front of the convex mirror.

🎯 Exam Tip: Remember that for a convex mirror, the focal length (f) is positive, and virtual image distance (v) is positive (behind the mirror).

Question 4. An object placed 20 cm in front of a convex mirror is found to have an image 15cm behind the mirror. Find the focal length of the mirror.
Solution:
Given: Object distance (u) = -20 cm
Image distance (u) = 15 cm
To find: focal length (f) = ?
Formula: \( \frac{1}{v} + \frac{1}{u} = \frac{1}{f} \)
Solution:
\( \frac{1}{v} + \frac{1}{u} = \frac{1}{f} \)

\( \implies \frac{1}{15} + \frac{1}{(-20)} = \frac{1}{f} \)

\( \implies \frac{1}{15} - \frac{1}{20} = \frac{1}{f} \)

\( \implies \frac{4-3}{60} = \frac{1}{f} \)

\( \implies \frac{1}{60} = \frac{1}{f} \)

\( f = 60 \text{ cm} \)
The focal length of the convex mirror is 60 cm.
In simple words: Applying the mirror formula with the given object and image distances, the focal length of the convex mirror is found to be 60 cm.

🎯 Exam Tip: Pay close attention to sign conventions: object distance (u) is negative, image distance (v) for a convex mirror is positive.

Numerical For Practice

Question 5. An object is placed at a distance of 36 cm from a concave mirror of focal length 12 cm. Find the image distance.
Answer:
-18 cm
In simple words: Given object distance and focal length for a concave mirror, the image distance is calculated as -18 cm using the mirror formula.

🎯 Exam Tip: For concave mirrors, both object distance and focal length are typically negative in sign convention if the object is in front.

Question 6. An arrow is placed at a distance of 25 cm from a diverging mirror of focal length 20 cm. Find the image distance.
Answer:
11.1 cm
In simple words: For a diverging (convex) mirror, given the object distance and focal length, the image distance is found to be 11.1 cm.

🎯 Exam Tip: Remember that "diverging mirror" means a convex mirror, which has a positive focal length.

Type - B

Question 1. An object 4cm in height is placed at a distance of 36 cm from a concave mirror. The image is formed 18 cm in the front of the mirror. Find the height of the image.
Solution:
Given: Object height (h₁) = 4 cm
Image distance (u) = -18 cm
Object distance (u) = -36 cm
To find: Height of image (h2) = ?
Formula: \( M = \frac{h_2}{h_1} = - \frac{v}{u} \) ... (As image is real.)
Solution:
\( \frac{h_2}{h_1} = - \frac{v}{u} \)

\( \implies h_2 = - \frac{h_1 v}{u} \)

\( \implies h_2 = - \frac{4 \times (-18)}{-36} \)

\( \implies h_2 = -2 \text{ cm} \)
The height of the image is 2 cm and it is inverted.
In simple words: Using the magnification formula and the given object/image distances and object height, the image height is found to be -2 cm, indicating an inverted image.

🎯 Exam Tip: Be careful with the negative sign in the magnification formula (\( -v/u \)) and the signs of v and u as per convention.

Question 2. An object 2 cm high is placed at a distance of 16cm from a concave mirror which produces a real image 3 cm high. Find the image distance.
Solution:
Given: Object height (h₁) = 2 cm
Object distance (u) = -16 cm
Image height (h2) = -3 cm
To find: Image distance (u) = ?
Formula: \( M = \frac{h_2}{h_1} = - \frac{v}{u} \) (As image is real.)
Solution:
\( \frac{h_2}{h_1} = - \frac{v}{u} \)

\( \implies \frac{-3}{2} = - \frac{v}{-16} \)

\( \implies v = \frac{-16 \times 3}{2} \)

\( v = -24 \text{ cm} \)
The image is formed at a distance of 24 cm in front of the mirror.
In simple words: Given object and image heights, along with object distance, the image distance is calculated to be -24 cm using the magnification formula. The negative sign indicates it's in front of the mirror.

🎯 Exam Tip: Remember that a real image is inverted, so its height (h2) should be taken as negative. This is critical for getting the correct sign for v.

Numericals For Practice

Question 3. An object 10cm in height is placed at a distance of 36 cm from a concave mirror. 1f the image is formed at a distance of 18 cm in front of the mirror, find the height of the image.
Answer:
-5cm
In simple words: With the object height, object distance, and image distance provided, the image height for the concave mirror is calculated as -5 cm.

🎯 Exam Tip: Ensure consistent use of sign conventions for object/image distances and heights to avoid errors.

