CBSE Class 8 Science Light Solved Examples

Question. How will the figure shown here  look if seen through the rear view mirror of a car?

Answer : A

Question. Four students, P, Q, R and S are standing in front of a mirror facing it. Their positions, as seen from the top, are shown below. Which of these students can see Q as well as Q's reflection in the mirror?

(a) only P
(b) only R and S
(c) only R
(d) only S

Answer : B

Question. A mirror is placed along the dotted line on the letter N. In which position will the image in the mirror coincide EXACTLY with the other half of the letter?

(a) 
(b) It is true only if the surface is CONCAVE or PLANE.
(c) It is true only if the surface is made of GLASS.
(d) It is true only if the surface is PLANE.

Answer : D

Question. When light is reflected from a smooth surface, the angle of incidence must be equal to the angle of reflection." Under which of the following situations is the above statement true?" It is always true.
(a) It is always true.
(b) It is true only if the surface is CONCAVE or PLANE.
(c) It is true only if the surface is made of GLASS.
(d) It is true only if the surface is PLANE.

Answer : A

Question. When we see with only one eye (by covering the other eye with our hands), what is the difference in what we see?
(a) There is no difference.
(b) We cannot distinguish colours.
(c) We cannot distinguish depths.
(d) We cannot distinguish heights.

Answer : C

Question. Shown here is the curved steel rod with a mirror attached to the end of it, that dentists use to examine our teeth. Identify the kind of mirror which is used in this instrument.

(a) Concave mirror
(b) Convex mirror
(c) Plane mirror
(d) Spherical mirror

Answer : A

Question. Two plane mirrors are inclined towards each other as shown in the figure. An object is placed between them.  Which of the following is true about the images that will be formed?

(a) Two images of the object are formed, one in each mirror.
(b) The number of images will depend on the angle between the mirrors.
(c) The number of images will depend on the length of the mirrors.
(d) The number of images will be determined by how big the object is.

Answer : B

Question. A RED coloured object in GREEN light looks BLACK because it absorbs all the green light. In BLUE light, how would a YELLOW object look?
(a) Red
(b) Magenta
(c) White
(d) Black

Answer : D

Question. Among RED, GREEN and BLUE, which colours would be ABSORBED by a CYAN object?
(a) Red
(b) Green
(c) Blue
(d) Yellow

Answer : A

Question. The arrangement shown below was used in 19th century theatres to create images of ghosts that could appear and disappear at will. What should be placed at X for the trick to work?

(a) a light source
(b) a mirror
(c) another actor
(d) a lens

Answer : A

Question. Shakuntala takes some amount of water in a measuring cylinder and reads the value keeping her eye at the level of water. Then she adds a little more water. For measuring the new level of water, she should

(a) keep her eye at the same level
(b) raise the level of her eye
(c) lower the level of her eye
(d) (Re-read the first level correctly)

Answer : B

Question. When light is reflected from a smooth surface, the angle of incidence must be equal to the angle of reflection." Under which of the following situations is the above statement true?
(a) It is always true.
(b) It is true only if the surface is CONCAVE or PLANE.
(c) It is true only if the surface is made of GLASS.
(d) It is true only if the surface is PLANE.

Answer : A

Question. Which box(es) of the table would best represent an image formed by a plane mirror?
Use the table given below to answer the question.

(a) 1 and 3 only
(b) 2 , 4 and 6 only
(c) 4 only
(d) 5 only

Answer : C

Question. If the image formed by a concave mirror (shown in the figure) fits into Box 1 of the table given above then the object is placed at _________.

Question. Use the table given below to answer the question.

(a) A
(b) B
(c) C
(d) D

Answer : C

Question. Any three colours of light, which produce white light when combined with the correct intensity, are called PRIMARY COLOURS of light. (Red, Green and Blue) Colours produced by the addition of equal intensities of two primary colours of light are called SECONDARY COLOURS of light. (Yellow, Cyan and Magenta) Any two colours of light, which produce white when combined, are said to be COMPLEMENTARY COLOURS of each other. Study the figure given here and answer the following question. Which of the following colours is a complementary colour of 'BLUE'?

