CBSE Class 10 Science Light Reflection and Refraction VBQs Set 05

This CBSE Class 10 Science VBQs page offers 18 real life and case based questions with answers on Chapter 9, Light – Reflection and Refraction. The questions connect mirrors and lenses to everyday uses such as rear-view and security mirrors, dentist and shaving mirrors, solar furnaces, magnifying glasses and slide projectors, with solved numericals and ray diagrams.

Real Life and Competency Based Questions

Question 1. A student held a concave mirror in his hand and directed its reflecting surface towards the sun. Then he directed the light reflected by the mirror on to a sheet of paper held close to the mirror. After that, he moves the sheet of paper back and forth gradually until he got a bright, sharp spot of light on the paper. He held the mirror and the paper in the same position till the paper started burning and producing smoke. Eventually, it can catch fire also. (A) What is the term used for the distance between the mirror and the paper? (B) The student wants to depict the above experiment as a ray diagram. Draw a ray diagram which he would have drawn. (C) Give reasons: Concave mirrors are commonly used: (i) in torches, search light and vehicle headlights (ii) as shaving mirrors (iii) by dentist (iv) in solar furnaces
Answer:
(A) The distance between the mirror and the paper (where the sharp bright spot is formed) is the focal length of the mirror, f = distance of the principal focus F from the pole P.
(B) Ray diagram: parallel rays from the Sun fall on the concave mirror and, after reflection, pass through the principal focus F, forming a real, inverted, point-sized image of the Sun there.

(C) Concave mirrors are used:
(i) in torches, search lights and headlights: a bulb placed at the focus gives, after reflection, a powerful parallel beam of light;
(ii) as shaving mirrors: when the face is between the pole and the focus, the image is erect, virtual and enlarged;
(iii) by dentists: they give a large, erect image of the teeth;
(iv) in solar furnaces: they converge the Sun's rays at the focus and produce a lot of heat.
In simple words: The distance from the mirror to the bright spot is the focal length; a concave mirror collects light and can also magnify.

Teacher's Note:
1. The paper burns because the Sun's energy is concentrated on a very small spot.
2. Never look at the Sun directly or into a mirror reflecting sunlight.

Question 2. There are numerous uses of concave mirrors in daily life. It is used in the aircraft landing to guide the airplane, it is used as a torch to reflect the light rays, it is used during shaving to get an erect and enlarged image of the face, etc. A student had a concave mirror with him and wanted to know its focal length. So, he performed an activity to find out the focal length of the concave mirror. In the activity, the mirror stand is at the 10 cm mark of the scale and the screen carrying the sharp image of a distant tree is at the 40 cm mark.
(A) What is the radius of curvature of the concave mirror used by the student in the activity? (B) What is the magnification produced by a concave mirror when object is placed between pole and focus? (C) Using a concave mirror of radius of curvature 40 cm, if the object is kept at 40 cm, what is the position and nature of image formed?
Answer:
(A) The image of a distant object is formed at the focus. Focal length f = 40 − 10 = 30 cm, so the radius of curvature R = 2f = 60 cm.
(B) When the object is between the pole and the focus, the image is virtual, erect and enlarged, so the magnification is positive and greater than 1 (m > +1).
(C) R = 40 cm, so f = −20 cm, and the object at 40 cm is at the centre of curvature (u = −40 cm). \( \dfrac{1}{v}=-\dfrac{1}{20}+\dfrac{1}{40}=-\dfrac{1}{40} \), so v = −40 cm.
The image is formed at C, 40 cm in front of the mirror; it is real, inverted and of the same size (m = −1).
In simple words: Distance to the screen minus mirror position gives f = 30 cm, so R = 60 cm; an object at C gives an image at C.

Teacher's Note:
1. An object at the centre of curvature gives a real, inverted image of the same size, not an erect one.
2. An image formed on a screen is always real.

Question 3. Sushma went with her mother to purchase some gold jewellery for her sister's marriage. The salesman asked them to check the hallmark and gold purity to ascertain the genuineness of the gold article. As the hallmark was too small to be seen by the naked eye, the salesman gave them a magnifying glass to see the hallmark. Which type of lens is used as a magnifying glass and why?
Answer:
A convex lens of short focal length is used as a magnifying glass.
When the object (the hallmark) is held between the optical centre and the principal focus, a convex lens forms a virtual, erect and enlarged image on the same side as the object. So the tiny hallmark appears large and can be read easily.
A concave lens cannot be used, as it always gives a diminished image.
In simple words: A convex lens held close to the object makes it look bigger and upright.