Question 4. A converging mirror forms a real image of height 4 cm of an object of height 1 cm placed 20 cm away from the mirror. Find the image distance.
Answer:
-80cm
In simple words: Using object/image heights and object distance for a converging mirror, the image distance is determined to be -80 cm.

🎯 Exam Tip: A "converging mirror" refers to a concave mirror. Remember that real images are inverted, so one of the heights must be negative in calculation.

Type - C

Question 1. Rajashree wants to get an inverted image of height 5 cm of an object kept at a distance of 30 cm from a concave mirror. The focal length of the mirror is 10 cm. At what distance from the mirror should she place the screen? What will be the type of the image, and what is the height of the object?
Solution:
Given:
Focal length = f = -10 cm,
Object distance = u = -30 cm
Height of the image = h2 = 7 cm
To find: Height of the object = h₁ = ?
Image distance = u =?
Formulae:
\( \frac{1}{v} + \frac{1}{u} = \frac{1}{f} \)
\( M = \frac{h_2}{h_1} = - \frac{v}{u} \)
Solution:
\( \frac{1}{v} + \frac{1}{u} = \frac{1}{f} \)

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

\( \implies \frac{1}{v} = \frac{1}{-10} - \frac{1}{-30} \)

\( \implies \frac{1}{v} = \frac{-1}{10} + \frac{1}{30} \)

\( \implies \frac{1}{v} = \frac{-3+1}{30} \)

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

\( \implies \frac{1}{v} = \frac{1}{-15} \text{ cm} \)

\( v = -15 \text{ cm} \)
Rajashree has to place the screen 15 cm to the left of the mirror.
Magnification formula
\( M = \frac{h_2}{h_1} = - \frac{v}{u} \)

\( \implies h_1 = - \frac{uh_2}{v} \)

\( \implies h_1 = - \frac{(-30) \times (-5)}{-15} \)

\( h_1 = (-2) \times (-5) \)

\( h_1 = 10 \text{ cm} \)
The height of the object is 10 cm. Thus, the image will be real and diminished.
In simple words: For a concave mirror with the given focal length and object distance, the screen should be placed at 15 cm in front of the mirror to get a real, inverted image. The object height is calculated to be 10 cm, resulting in a diminished image.

🎯 Exam Tip: This is a multi-step numerical problem. First, calculate the image distance using the mirror formula, then use the magnification formula to find the object height. Remember to use correct signs for image height based on "inverted" (negative).

Question 2. A 10 cm long stick is kept horizontally in front of the concave mirror having focal length of 10 cm in such a way that the end of the stick closest to the pole is at a distance of 20 cm. What will be the length of the image?
Answer: Solution: The stick is kept parallel to the Principal axis. Distance between A and P is 20 cm. Say \(u_1 = 20\) cm. Hence, the other end of the stick is at distance, \(u_2 = (u_1 + 10) = 30\) cm from pole of the mirror.
ℹ️ चित्र व्याख्या (Diagram Explanation): यह चित्र एक अवतल दर्पण के सामने रखी एक वस्तु को दर्शाता है। वस्तु (एक छड़ी) को मुख्य अक्ष के समानांतर रखा गया है, जिसमें एक सिरा (A) दर्पण से 20 सेमी दूर और दूसरा सिरा (B) 10 सेमी दूर है, जो कुल 10 सेमी की लंबाई दर्शाता है।
Using mirror formula for concave mirror,
\[ \frac{1}{v_1} + \frac{1}{u_1} = \frac{1}{f} \]
\[ \frac{1}{v_1} = \frac{1}{f} - \frac{1}{u_1} \]
\[ \frac{1}{v_1} = \frac{1}{-10} - \frac{1}{-20} \]
\[ \frac{1}{v_1} = - \frac{1}{10} + \frac{1}{20} \]
\[ \frac{1}{v_1} = \frac{-2+1}{20} \]
\[ \frac{1}{v_1} = - \frac{1}{20} \]
\( \implies \) \(v_1 = -20\) cm
Also, for other end of the stick,
\[ \frac{1}{v_2} + \frac{1}{u_2} = \frac{1}{f} \]
\[ \frac{1}{v_2} = \frac{1}{f} - \frac{1}{u_2} \]
\[ \frac{1}{v_2} = \frac{1}{-10} - \frac{1}{-30} \]
\[ \frac{1}{v_2} = - \frac{1}{10} + \frac{1}{30} \]
\[ \frac{1}{v_2} = \frac{-3+1}{30} \]
\[ \frac{1}{v_2} = - \frac{2}{30} \]
\[ \frac{1}{v_2} = - \frac{1}{15} \]
\( \implies \) \(v_2 = -15\) cm
Here, negative signs indicate that images are formed on the left of the mirror.
ℹ️ चित्र व्याख्या (Diagram Explanation): यह चित्र एक अवतल दर्पण के सामने रखी वस्तु (छड़ी AB) और उसके द्वारा बनने वाले प्रतिबिंब (A'B') को दर्शाता है। वस्तु 10 सेमी लंबी है, जिसका सिरा A दर्पण से 20 सेमी दूर और सिरा B दर्पण से 30 सेमी दूर है। प्रतिबिंब के सिरे A' और B' क्रमशः 20 सेमी और 15 सेमी की दूरी पर स्थित हैं, जो दर्शाता है कि प्रतिबिंब छोटा और उल्टा है।
The length of the image formed is given by, \(|v_2 - v_1| = |-15 - (-20)| = |-15 + 20| = 5\) cm. The length of the image is 5 cm.
In simple words: To find the image length of an extended object in a concave mirror, we calculate the image positions for both ends of the object using the mirror formula. The difference between these image positions gives the length of the image.