(a) Cyan
(b) Magenta
(c) Red
(d) Yellow

Answer : D
 

Solved Examples

EXERCISES

Ex.1: Where is the image of an object placed 7 centimeters away from a 5 centimeter focal length of a Convex lens? Concave mirror? Are the images real or virtual in each case? [3 Marks]

Answer:
1. A convex lens has a positive focal length, so \(f\) = 5 cm and object distance \(u\) = -7 cm.
2. Applying the lens equation \(\frac{1}{f} = \frac{1}{v} - \frac{1}{u}\), we get \(\frac{1}{5} = \frac{1}{v} - \frac{1}{-7}\), which gives \(\frac{1}{5} = \frac{1}{v} + \frac{1}{7}\).
3. Solving for \(v\) gives \(v\) = 17.5 cm, and the image is real since \(v\) is positive.
4. A concave mirror with a focal length of 5 cm and object distance of -7 cm yields the same result, forming a real image 17.5 cm in front of the mirror.

Teacher's Note:
a) The lens equation connects focal length, image distance, and object distance.
b) Sign convention must be applied carefully to object distances.
c) A positive image distance for a lens indicates a real image formed on the opposite side.
d) Both convex lenses and concave mirrors can form real images when the object is placed beyond the focal length.

Ex. 2: A laserbeam strikes a vertical surface at an angle of 48°. The reflected beam can be seen as a spot on a horizontal surface. The spot is 10 meters away from the point of incidence on the vertical surface. How far is the horizontal distance from the spot to the vertical surface? [3 Marks]

Answer:
1. The angle of reflection equals the angle of incidence, so the angle of reflection is also 48°.
2. The angle between the vertical surface and the reflected beam is 90° - 48° = 42°.
3. The reflected beam acts as the hypotenuse of length 10 meters.
4. The horizontal distance is the horizontal projection given by 10 sin(42°) = 6.7 meters.

Teacher's Note:
a) The law of reflection states that the angle of incidence equals the angle of reflection.
b) Trigonometric ratios help resolve the components of right-angled triangles.
c) Make sure to calculate complementary angles with respect to surfaces correctly.

Ex.3: A convex lens of focal length 25 cm and a concave lens of focal length 10 cm are placed in close contact with each other.
i) What is the power of this combination?
ii) What is the focal length of this combination? [4 Marks]

Answer:
1. Focal length of the convex lens \(f_1\) = + 25 cm = 0.25 m, so power \(P_1 = \frac{1}{0.25}\) = 4 D.
2. Focal length of the concave lens \(f_2\) = - 10 cm = - 0.1 m, so power \(P_2 = \frac{1}{-0.1}\) = - 10 D.
3. The power of the combination is \(P = P_1 + P_2\) = 4 D - 10 D = -6 D.
4. The focal length of the combination is \(f = \frac{1}{P} = \frac{1}{-6}\) m = -16.6 cm.

Teacher's Note:
a) Power of a lens is expressed in dioptres and is the reciprocal of focal length in meters.
b) Powers of lenses in contact are added algebraically.
c) Convex lenses have positive power while concave lenses have negative power.
d) Verify the final focal length by converting back from total power.

Ex. 4: The speed of light in air is \(3 \times 10^8\) m/s and the speed of light in water is \(2.26 \times 10^8\) m/s. Find the refractive index of water. [2 Marks]

Answer:
1. Speed of light in air \(c = 3 \times 10^8\) m/s and speed in water \(v = 2.26 \times 10^8\) m/s.
2. Refractive index \(n = \frac{c}{v} = \frac{3 \times 10^8}{2.26 \times 10^8}\) = 1.33.

Teacher's Note:
a) Refractive index is the ratio of the speed of light in a vacuum or air to its speed in a medium.
b) It is a dimensionless quantity having no units.

Ex.5: A convex mirror used for rear-view on an automobile has a radius of curvature of 3.00 m. If a bus is located at 5.00 m from this mirror, find the position, nature and size of the image. [4 Marks]

Answer:
1. Given radius of curvature, R = + 3.00 m, object-distance, u = - 5.00 m.
2. Focal length, f = R / 2 = \(\frac{3.00}{2}\) = + 1.50 m.
3. Using mirror formula \(\frac{1}{v} + \frac{1}{u} = \frac{1}{f}\), substituting values gives \(\frac{1}{v} + \frac{1}{-5.00} = \frac{1}{1.50}\).
4. Solving for image distance gives v = + 1.15 m, meaning the image is 1.15 m behind the mirror.
5. Magnification m = \(\frac{-v}{u}\) = \(\frac{-1.15}{-5.00}\) = + 0.23.
6. The image is virtual, erect and smaller in size by a factor of 0.23.