Teacher's Note:
1. The object must be within the focal length of the lens.
2. A smaller focal length gives greater magnification.

Question 4. Search mirrors are mirrors that are used to look for hidden objects underneath the cars as shown. The hidden objects can be easily spotted as the mirror provides a wider field of view. (A) What type of mirrors are generally used to make search mirrors? (B) With the help of a ray diagram describe the nature of image formed by the type of mirror identified in (A).
Answer:
(A) A convex mirror is used, because it is a diverging mirror and gives a much wider field of view than a plane or concave mirror.
(B) A convex mirror always forms a virtual, erect and diminished image behind the mirror, between its pole and focus, for every position of the object.

In simple words: A convex mirror shows a large area as a small upright image.

Teacher's Note:
1. Wider field of view is the key property of a convex mirror.
2. The same property is used in vehicle rear-view mirrors and shop security mirrors.

Question 5. Rohit wants to have an erect image of an object using a converging mirror of focal length 40 cm. (A) Help Rohit by specifying the range of distance where the object can be placed in front of the mirror. Justify by providing a reason for the same. (B) Draw a ray diagram to show image formation in this case. (C) State one use of the mirror based on the above kind of image formation.
Answer:
(A) A concave (converging) mirror gives an erect image only when the object is between the pole and the focus. So the object must be placed at a distance between 0 and 40 cm from the mirror. In this region the image is virtual, erect and enlarged, and it forms behind the mirror.
(B) Ray diagram (object 20 cm from the mirror):

(C) Such a mirror is used as a shaving mirror (or by dentists and ENT doctors) to see an enlarged, erect image.
In simple words: Hold the object closer than 40 cm and the concave mirror shows a large upright image.

Teacher's Note:
1. Within the focal length the reflected rays diverge, so the image is virtual.
2. The image in this range is virtual, so it cannot be caught on a screen.

Question 6. Slides are small transparencies mounted in sturdy frames ideally suited to magnification and projection, since they have a very high resolution and a high image quality. There is a tray where the slides are to be put into a particular orientation so that the viewers can see the enlarged erect images of the transparent slides. This means that the slides will have to be inserted upside down in the projector tray. To show her students the images of insects that she investigated in the lab, Mrs. Iyer brought a slide projector. Her slide projector produced a 500 times enlarged and inverted image of a slide on a screen 10 m away. (A) Based on the text and data given in the above paragraph, what kind of lens must the slide projector have? (B) If v is the symbol used for image distance and u for object distance then with one reason state what will be the sign for v/u in the given case? (C) (i) A slide projector has a convex lens with a focal length of 20 cm. The slide is placed upside down 21 cm from the lens. How far away should the screen be placed from the slide projector's lens so that the slide is in focus? (ii) When a slide is placed 15 cm behind the lens in the projector, an image is formed 3 m in front of the lens. If the focal length of the lens is 14 cm, draw a ray diagram to show image formation (not to scale).
Answer:
(A) The slide projector forms a real, inverted and enlarged image on a screen, so it must have a convex (converging) lens.
(B) The magnification is m = v/u. The image is inverted and 500 times larger, so m = −500. Therefore v/u is negative (−500): u is negative and v is positive, so they have opposite signs.
(C)(i) f = +20 cm, u = −21 cm. \( \dfrac{1}{v}=\dfrac{1}{f}+\dfrac{1}{u}=\dfrac{1}{20}-\dfrac{1}{21}=\dfrac{1}{420} \), so the screen must be 420 cm (4.2 m) from the lens.
(ii) The slide (object) is just beyond the focus (15 cm against f ≈ 14 cm), so a real, inverted and highly enlarged image forms far away. (With u = −15 cm and v = +300 cm the lens formula gives f = 14.3 cm, i.e. about 14 cm; the diagram below is only a schematic, not to scale: the slide is just beyond F₁ and the image is far beyond 2F₂.)

In simple words: A projector uses a convex lens; the slide sits just outside the focus and the screen is far away.

Teacher's Note:
1. A real image from a lens has v positive and u negative, so v/u is negative.
2. The closer the slide is to the focus, the farther the screen and the larger the picture.