🎯 Exam Tip: Remember to use the Cartesian sign conventions consistently for object distance, image distance, and focal length when solving numerical problems involving mirrors. Pay close attention to negative signs indicating distances measured against the direction of incident light or inverted images.

Question 3. An object 2 cm in height is placed at a distance of 16 cm from a concave mirror. If the focal length of the mirror is 9.6 cm., find the image distance, nature and size of the image.
Answer: \(u = -24\) cm, \(h_2 = -3\) cm; real, inverted and enlarged.
In simple words: For a concave mirror, when an object is placed between the focal point and the center of curvature, a real, inverted, and enlarged image is formed beyond the center of curvature.

🎯 Exam Tip: When given an object's height, distance, and focal length for a mirror, always use the mirror formula and magnification formula to find the image distance, height, nature, and size. Ensure correct sign conventions for each quantity. The nature (real/virtual) and orientation (inverted/erect) are directly related to the signs of image distance and magnification.

Question 4. An arrow of 2.5cm height is placed at a distance of 25 cm from a diverging mirror of focal length 20 cm. Find the nature, position and size of the image formed.
Answer: \(v = 11.1\)cm, \(h_1 = 1.1\)cm; virtual and in diminished form.
In simple words: For a diverging mirror (convex mirror), an object always forms a virtual, erect, and diminished image, regardless of the object's position.

🎯 Exam Tip: Remember that convex mirrors always produce virtual, erect, and diminished images. The focal length of a convex mirror is positive, and the image is always formed behind the mirror. Be meticulous with sign conventions, especially for diverging mirrors.

Define the following:

Question 1. Centre of curvature of mirror (C)
Answer: The centre of the sphere of which the mirror is a part is called the centre of curvature of the mirror.
In simple words: The center of curvature is the center of the imaginary sphere from which a spherical mirror is cut.

🎯 Exam Tip: Understand that the center of curvature is a geometric property of the spherical mirror, not a point on the mirror itself. It's crucial for drawing ray diagrams and understanding mirror optics.

Question 2. The radius of curvature (R)
Answer: The radius of the sphere of which the mirror is a part, is called the radius of curvature of the mirror.
In simple words: The radius of curvature is the radius of the imaginary sphere from which a spherical mirror is part, and it's twice the focal length (R=2f).

🎯 Exam Tip: The radius of curvature (R) is a fundamental parameter for spherical mirrors, directly related to the focal length (f). Accurate knowledge of R helps in locating the center of curvature and drawing precise ray diagrams.

Question 3. Pole (P)
Answer: The centre of the mirror surface is called its pole.
In simple words: The pole is the central point on the reflecting surface of a spherical mirror.

🎯 Exam Tip: The pole serves as the origin for measuring all distances in the Cartesian sign convention. It's a key reference point for mirror formulas and ray diagrams.

Question 4. The principal axis of a mirror
Answer: The straight line passing through the pole and centre of curvature of the mirror is called its principal axis.
In simple words: The principal axis is an imaginary straight line that passes through the pole and the center of curvature of a spherical mirror.

🎯 Exam Tip: The principal axis is a crucial reference line for understanding the geometry of spherical mirrors and for drawing ray diagrams accurately. It defines the orientation for measuring heights and distances.

