CBSE Class 10 Science Light reflection and refraction Sure Shot Questions Set 01

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CHAPTER - 10
LIGHT - REFLECTION AND REFRACTION
ASSIGNMENT QUESTIONS SET - 1

1. Define refraction.

Answer:
1. Refraction is the bending of light when it passes obliquely from one transparent medium to another.
2. It happens because the speed of light is different in the two media.

Teacher's Note:
a) Always say "obliquely"; a ray along the normal does not bend.
b) The cause is the change in speed of light.

2. Define refractive index.

Answer:
1. The refractive index of a medium is the ratio of the speed of light in vacuum to the speed of light in the medium.
2. n = \( \frac{c}{v} \), where c is the speed of light in vacuum and v is the speed of light in the medium.
3. For two media, n₂₁ = \( \frac{\sin i}{\sin r} \).

Teacher's Note:
a) Refractive index tells how much a medium bends light.
b) It is a ratio, so it has no unit.
c) A higher refractive index means a denser optical medium.

3. What is the unit of refractive index?

Answer:
1. Refractive index is the ratio of two speeds (or of two sines).
2. So it has no unit.

Teacher's Note:
a) A ratio of like quantities is unitless.
b) Say "no unit" and give the reason.

4. List out the factors on which the refractive index of a medium depends.

Answer:
1. The nature (material) of the medium.
2. The wavelength (colour) of the light used.
3. The temperature of the medium.

Teacher's Note:
a) Colour matters because violet bends more than red.
b) It does not depend on the angle of incidence.

5. Define angle of incidence.

Answer:
1. The angle of incidence is the angle between the incident ray and the normal drawn at the point of incidence.

Teacher's Note:
a) The angle is measured from the normal, not from the surface.
b) Write it as i in diagrams.

6. What is the angle of incidence if a ray of light is incident normal to the surface separating the two media?

Answer:
1. A ray that falls normally travels along the normal.
2. So the angle between the ray and the normal is 0°.
3. Hence, the angle of incidence is 0° and the ray goes straight without bending.

Teacher's Note:
a) Normal incidence means 0°, not 90°.
b) The angle of refraction is also 0°.

7. What is a lens?

Answer:
1. A lens is a transparent material (such as glass) bounded by two surfaces, of which at least one is spherical.
2. It refracts light to form images.

Teacher's Note:
a) Lenses are of two types: convex and concave.
b) At least one surface must be curved.

8. What is a concave lens?

Answer:
1. A concave lens is a lens that is thinner in the middle and thicker at the edges.
2. It is bounded by two inward-curved (or one plane and one inward-curved) surfaces.
3. It diverges a parallel beam of light, so it is also called a diverging lens.

Teacher's Note:
a) Compare its shape with a convex lens.
b) Remember: concave means diverging.

9. What is the nature of the focus of a concave lens?

Answer:
1. Rays parallel to the principal axis diverge after passing through a concave lens.
2. They appear to come from a point on the same side as the incident light.
3. Hence, the focus of a concave lens is virtual.

Teacher's Note:
a) A virtual focus cannot be obtained on a screen.
b) Its focal length is taken as negative.

10. What type of image is formed by a concave lens?

Answer:
1. A concave lens always forms a virtual, erect and diminished image.
2. The image is formed on the same side of the lens as the object, between the optical centre and the focus.

Teacher's Note:
a) The image does not change nature with the object position.
b) The magnification is positive and less than 1.

11. A thin lens has a focal length f = -12 cm. Is it convex or concave lens?

Answer:
1. The focal length is negative.
2. A concave lens has a negative focal length, and a convex lens has a positive one.
3. Hence, the lens is a concave lens.

Teacher's Note:
a) Sign of f identifies the lens.
b) Convex: positive; concave: negative.

12. A lens forms an erect image for all positions of the object in front of it. Is the lens convex or concave?

Answer:
1. A convex lens forms an erect image only when the object is within the focal length.
2. A concave lens forms an erect, virtual image for every position of the object.
3. Hence, the lens is a concave lens.

Teacher's Note:
a) Only a concave lens gives an erect image everywhere.
b) A convex lens gives an inverted real image beyond F.

13. Where should an object be placed so that a real and inverted image of same size is obtained using a convex lens?

Answer:
1. A real, inverted image of the same size is formed when the object is placed at 2F (the centre of curvature).
2. The image is then also formed at 2F on the other side of the lens.

Teacher's Note:
a) Use u = v = 2f for the same-size image.
b) The magnification is -1.