Question 7. (a) A security mirror used in a big showroom has radius of curvature 5 m. If a customer is standing at a distance of 20 m from the cash counter, find the position, nature and size of the image formed in the security mirror. (b) Neha visited a dentist in his clinic. She observed that the dentist was holding an instrument fitted with a mirror. State the nature of this mirror and reason for its use in the instrument used by dentist.
Answer:
(a) A security mirror is a convex mirror. R = +5 m, so f = +2.5 m; u = −20 m.
\( \dfrac{1}{v}=\dfrac{1}{f}-\dfrac{1}{u}=\dfrac{1}{2.5}+\dfrac{1}{20}=\dfrac{9}{20} \), so v = 20/9 ≈ +2.22 m (behind the mirror).
m = −v/u = −(2.22)/(−20) ≈ +0.11 (about 1/9). The image is virtual, erect and diminished to about one-ninth of the customer's size.
(b) The dentist's mirror is a concave mirror. When the teeth are held between the pole and the focus, it forms a virtual, erect and enlarged image, so the dentist can see a large clear image of the teeth.
In simple words: A convex security mirror shows a wide view with a small image; a dentist's concave mirror enlarges the teeth.

Teacher's Note:
1. R = 2f, with a positive sign for a convex mirror.
2. Both mirror types are used because of the kind of image they give.

Question 8. Rishi went to a palmist to show his palm. The palmist used a special lens for this purpose. (i) State the nature of the lens and reason for its use. (ii) Where should the palmist place/hold the lens so as to have a real and magnified images of an object? (iii) If the focal length of this lens is 10 cm and the lens is held at a distance of 5 cm from the palm, use lens formula to find the position and size of the image.
Answer:
(i) The lens is a convex lens. It is used as a magnifying glass, because when an object is held within its focal length it gives an erect, virtual and enlarged image.
(ii) For a real and magnified image the object must be placed between F and 2F of the lens (that is, at a distance greater than f but less than 2f).
(iii) f = +10 cm, u = −5 cm: \( \dfrac{1}{v}=\dfrac{1}{10}-\dfrac{1}{5}=-\dfrac{1}{10} \), so v = −10 cm: the image is formed 10 cm from the lens on the same side as the palm.
m = v/u = (−10)/(−5) = +2, so the image is virtual, erect and twice the size of the object.
In simple words: Held within its focal length, the convex lens shows the palm twice as big and upright.

Teacher's Note:
1. A virtual image needs the object between the optical centre and F.
2. m = +2 means erect and double in size.

Question 9. A truck uses a convex mirror as view finder whose radius of curvature is 2.0 m. A maruti car is coming behind the truck at a distance of 10 m. What will be the position of the image of the car and size of the image of the car when observed by the driver of the truck through the convex mirror?
Answer:
R = +2.0 m, so f = +1.0 m. Object distance u = −10 m.
\( \dfrac{1}{v}=\dfrac{1}{f}-\dfrac{1}{u}=1+\dfrac{1}{10}=\dfrac{11}{10} \), so v = +0.91 m (10/11 m), i.e. the image is about 0.9 m behind the mirror.
m = −v/u = −(10/11)/(−10) = +1/11 ≈ 0.09.
The image is virtual, erect and about one-eleventh (≈ 0.09 times) the size of the car.
In simple words: The car's image is a small upright image about 0.9 m behind the mirror and one-eleventh of its real size.

Teacher's Note:
1. Convex mirrors always give diminished images, which is why vehicles use them.
2. The magnification here is positive, which shows an erect image.

Question 10. A convex mirror used on a car has a radius of curvature 3 m. If a bus is located 5 m from this mirror, find the position, nature and size of the image.
Answer:
R = +3 m, so f = +1.5 m; u = −5 m.
\( \dfrac{1}{v}=\dfrac{1}{f}-\dfrac{1}{u}=\dfrac{1}{1.5}+\dfrac{1}{5}=\dfrac{13}{15} \), so v ≈ +1.15 m (behind the mirror).
m = −v/u = −1.15/(−5) ≈ +0.23.
The image is virtual, erect and diminished (about 0.23 times, nearly one-fourth, the size of the bus).
In simple words: The bus appears as a small upright image about 1.15 m behind the mirror.

Teacher's Note:
1. Use f = R/2 with a positive sign for the convex mirror.
2. Magnification between 0 and 1 means a diminished erect image.