Question 5. The focus of a concave mirror (F)
Answer: Incident rays which are parallel to the principal axis of a concave mirror, after reflection from the mirror, meet at a particular point in front of the mirror on the principal axis. This point (F) is called the principal focus of the concave mirror.
In simple words: The principal focus of a concave mirror is the point on the principal axis where parallel incident rays converge after reflection.

🎯 Exam Tip: For concave mirrors, the principal focus is real and lies in front of the mirror. This property is used in applications like solar cookers to concentrate light.

Question 6. Focus of a convex mirror (F)
Answer: Incident rays parallel to the principal axis, after reflection, appear to come from a particular point behind the mirror lying along the principal axis. This point is called the principal focus of the convex mirror.
In simple words: The principal focus of a convex mirror is the point on the principal axis from which parallel incident rays appear to diverge after reflection.

🎯 Exam Tip: For convex mirrors, the principal focus is virtual and lies behind the mirror. This diverging property is utilized in applications like rear-view mirrors to spread light and provide a wider field of view.

Question 7. Focal length of a mirror (f)
Answer: The distance (f) between the pole and the principal focus of the mirror is called the focal length. This distance is half of the radius of curvature of the mirror. \(f = \frac{R}{2}\)
In simple words: Focal length is the distance between the mirror's pole and its principal focus, and it is half the radius of curvature.

🎯 Exam Tip: The focal length is a critical parameter in the mirror formula. Remember that for concave mirrors, f is negative, and for convex mirrors, f is positive, according to the Cartesian sign convention.

Answer the following in short:

Question 1. What are the rules for drawing ray diagrams for the formation of image by spherical mirror?
Answer: The rules are as follows:
• If an incident ray is parallel to the principal axis, then the reflected ray passes through the principal focus.
• If an incident ray passes through the principal focus of the mirror, the reflected ray is parallel to the principal axis.
• If an incident ray passes through the centre of curvature of the mirror, the reflected ray traces the same path back.
In simple words: The three main rules for drawing ray diagrams involve parallel rays reflecting through the focus, rays passing through the focus reflecting parallel, and rays passing through the center of curvature reflecting back along the same path.

🎯 Exam Tip: Mastering these three ray-tracing rules is essential for accurately constructing image formation diagrams for both concave and convex mirrors. Practice drawing various object positions to understand image characteristics.

Distinguish between:

Question 1. Convex mirror and Concave mirror
Answer:

Convex mirror	Concave mirror
(i) In a convex mirror, the reflecting surface is on the outer side.	(i) In a concave mirror, the reflecting surface is on the inner side.
(ii) It is called a diverging mirror.	(ii) It is called a converging mirror.
(iii) The focus of a convex mirror is virtual.	(iii) The focus of a concave mirror is real.
(iv) It can form only a virtual image.	(iv) It can form a real as well as a virtual image.
(v) It can form only a diminished image.	(v) It can form an enlarged, diminished as well as the same size image.

In simple words: Convex mirrors diverge light, have a virtual focus, and always form virtual, erect, and diminished images, while concave mirrors converge light, have a real focus, and can form both real/virtual and magnified/diminished/same-size images depending on object position.

🎯 Exam Tip: This distinction is fundamental. Focus on the reflecting surface, whether light converges or diverges, the nature of the focus (real/virtual), and the range of image types each mirror can form. This table is a concise summary for quick revision.

Question 2. Real image and Virtual image
Answer:

Real image	Virtual image
(i) A real image is formed only when the reflected rays actually meet at a point.	(i) A virtual image is formed only when the reflected rays appear to meet at a point.
(ii) Real images can be obtained on a screen.	(ii) Virtual images cannot be obtained on a screen.
(iii) All real images are inverted.	(iii) All virtual images are erect.

In simple words: Real images are formed by the actual intersection of reflected rays, can be projected onto a screen, and are always inverted; virtual images are formed when reflected rays only appear to intersect, cannot be projected, and are always erect.

🎯 Exam Tip: The ability to be projected onto a screen is the key differentiator between real and virtual images. Also, remember the consistent correlation: real images are inverted, and virtual images are erect.

Answer the following questions:

Question 1. If we keep the mirrors parallel to each other, how many images will we see?
Answer: When two mirrors are kept parallel to each other, infinite images are formed, this is because light gets reflected infinite times.
In simple words: When two mirrors face each other in parallel, light bounces back and forth indefinitely, creating an endless series of images.