Teacher's Note:
a) Convex mirrors always produce virtual, erect and diminished images regardless of object position.
b) Remember to use proper sign conventions: radius and focal length of convex mirror are positive, while object distance is always negative.
c) The positive sign of magnification indicates that the image formed is erect.
d) Always verify the calculated image distance by substituting it back into the mirror formula.

Ex.6: A 2.0 cm tall object is placed perpendicular to the principal axis of a convex lens of focal length 10 cm. The distance of the object from the lens is 15 cm. Find the nature, position and size of the image. Also find its magnification. [4 Marks]

Answer:
1. Given height of object h = + 2.0 cm, focal length f = + 10 cm, object-distance u = - 15 cm.
2. Using lens formula \(\frac{1}{v} - \frac{1}{u} = \frac{1}{f}\), rearranging gives \(\frac{1}{v} = \frac{1}{f} + \frac{1}{u}\).
3. Substituting values: \(\frac{1}{v} = \frac{1}{10} + \frac{1}{-15}\) = \(\frac{3 - 2}{30}\) = \(\frac{1}{30}\).
4. Solving for image distance gives v = + 30 cm, so the image is formed at a distance of 30 cm on the other side of the lens.
5. Magnification m = \(\frac{v}{u}\) = \(\frac{30}{-15}\) = - 2.
6. Height of image h' = m × h = (- 2) × 2.0 cm = - 4.0 cm, meaning a real, inverted image of height 4.0 cm is formed.

Teacher's Note:
a) A convex lens has a positive focal length, and real objects placed in front have a negative object distance.
b) A negative sign for image height and magnification indicates that the formed image is real and inverted.
c) The lens formula differs from the mirror formula by a negative sign between the inverse of distance terms.
d) Check units and dimensions at every step to ensure dimensional consistency in optical problems.

Basic Level assignment

1. An object is kept in front of a concave mirror of focal length 15 cm. The image formed is three times the size of the object. Calculate the two possible distances of the object from the mirror. [3 Marks]

Answer:
1. Given: Focal length f = -15 cm.
2. Case I: For a real and inverted image, magnification m = -3.
3. Using m = -v / u, we get -3 = -v / u, so v = 3u.
4. Using mirror formula \(\frac{1}{f} = \frac{1}{v} + \frac{1}{u}\), we have \(\frac{1}{-15} = \frac{1}{3u} + \frac{1}{u}\).
5. \(\frac{1}{-15} = \frac{4}{3u}\), which gives u = -20 cm.
6. Case II: For a virtual and erect image, magnification m = +3.
7. Using m = -v / u, we get 3 = -v / u, so v = -3u.
8. Using mirror formula, \(\frac{1}{-15} = \frac{1}{-3u} + \frac{1}{u}\).
9. \(\frac{1}{-15} = \frac{2}{3u}\), which gives u = -10 cm.
10. Thus, the two possible object distances are 20 cm and 10 cm in front of the mirror.

Teacher's Note:
a) Concave mirrors can form both real and virtual images depending on the object position.
b) Always consider both positive and negative magnification values when two possible distances are asked.
c) Sign convention must be strictly followed throughout the mirror formula calculation.

2. A concave mirror is placed in water. Will there be any change in the focal length? Give reason. [2 Marks]

Answer:
1. There will be no change in the focal length of a concave mirror when placed in water.
2. The focal length of a mirror depends only on the radius of curvature of its reflecting surface (f = R / 2).
3. It does not depend on the surrounding medium because reflection of light does not involve refraction.

Teacher's Note:
a) Focal length of lenses depends on the refractive index of the surrounding medium, but mirrors do not.
b) Students often confuse lenses and mirrors regarding the effect of water medium on focal length.

3. Calculate the speed of light in a medium, whose critical angle is 45°. [3 Marks]

Answer:
1. Given critical angle \(C = 45^\circ\).
2. The refractive index of the medium is given by \(\mu = \frac{1}{\sin C}\).
3. \(\mu = \frac{1}{\sin 45^\circ} = \frac{1}{1 / \sqrt{2}} = \sqrt{2} \approx 1.414\).
4. Refractive index is also related to the speed of light by \(\mu = \frac{c}{v}\), where c is the speed of light in vacuum.
5. \(v = \frac{c}{\mu} = \frac{3 \times 10^8}{\sqrt{2}} \approx 2.12 \times 10^8\) m/s.