Case Based Questions

Question 11. We know that the characteristics of image formed by a concave mirror depend on the position of the object with respect to the mirror. When an object is placed between F and infinity, the image formed is real and inverted. But when the object is placed between F and mirror it cannot be obtained on the screen. The image formed in this case is virtual, erect and magnified. Such image may be seen by looking in the mirror directly. When the object is moved from focus towards infinity, the image moves from infinity towards focus and its size decreases. When object is placed at 2F (the centre of curvature) image of the same size is formed at 2F, itself.
(i) If an object is placed 10 cm in front of a concave mirror of focal length 20 cm, the image will be (a) real, erect, magnified (b) real, inverted, diminished (c) virtual, erect and magnified (d) virtual, erect and diminished
(ii) The minimum distance between the object and its real image for concave mirror is (a) zero (b) f (c) 2f (d) 4f
(iii) An object is placed at the centre of curvature of a concave mirror. The distance between its image and the pole is (a) equal to f (b) equal to 2f (c) greater than 2f (d) between f and 2f
(iv) A candle flame 3 cm high is placed at a distance of 3 m from a wall. How far from the wall must a concave mirror be placed in order that it may form an image of the flame 9 cm high on the wall? (a) 1.5 m (b) 2.5 m (c) 3 m (d) 4.5 m
Answer:
(i) (c) The object (10 cm) is inside the focal length (20 cm), so the image is virtual, erect and magnified. Check: v = +20 cm, m = +2.
(ii) (a) zero. For a real image the object and image distances are least when the object is at the centre of curvature: the image is formed at the same point, so the separation is zero.
(iii) (b) equal to 2f. The image of an object at C is formed at C, and the distance of C from the pole is R = 2f.
(iv) (d) 4.5 m. The image is real, so m = −3 and v = 3u (both negative). The wall is farther than the candle: |v| − |u| = 3 m, so 3|u| − |u| = 3 and |u| = 1.5 m. The wall carries the real image, so the mirror must be 4.5 m from the wall.
In simple words: Use m = −3 for a real image three times as tall, and the 3 m gap between candle and wall to find the distances.

Teacher's Note:
1. Check the position of the object against f and 2f before choosing the type of image.
2. For the wall problem, m = −v/u fixes the ratio of the two distances.

Teacher's Note:
1. The object at 2F is the key clue to find f quickly from a table.
2. The minimum object-to-real-image distance for a convex lens is 4f.

Question 13. If rays parallel to the axis fall on a thin lens, they will be focused to a point called the focal point, F. This will not be precisely true for a lens with spherical surfaces. But it will be very nearly true; i.e., parallel rays will be focused to a tiny region that is nearly a point, if the diameter of the lens is small compared to radii of curvature of the two lens surfaces. This condition is satisfied by a thin lens. To find the image point by drawing rays would be difficult if we had to determine all the refractive angles. So, to find the image point, we need to consider only three rays: Ray 1 is drawn parallel to the axis; therefore it is refracted by the lens so that it passes along a line through the focal point F. Ray 2 is drawn on a line, passing through the other focal point F and emerges from the lens parallel to the axis. Ray 3 is directed towards the very centre of the lens where the two surfaces are essentially parallel to each other; this ray therefore emerges from the lens at the same angle as it entered. Any two of these rays will suffice to locate the image point, which is the point where they intersect.
(i) How will the image formed by convex lens be affected if the upper half of the lens is wrapped with the black paper? (a) There will be no effect. (b) The brightness of the image will be reduced. (c) The upper half of the image will be absent. (d) The size of the image is reduced to one-half.
(ii) The refractive index of glass with respect to air is 3/2 and the refractive index of water with respect to air is 4/3. The refractive index of glass with respect to water will be (a) 1.125 (b) 1.225 (c) 1.425 (d) 1.525
(iii) To form a real image twice the size of the object, using a convex lens of focal length 20 cm, the object distance must be (a) less than 20 cm (b) greater than 40 cm (c) between 20 cm and 40 cm (d) none of these
(iv) Two thin lenses of power +3.5 D and – 2.5 D are placed in contact, then power and focal length of the lens combination is (a) +1 D, +100 cm (b) +2 D, +150 cm (c) +1 D, +200 cm (d) +2 D, +100 cm
(v) Convex lens and concave lens are also known as (a) converging lens and converging lens (b) converging lens and diverging lens (c) diverging lens and converging lens (d) diverging lens and diverging lens
Answer:
(i) (b). Each part of the lens forms the whole image, so the image stays complete and of the same size, but less light passes through, so the brightness is reduced.
(ii) (a) 1.125. n_gw = n_ga/n_wa = (3/2)/(4/3) = 9/8 = 1.125.
(iii) (c) between 20 cm and 40 cm. For a real image with m = −2: v = −2u and \( \dfrac{1}{v}-\dfrac{1}{u}=\dfrac{1}{20} \) gives \( -\dfrac{1}{2u}-\dfrac{1}{u}=\dfrac{1}{20} \), so u = −30 cm, which lies between f and 2f. (The question asks for a real image; a virtual image twice the size would need u = −10 cm.)
(iv) (a) +1 D, +100 cm. P = +3.5 + (−2.5) = +1 D and f = 1/P = 1 m = +100 cm.
(v) (b). A convex lens converges a parallel beam; a concave lens diverges it.
In simple words: A half-covered lens still gives a complete but dimmer image; powers of lenses in contact add up.