🎯 Exam Tip: This is a classic concept related to multiple reflections. The formula for the number of images formed by two inclined mirrors is \(n = \frac{360^\circ}{A} - 1\). When A = \(0^\circ\) (parallel mirrors), this formula approaches infinity, indicating an infinite number of images.

Answer in detail:

Question 1. What sign conventions are used for reflection from a spherical mirror?
Answer:
ℹ️ चित्र व्याख्या (Diagram Explanation): यह चित्र गोलीय दर्पणों के लिए कार्टेशियन चिह्न परिपाटी को दर्शाता है। इसमें ध्रुव को मूल बिंदु (origin) माना गया है। मुख्य अक्ष को X-अक्ष के रूप में दिखाया गया है, जिस पर दाईं ओर की दूरियाँ धनात्मक और बाईं ओर की दूरियाँ ऋणात्मक होती हैं। Y-अक्ष पर, ऊपर की ओर की ऊँचाइयाँ धनात्मक और नीचे की ओर की ऊँचाइयाँ ऋणात्मक होती हैं। आपतित किरण की दिशा दाईं ओर इंगित की गई है।
According to the Cartesian sign convention, the pole of the mirror is taken as the origin. The principal axis is taken as the X-axis of the frame of reference. The sign conventions are as follows.
• The object is always kept on the left of the mirror. All distances parallel to the principal axis are measured from the pole of the mirror.
• All distances measured towards the right of the pole are taken to be positive, while those measured towards the left are taken to be negative.
• Distance measured vertically upwards from the principal axis are taken to be positive.
• Distance measured vertically downwards from the principal axis are taken to be negative.
• The focal length of a concave mirror is negative while that of a convex mirror is positive.
In simple words: The Cartesian sign convention defines how to measure distances and heights for spherical mirrors, using the pole as the origin, the principal axis as the X-axis, and assigning positive or negative signs based on direction relative to the incident light and principal axis.

🎯 Exam Tip: Correct application of the Cartesian sign convention is crucial for accurately solving numerical problems involving mirrors. A common mistake is not consistently applying signs for object distance, image distance, focal length, and heights. Practice with various scenarios to internalize these rules.

Question 2. Draw ray diagrams for the image obtained in convex mirrors.
Answer:
ℹ️ चित्र व्याख्या (Diagram Explanation): यह चित्र एक उत्तल दर्पण के लिए किरण आरेख को दर्शाता है, जहाँ वस्तु अनंत और ध्रुव (P) के बीच कहीं रखी है। आपतित किरणें दर्पण से परावर्तन के बाद इस प्रकार अपसरित होती हैं कि वे दर्पण के पीछे स्थित फोकस (F) से आती हुई प्रतीत होती हैं। इससे दर्पण के पीछे एक आभासी, सीधा और छोटा प्रतिबिंब बनता है।

Image position	Nature of image
Behind the mirror.	(A) Virtual, (B) Erect (C) Diminished

In simple words: Convex mirrors always form virtual, erect, and diminished images, located behind the mirror between the pole and the principal focus, regardless of where the object is placed in front of the mirror.

🎯 Exam Tip: For convex mirrors, there's only one general case for ray diagrams: when the object is anywhere between infinity and the pole. This always results in a virtual, erect, and diminished image formed behind the mirror. This consistency simplifies drawing and understanding.

Question 3. In order to see the full image of a person standing in front of a mirror, the minimum height of the mirror must be half the height of the person. Explain.
Answer:
ℹ️ चित्र व्याख्या (Diagram Explanation): यह चित्र एक व्यक्ति (HF) को समतल दर्पण (PQ) के सामने खड़ा दिखाता है। दर्पण की न्यूनतम ऊँचाई (PQ) व्यक्ति की पूर्ण छवि (H'F') देखने के लिए आवश्यक है, जहाँ R और S क्रमशः व्यक्ति की आँखों (E) और पैरों के मध्य-बिंदु हैं। किरणें व्यक्ति के सिर (H) और पैरों (F) से आती हैं, दर्पण से परावर्तित होती हैं, और आँखों (E) तक पहुँचती हैं, जिससे पूर्ण प्रतिबिंब बनता है।
Proof:
1. In the figure, the point at the top of the head, the eyes and a point at the feet of a person are indicated by H, E and F respectively.
2. R and S are midpoints of HE and EF respectively.
3. The mirror PQ is at a height of NQ from the ground and is perpendicular to it. PQ is the minimum height of the mirror in order to obtain the full image of the person.
For this, RP and QS must be perpendicular to the mirror.
Minimum height of the mirror
PQ = RS
= RE + ES
= \( \frac{HE}{2} + \frac{EF}{2} = \frac{HF}{2} \)
= Half of the person's height.
In simple words: To see your entire reflection in a plane mirror, the mirror's vertical length only needs to be half of your height, because of the geometry of light reflection.