Teacher's Note:
a) Critical angle is the angle of incidence in the denser medium for which the angle of refraction in the rarer medium is \(90^\circ\).
b) Standard value of speed of light in vacuum is \(3 \times 10^8\) m/s.
c) Rationalize the denominator or use numerical approximation for \(\sqrt{2}\) to simplify the final velocity.

4. A beaker is filled with water to a height of 12.5 cm. The apparent depth of a needle lying at the bottom of the beaker is measured to be 9.4 cm. What is the refractive index of water? [3 Marks]

Answer:
1. Given real depth (d) = 12.5 cm.
2. Apparent depth (d') = 9.4 cm.
3. Refractive index of water \(\mu = \frac{\text{Real depth}}{\text{Apparent depth}}\).
4. \(\mu = \frac{12.5}{9.4}\).
5. \(\mu \approx 1.33\).

Teacher's Note:
a) Refractive index measures the bending of light when passing from one medium to another.
b) Division must be carried out carefully to two decimal places.
c) The refractive index of water is a standard dimensionless quantity approximately equal to \(\frac{4}{3}\).

5. Magnification m = + 1 for a plane mirror. What is the significance of m = 1 and the + sign of m? [2 Marks]

Answer:
1. The magnitude 1 indicates that the size of the image is exactly equal to the size of the object.
2. The positive sign indicates that the image formed is virtual and erect, lying above the principal axis.

Teacher's Note:
a) Magnification is defined as the ratio of image height to object height.
b) Plane mirrors always form virtual, erect images of the same size as the object.

6. An object is placed at 0.06 m from a convex lens of focal length 0.10 m. Calculate the position of the image. [3 Marks]

Answer:
1. Given object distance u = -0.06 m = -6 cm.
2. Focal length f = +0.10 m = +10 cm.
3. Using lens formula \(\frac{1}{f} = \frac{1}{v} - \frac{1}{u}\), we get \(\frac{1}{v} = \frac{1}{f} + \frac{1}{u}\).
4. \(\frac{1}{v} = \frac{1}{10} + \frac{1}{-6} = \frac{1}{10} - \frac{1}{6}\).
5. \(\frac{1}{v} = \frac{3 - 5}{30} = \frac{-2}{30} = \frac{-1}{15}\).
6. v = -15 cm = -0.15 m.
7. The image is formed at a distance of 0.15 m on the same side as the object.

Teacher's Note:
a) Convert meters to centimeters for easier arithmetic calculation if required.
b) Lens formula has a minus sign between terms, unlike the mirror formula.
c) A negative image distance for a convex lens indicates a virtual image formed on the object side.

7. Two thin lenses of focal lengths + 10 cm and - 5 cm. are kept in contact. What is the
(a) focal length (b) power of the combination? [3 Marks]

Answer:
1. Given focal lengths \(f_1 = +10\) cm = +0.1 m, and \(f_2 = -5\) cm = -0.05 m.
2. (a) Combined focal length F is given by \(\frac{1}{F} = \frac{1}{f_1} + \frac{1}{f_2}\).
3. \(\frac{1}{F} = \frac{1}{10} + \frac{1}{-5} = \frac{1}{10} - \frac{2}{10} = \frac{-1}{10}\).
4. F = -10 cm = -0.1 m.
5. (b) Power of the combination \(P = P_1 + P_2 = \frac{1}{f_1(\text{in meters})} + \frac{1}{f_2(\text{in meters})}\).
6. \(P = \frac{1}{0.1} + \frac{1}{-0.05} = 10 - 20 = -10\) D.

Teacher's Note:
a) Powers of lenses in contact are directly additive.
b) Focal length must be expressed in meters when calculating power in dioptres.
c) The negative sign of the equivalent focal length indicates that the combination behaves as a diverging lens.

8. If the power of lens is +5 dioptre, what is its focal length? [1 Mark]

Answer:
1. Power P = +5 D.
2. Focal length f = \(\frac{1}{P}\) = \(\frac{1}{5}\) m = +0.2 m = +20 cm.

Teacher's Note:
a) Power and focal length are inversely related by the formula P = 1 / f (with f in meters).
b) Positive power indicates a converging (convex) lens.

9. Two thin lenses of power +5 D and - 3 D are in contact. What is the focal length of the combination? [2 Marks]

Answer:
1. Given powers \(P_1 = +5\) D and \(P_2 = -3\) D.
2. Net power P = \(P_1 + P_2 = +5 + (-3) = +2\) D.
3. Focal length of the combination f = \(\frac{1}{P} = \frac{1}{2}\) m = +0.5 m = +50 cm.