Teacher's Note:
1. Check each sub-part with the matching formula: n = n₁/n₂, P = P₁ + P₂, 1/v − 1/u = 1/f.
2. Covering part of a lens reduces brightness, not the completeness of the image.

Question 14. The radius of curvature of a convex mirror of a moving automobile is 2.0 m. A truck is coming behind and is at a distance of 3.5 m.
1. What is the image distance behind the mirror? (a) 0.48 m (b) 0.24 m (c) 0.78 m (d) 0.84 m
2. The nature of the image is (a) Diminished and inverted (b) Virtual and erect (c) Virtual and inverted (d) Real and erect
3. The size of the image relative to the size of the truck is (a) 22% (b) 24% (c) 26% (d) 28%
Answer:
R = +2.0 m, so f = +1.0 m; u = −3.5 m.
1. (c) 0.78 m. \( \dfrac{1}{v}=\dfrac{1}{1}-\dfrac{1}{-3.5}=\dfrac{1}{1}+\dfrac{1}{3.5}=\dfrac{4.5}{3.5} \), so v = 3.5/4.5 ≈ +0.78 m (behind the mirror).
2. (b) Virtual and erect. A convex mirror always forms a virtual, erect and diminished image of a real object.
3. (a) 22%. m = −v/u = 0.778/3.5 ≈ 0.22, so the image is about 22% of the size of the truck.
In simple words: The truck's image is virtual, upright, 0.78 m behind the mirror and about 22% of its size.

Teacher's Note:
1. Positive v means the image is behind the mirror.
2. Magnification between 0 and 1 means a diminished image.

Question 15. 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.
1. The image distance is (a) –80 cm (b) –90 cm (c) –100 cm (d) –110 cm
2. Magnitude of the focal length of the mirror is: (a) 15 cm (b) 16 cm (c) 17 cm (d) 18 cm
3. The mirror used here is a ______ mirror. (a) concave (b) convex (c) plano concave (d) plano convex
Answer:
The image is real and inverted, so h′ = −4 cm and m = h′/h = −4. Also u = −20 cm.
1. (a) −80 cm. m = −v/u gives −4 = −v/(−20), so v = −80 cm.
2. (b) 16 cm. \( \dfrac{1}{f}=\dfrac{1}{v}+\dfrac{1}{u}=-\dfrac{1}{80}-\dfrac{1}{20}=-\dfrac{5}{80} \), so f = −16 cm; the magnitude is 16 cm.
3. (a) concave. A converging mirror with a negative focal length is a concave mirror.
In simple words: A real image four times taller means m = −4, which puts the image 80 cm away and f at 16 cm.

Teacher's Note:
1. A converging mirror is the same as a concave mirror.
2. Use m = −v/u first, then the mirror formula.

Question 16. When the rays of light travel from one transparent medium to another, the path of light are deviated. This phenomenon is called the refraction of light. The bending of light depends on the optical density of the medium through which the light pass. The speed of light varies from medium to medium. A medium in which the speed of light is more is an optically rarer medium whereas in which the speed of light is less is an optically denser medium. Whenever light goes from one medium to another, the frequency of light does not change however, speed and wavelength change. It concluded that change in the speed of light is the basic cause of refraction.
1. When light travels from air to glass, the ray of light bends: (a) towards the normal (b) away from normal (c) anywhere (d) none of these
2. A ray of light passes from medium A to another medium B. No bending of light occurs if the ray of light hits the boundary of medium B at an angle of: (a) 0° (b) 45° (c) 90° (d) 120°
3. When light passes from one medium to another, the frequency of light: (a) increases (b) decreases (c) remains same (d) none of these
4. When light passes from glass to water, the speed of light: (a) increases (b) decreases (c) remains same (d) first increases then decrease
5. The bottom of a pool filled with water appears to be ______ due to the refraction of light. (a) shallower (b) deeper (c) at same depth (d) empty
Answer:
1. (a) towards the normal. Glass is optically denser than air, so the ray bends towards the normal.
2. (a) 0°. A ray along the normal (angle of incidence 0°) passes straight into the second medium without bending.
3. (c) remains same. The frequency is fixed by the source; only the speed and wavelength change.
4. (a) increases. Glass (n ≈ 1.5) is denser than water (n ≈ 1.33), so the light speeds up when it goes from glass to water.
5. (a) shallower. Rays from the bottom bend away from the normal on leaving water, so the bottom appears raised and the pool looks less deep.
In simple words: Refraction occurs because the speed of light changes; frequency stays the same.