🎯 Exam Tip: This is a classic concept that demonstrates the laws of reflection. The key insight is that light from your feet and head only needs to reach specific points on the mirror to reflect into your eyes. This understanding is often tested to gauge comprehension of reflection principles.

Question 4. Determine the sign of magnification in each of the 6 cases in the table and verify that they are same using formulae
\(M = \frac{h_2}{h_1}\) and \(M = -\frac{v}{u}\)
Answer:

No.	Position of object (u)	Position of image (v)	Name of image	\(M = \frac{h_2}{h_1}\)	\(M = -\frac{v}{u}\)
1.	Between pole and focus u is (-ve)	Behind the mirror v is (+ve)	Erect and virtual \(h_2\) is (+ve)	\(M = \frac{+\text{ve}}{+\text{ve}} = +\text{ve}\)	\(M = -\frac{(+\text{ve})}{(-\text{ve})} = +\text{ve}\)
2.	At the focus. u is (-ve)	At infinity v is (-ve)	Inverted and real \(h_2\) is (-ve)	\(M = \frac{-\text{ve}}{+\text{ve}} = -\text{ve}\)	\(M = -\frac{(-\text{ve})}{(-\text{ve})} = -\text{ve}\)
3.	Between focus and centre of curvature u is (-ve)	Beyond the centre of curvature v is (-ve)	Inverted and real \(h_2\) is (-ve)	\(M = \frac{-\text{ve}}{+\text{ve}} = -\text{ve}\)	\(M = -\frac{(-\text{ve})}{(-\text{ve})} = -\text{ve}\)
4.	At the centre of curvature u is (-ve)	At the centre of curvature v is (-ve)	Inverted and real \(h_2\) is (-ve)	\(M = \frac{-\text{ve}}{+\text{ve}} = -\text{ve}\)	\(M = -\frac{(-\text{ve})}{(-\text{ve})} = -\text{ve}\)
5.	Beyond the centre of curvature u is (-ve)	Between the centre of curvature and focus v is (-ve)	Inverted and real \(h_2\) is (-ve)	\(M = \frac{-\text{ve}}{+\text{ve}} = -\text{ve}\)	\(M = -\frac{(-\text{ve})}{(-\text{ve})} = -\text{ve}\)
6.	At very large distance (infinity) u is (-ve)	At focus v is (-ve)	Inverted and real \(h_2\) is (-ve)	\(M = \frac{-\text{ve}}{+\text{ve}} = -\text{ve}\)	\(M = -\frac{(-\text{ve})}{(-\text{ve})} = -\text{ve}\)

In simple words: Magnification's sign tells us if an image is erect (positive M) or inverted (negative M); by comparing the formulas, we see that positive magnification corresponds to virtual and erect images (positive \(h_2\), \(v\) positive, \(u\) negative), while negative magnification corresponds to real and inverted images (negative \(h_2\), \(v\) negative, \(u\) negative).

🎯 Exam Tip: The sign of magnification is a quick indicator of the image's orientation (erect for positive, inverted for negative). Always ensure consistency between the sign derived from \(h_2/h_1\) and \(-v/u\). This verification helps in cross-checking numerical answers and understanding image properties.

Question 5. Explain the images formed by concave mirrors with respect to position of the image and object and also the Nature and size of image
Answer:

No.	Position of the object	Position of the image	Nature of image	Size of the image
1	Between pole and focus	Behind the mirror	Erect, virtual	Magnified
2	At the focus	At infinity	Inverted, real	Very large
3	Between focus and centre of curvature	Beyond the centre of curvature	Inverted, real	Magnified
4	At the centre of curvature	At the centre of curvature	Inverted, real	Same as the object
5	Beyond the centre of curvature	Between the centre of curvature and focus	Inverted, real	Diminished
6	At a very large (infinite) distance	At focus	Inverted, real	Point image

In simple words: Concave mirrors produce a variety of images depending on object position: highly magnified and virtual when close, real and magnified at center of curvature, real and diminished further away, and a point image at focus when at infinity.

🎯 Exam Tip: Memorize the six key cases for image formation by concave mirrors, focusing on the relationship between object position, image position, nature (real/virtual), and size (magnified/diminished/same size). This table is an excellent summary for quick recall and ray diagram analysis.