Teacher's Note:
a) Net power of lenses in contact is the algebraic sum of individual powers.
b) Always convert the final power in dioptres to centimeters or meters as appropriate for focal length.

10. Where should an object be placed from a converging lens of focal length 20 cm, so as to obtain a real image of magnification 2? [3 Marks]

Answer:
1. Given focal length f = +20 cm for a converging (convex) lens.
2. For a real image, magnification m = -2.
3. Using m = v / u, we get -2 = v / u, so v = -2u.
4. Using lens formula \(\frac{1}{f} = \frac{1}{v} - \frac{1}{u}\), we substitute v = -2u.
5. \(\frac{1}{20} = \frac{1}{-2u} - \frac{1}{u} = \frac{-1 - 2}{2u} = \frac{-3}{2u}\).
6. 2u = -60 cm, which gives u = -30 cm.
7. The object should be placed at a distance of 30 cm in front of the converging lens.

Teacher's Note:
a) Real images formed by convex lenses are always inverted, hence magnification is negative.
b) Substituting v in terms of u simplifies the lens formula into a linear equation in one variable.
c) Verify the result: object at 30 cm (between F and 2F) forms a real, enlarged image beyond 2F.

11. A converging lens has a focal length of 20 cm in air. It is made of a material of refractive index 1.6. If it is immersed in a liquid of refractive index 1.3, what will be its new focal length? [3 Marks]

Answer:
1. Given data: Refractive index of lens \(n_g\) = 1.6, Refractive index of air \(n_a\) = 1.0, Focal length in air \(f_a\) = 20 cm, Refractive index of liquid \(n_l\) = 1.3.
2. Using Lens Maker's Formula for air: \(\frac{1}{f_a} = (n_{ga} - 1)(\frac{1}{R_1} - \frac{1}{R_2})\), where \(n_{ga} = \frac{1.6}{1.0} = 1.6\). Thus, \(\frac{1}{20} = (1.6 - 1) K\), which gives \(K = \frac{1}{20 \times 0.6} = \frac{1}{12}\).
3. Using Lens Maker's Formula for liquid: \(\frac{1}{f_l} = (n_{gl} - 1)K\), where \(n_{gl} = \frac{n_g}{n_l} = \frac{1.6}{1.3} = \frac{16}{13}\).
4. Substituting the values: \(\frac{1}{f_l} = (\frac{16}{13} - 1) \times \frac{1}{12} = \frac{3}{13} \times \frac{1}{12} = \frac{1}{52}\).
5. Therefore, the new focal length in the liquid \(f_l\) = 52 cm.

Teacher's Note:
a) The focal length of a lens depends on the refractive index of the surrounding medium relative to the lens material.
b) Students often forget to use the relative refractive index \(\frac{n_g}{n_l}\) when the lens is immersed in a liquid.
c) Since the refractive index of the liquid is less than that of the glass, the focal length increases.

12. Red light is incident on a thin converging lens of focal length ‘f’. Briefly explain how the focal length of the lens will change, if red light is replaced with blue light. [2 Marks]

Answer:
1. The focal length of a lens is inversely proportional to the refractive index of its material minus one, as given by the Lens Maker's Formula.
2. The refractive index of the lens material is greater for blue light than for red light (\(n_b \gt n_r\)) because refractive index decreases with an increase in wavelength.
3. Since \(n\) is greater for blue light, the term \((n - 1)\) increases, which means the focal length will decrease.

Teacher's Note:
a) Recall Cauchy's relation or the dependence of refractive index on wavelength (smaller wavelength means larger refractive index).
b) A common mistake is assuming focal length is constant for all colors of light.

13. Name the type of mirror used in the following situations.
(a) Headlights of a car. (b) side/rear-view mirror of a vehicle. (c) Solar furnace.
Support your answer with reason. [3 Marks]

Answer:
1. (a) Headlights of a car: Concave mirror, because when a light source is placed at the focus of a concave mirror, it produces a powerful parallel beam of light.
2. (b) Side or rear-view mirror of a vehicle: Convex mirror, because it always gives an erect, diminished image and provides a wider field of view.
3. (c) Solar furnace: Concave mirror, because it converges a parallel beam of sun rays to a sharp focal point, producing high heat.