Teacher's Note:
1. Rarer to denser: bends towards the normal; denser to rarer: bends away.
2. The apparent raising of the bottom is also why a coin in water looks closer.

Question 17. A solar furnace is used to generate heat using sun radiation. The solar furnace consists of flat solar tracking heliostat, a parabollic collecting mirror, an attenuator or shutter, and the test zone area.
(i) What is solar furnace? (a) Uses concentrated solar power to produce high temperatures for industry (b) Uses solar energy to evaporate water and collect it within the same closed system (c) Uses solar energy to dry substances (d) Uses solar energy to dry liquids
(ii) Select the material that can be used as a heliostat surface to reflect maximum radiation (a) Ribbon silicon (b) Polycrystalline silicon (c) Convex mirror (d) Plane mirror
(iii) The device that can control the amount of light entering the furnace (a) Concentrator (b) Heliostat (c) Attenuator (d) None of the above
(iv) Name the Parabolic mirror used in this process (a) Convex mirror (b) Plane mirror (c) Concave mirror (d) Combination of convex and concave mirror
Answer:
(i) (a). A solar furnace concentrates sunlight to produce very high temperatures.
(ii) (d) Plane mirror. The heliostat is a flat mirror that tracks the Sun and reflects the sunlight to the collecting (concentrating) mirror.
(iii) (c) Attenuator. The attenuator (shutter) controls how much light enters the furnace.
(iv) (c) Concave mirror. A parabolic collecting mirror is a concave (converging) mirror that brings the parallel rays of the Sun to its focus.
In simple words: A flat mirror follows the Sun and a big concave mirror focuses the light to make intense heat.

Teacher's Note:
1. Concave mirrors converge parallel rays at the focus.
2. The temperature rises where the rays are concentrated.

Question 18. Rear view mirror is a device that allows the driver to see rearward. It usually finds its place at the top of windscreen in side of the cabin. This device is one of the most basic but essential safety devices in the vehicle. It provides assistance to the driver during overtaking, parking in reverse gear etc. Generally, vehicles also have a pair of mirrors attached to the body from outside. They are popular as ‘side mirrors’ or Outer Rear View Mirrors (ORVM) which serve the same purpose.
(i) For a real object, which of the following can produce a real image? (a) Plane mirror (b) Concave mirror (c) Convex mirror (d) None of the above
(ii) An object at a distance of +15 cm is slowly moved towards the pole of a convex mirror. The image will get... (a) shortened and real (b) enlarged and real (c) enlarge and virtual (d) diminished and virtual
(iii) A convex mirror is used: (a) by a dentist (b) for shaving (c) as a rear view mirror in vehicles (d) as a light reflector for obtaining a parallel beam of light
(iv) The word ‘AMBULANCE’ is written on the vehicle as: (a) CNALUBMA (b) AMBULANCE (c) ECNALUBMA (d) None of the above
(v) Mark the correct statement: (a) Convex mirror conform images of objects spread over a large area. (b) Convex mirrors are used by dentist. (c) In convex mirror image is formed larger in size, erect and real. (d) Convex mirror forms real image.
Answer:
(i) (b) Concave mirror. For a real object, only a concave mirror can form a real image (when the object is beyond its focus); plane and convex mirrors give virtual images only.
(ii) (d) diminished and virtual. A convex mirror always forms a virtual, erect and diminished image; as the object nears the pole the image gets larger but stays smaller than the object.
(iii) (c) as a rear view mirror in vehicles, since it gives an erect, diminished image with a wide field of view.
(iv) (b). The word is written laterally inverted (each letter reversed) in front of the ambulance, so that a driver looking in the rear-view mirror reads AMBULANCE correctly.
(v) (a). A convex mirror forms images of objects spread over a large area (a wide field of view); its image is always virtual and diminished, and dentists use concave mirrors.
In simple words: Convex mirrors show a wide view with small upright images, and ambulance is written reversed so that the rear-view mirror shows it correctly.

Teacher's Note:
1. A convex mirror cannot form a real image of a real object.
2. Lateral inversion in the mirror turns the reversed word back into a readable one.