Teacher's Note:
a) Identify the property of convergence and divergence for concave and convex mirrors respectively.
b) Always provide the specific reason based on image characteristics or ray behavior for full credit.
c) Do not confuse the use of concave mirrors in headlights with rear-view mirrors.

14. One-half of a convex lens is covered with a black paper. Will this lens produce a complete image of the object? Verify your answer experimentally. Explain your observations. [3 Marks]

Answer:
1. Yes, the lens will still produce a complete image of the object.
2. Experimental verification: When a convex lens placed in front of a lighted candle is half-covered with black paper, a full image of the candle is observed on the screen placed on the other side.
3. Explanation: Light rays coming from all parts of the object still refract through the uncovered half of the lens and form a complete image, though the intensity or brightness of the image is reduced because fewer light rays pass through it.

Teacher's Note:
a) Every part of a lens forms a complete image, not just one half.
b) Students often incorrectly write that half of the image will disappear.
c) Emphasize that covering half the lens only decreases the brightness (intensity) of the image.

15. An object 5.0 cm in length is placed at a distance of 20 cm in front of a convex mirror of radius of curvature 30 cm. Find the position of the image, its nature and size. [3 Marks]

Answer:
1. Given data: Object height \(h = +5.0\) cm, Object distance \(u = -20\) cm, Radius of curvature \(R = +30\) cm.
2. Focal length \(f = \frac{R}{2} = \frac{30}{2} = +15\) cm.
3. Using mirror formula: \(\frac{1}{f} = \frac{1}{v} + \frac{1}{u} \Rightarrow \frac{1}{15} = \frac{1}{v} + \frac{1}{-20} \Rightarrow \frac{1}{v} = \frac{1}{15} + \frac{1}{20} = \frac{4 + 3}{60} = \frac{7}{60}\).
4. Image position \(v = \frac{60}{7} \approx +8.57\) cm behind the mirror.
5. Magnification \(m = -\frac{v}{u} = \frac{h'}{h} \Rightarrow -\frac{60/7}{-20} = \frac{3}{7}\). Height of image \(h' = \frac{3}{7} \times 5 = +\frac{15}{7} \approx +2.14\) cm.
6. Nature: Virtual and erect. Size: 2.14 cm.

Teacher's Note:
a) Sign convention rules: object distance is always negative, focal length of a convex mirror is positive.
b) A positive sign for image distance \(v\) indicates that the image is formed behind the mirror.
c) Check that the magnification is less than 1, confirming a diminished image.

16. An object of size 7.0 cm is placed at 27 cm in front of a concave mirror of focal length 18 cm. At what distance from the mirror should a screen be placed, so that a sharp focussed image can be obtained? Find the size and the nature of the image. [3 Marks]

Answer:
1. Given data: Object size \(h = +7.0\) cm, Object distance \(u = -27\) cm, Focal length \(f = -18\) cm.
2. Using mirror formula: \(\frac{1}{f} = \frac{1}{v} + \frac{1}{u} \Rightarrow \frac{1}{-18} = \frac{1}{v} + \frac{1}{-27} \Rightarrow \frac{1}{v} = -\frac{1}{18} + \frac{1}{27} = \frac{-3 + 2}{54} = -\frac{1}{54}\).
3. Image distance \(v = -54\) cm. The screen should be placed at a distance of 54 cm in front of the mirror.
4. Magnification \(m = -\frac{v}{u} = -\frac{-54}{-27} = -2\).
5. Image size \(h' = m \times h = -2 \times 7.0 = -14.0\) cm.
6. Nature: Real and inverted, with a size of 14 cm.

Teacher's Note:
a) A real image can be caught on a screen, so the screen position corresponds to the image distance \(v\).
b) Both object distance and focal length for a concave mirror are negative according to the sign convention.
c) The negative sign of the image height confirms that the image is inverted and real.

17. A doctor has prescribed a corrective lens of power + 1.5 D. Find the focal length of the lens. Is the prescribed lens diverging or converging? [2 Marks]

Answer:
1. Power of the lens \(P = +1.5\) D.
2. Focal length \(f = \frac{1}{P} = \frac{1}{+1.5} = \frac{10}{15} = +\frac{2}{3}\) m = \(+0.67\) m or \(+67.7\) cm.
3. Since the power and focal length are positive, the lens is a converging (convex) lens.

Teacher's Note:
a) Use the standard relation \(P = \frac{1}{f(\text{in meters})}\).
b) A positive power always corresponds to a converging lens used to correct hypermetropia.

18. State the New Cartesian Sign Convention for lenses. [2 Marks]

Answer:
1. The object is always placed to the left of the lens, so light rays fall on the lens from the left.
2. All distances are measured from the optical center of the lens. Distances measured in the direction of incident light are positive, while those measured against the direction of incident light are negative.
3. Heights measured upwards and perpendicular to the principal axis are positive, and downward heights are negative.

Teacher's Note:
a) Emphasize that the optical center is taken as the origin of the coordinate system.
b) Clear sign convention rules are essential for avoiding numerical sign errors.

19. Write down the magnification formula for a lens in terms of object distance and image distance. How does it differ from the corresponding formula for a mirror? [2 Marks]

Answer:
1. Magnification formula for a lens in terms of image distance (\(v\)) and object distance (\(u\)) is \(m = \frac{v}{u}\).
2. The corresponding formula for a mirror is \(m = -\frac{v}{u}\). Thus, the lens formula has a positive sign, whereas the mirror formula has a negative sign.

Teacher's Note:
a) Do not confuse the magnification formulas for lenses and mirrors.
b) In both cases, magnification in terms of heights is the same: \(m = \frac{h'}{h}\).

20. A convex lens produces an inverted image magnified three times of an object placed at a distance of 15 cm from it. Calculate focal length of the lens. [3 Marks]

Answer:
1. Given data: Object distance \(u = -15\) cm, Magnification \(m = -3\) (since the image is inverted, magnification is negative).
2. Using magnification formula for lenses: \(m = \frac{v}{u} \Rightarrow -3 = \frac{v}{-15} \Rightarrow v = +45\) cm.
3. Using lens formula: \(\frac{1}{f} = \frac{1}{v} - \frac{1}{u} \Rightarrow \frac{1}{f} = \frac{1}{45} - \frac{1}{-15} = \frac{1}{45} + \frac{1}{15} = \frac{1 + 3}{45} = \frac{4}{45}\).
4. Focal length \(f = \frac{45}{4} = +11.25\) cm.

Teacher's Note:
a) For an inverted image, the magnification is always negative.
b) Pay close attention to the minus sign in the lens formula: \(\frac{1}{f} = \frac{1}{v} - \frac{1}{u}\).
c) A positive focal length confirms that the lens is a convex lens.

21. a) Name the type of mirror used in (i) headlight of a car, (ii) rear view mirror of a bus. Justify your choice by giving reason.
b) When sunlight is concentrated at a paper placed at the principal focus of a convex lens, what happens to the paper? Why?
c) Observe carefully the figures (a) and (b) and tell which media is optically denser and why? [3 Marks]

Answer:
1. (a) (i) Headlight of a car uses a concave mirror to produce a strong parallel beam of light. (ii) Rear-view mirror uses a convex mirror because it provides a wide field of view and erect, diminished images.
2. (b) The paper starts burning after some time. This happens because the convex lens converges the parallel rays of sunlight to a sharp point (focus), concentrating the heat energy onto the paper.
3. (c) In figure (a), the refracted ray bends towards the normal (\(r \lt i\)), so the second medium is optically denser than the first medium. In figure (b), the refracted ray bends away from the normal (\(r \gt i\)), so the first medium is optically denser than the second medium.

Teacher's Note:
a) This is a multi-part conceptual question testing mirrors, lens focusing properties, and Snell's law/refraction rules.
b) Ensure all three sub-parts are answered clearly with proper scientific justifications.
c) Mentioning angle comparison (\(i\) and \(r\)) is crucial for part (c).

In the following videos we have explained you about the concept of Light and its properties. Watch this video to understand the concepts and also post your questions and comments in the comments section.

ANSWERS (Basic Level)
1. 10 cm & 20 cm               2. No

3. 2.12 × 108 m/sec          4. 1.33

6. 15 cm                          7. (a) –10 cm (b) –10D

8. 20 cm                          9. 50 cm 

10. 30 cm                        11. 52 cm

12. Fr > Fb i.e., focal length decreases.

13. (a) Concave, (b) Convex (c) Concave.

14. Yes                           

15.8.6 cm, virtual and erect and diminished (2.2 cm)

16. 54 cm, 14 cm, magnified, real and inverted

17. + 0.67 m, convex lens. 20. 11.25 cm