14. Write the relation between u,v and f of a thin lens.

Answer:
1. The relation is the lens formula: \( \frac{1}{v} - \frac{1}{u} = \frac{1}{f} \).
2. Here u is the object distance, v is the image distance and f is the focal length, all measured from the optical centre with the sign convention.

Teacher's Note:
a) Do not mix up with the mirror formula, which has a plus sign.
b) Put the signs of u, v and f before substituting.

15. What is the sign of u, v and f for a convex lens according to Cartesian sign convention?

Answer:
1. The object is always on the left, so u is negative.
2. For a real image on the right of the lens, v is positive (v is negative for a virtual image on the left).
3. The focal length of a convex lens is positive.

Teacher's Note:
a) Distances against the direction of the incident light are negative.
b) The focal length of a convex lens is positive.

16. An object of height 1m is placed at a distance of 2f from a convex lens. What is the height of the image formed?

Answer:
1. For u = -2f, the image distance is v = +2f.
2. Magnification m = \( \frac{v}{u} = \frac{2f}{-2f} \) = -1.
3. Height of the image = |m| × 1 m = 1 m.
4. Hence, the image is 1 m high, real and inverted.

Teacher's Note:
a) At 2f the image is the same size as the object.
b) The negative sign of m only shows that the image is inverted.

17. Define power.

Answer:
1. The power of a lens is the reciprocal of its focal length in metres.
2. P = \( \frac{1}{f} \) (f in metres).
3. Its SI unit is dioptre (D).

Teacher's Note:
a) Convex lens: positive power; concave lens: negative power.
b) Convert the focal length into metres first.

18. What is least distance of distinct vision?

Answer:
1. It is the minimum distance from the eye at which an object can be seen clearly without strain.
2. For a normal eye it is about 25 cm.

Teacher's Note:
a) It is also called the near point of the eye.
b) The value is 25 cm for a normal eye.

19. What happens when a ray of light passes through the optical centre of a lens?

Answer:
1. A ray passing through the optical centre of a thin lens goes straight.
2. It emerges without any deviation, along the same direction.

Teacher's Note:
a) The lens acts like a thin glass plate at its centre.
b) This ray is one of the three rays used for image construction.

20. State the laws of refraction.

Answer:
1. First law: the incident ray, the refracted ray and the normal at the point of incidence all lie in the same plane.
2. Second law (Snell's law): for light of a given colour and for a given pair of media, the ratio of the sine of the angle of incidence to the sine of the angle of refraction is constant.
3. That is, \( \frac{\sin i}{\sin r} \) = constant.

Teacher's Note:
a) The constant is the refractive index of the second medium with respect to the first.
b) Write both laws, not only Snell's law.

21. Diagrammatically represent the refraction of light through a rectangular glass slab.

Answer:
1. In the figure, a ray enters the glass slab at the top face. It bends towards the normal, because glass is denser than air.
2. It leaves from the opposite face and bends away from the normal.
3. The emergent ray is parallel to the incident ray, but it is shifted sideways by the lateral shift d.

Teacher's Note:
a) Label i, r, e, the normals and the lateral shift.
b) The emergent ray is parallel to the incident ray, since the faces are parallel.

22. Define convex lens. Why is it referred to as converging lens?

Answer:
1. A convex lens is a lens that is thicker at the centre and thinner at the edges.
2. A beam of light parallel to the principal axis, after refraction through a convex lens, converges to a point on the axis (the principal focus).
3. Hence, it is called a converging lens.

Teacher's Note:
a) Convex means thick in the middle.
b) Converging: it brings parallel rays together.

23. Draw a diagram to show the second principal focus of a convex lens.

Answer:
1. In the figure, rays parallel to the principal axis fall on a convex lens from the left.
2. After refraction, they all meet at one point F2 on the axis, on the other side of the lens.
3. This point F2 is called the second principal focus of the lens.

Teacher's Note:
a) Label F1, F2, the optical centre and the principal axis.
b) The distance OF2 is the focal length.

24. Distinguish between a convex and a concave lens.

Answer:

Teacher's Note:
a) Compare shape, action on light, focus, sign of focal length and images.
b) Write both columns for every point.

25. Draw a ray diagram to show the refraction of light when it passes through the optic centre of a convex lens.

Answer:
1. A ray of light directed towards the optical centre O of a convex lens passes through the lens without any deviation.
2. The emergent ray is in the same direction as the incident ray, as shown in the figure.

Teacher's Note:
a) Mark the arrows on the ray and label O.
b) The lens is thin at O, so the ray is only displaced a very small amount.

26. List out the uses of convex lenses.

Answer:
1. Convex lenses are used as a magnifying glass (simple microscope) and in compound microscopes and telescopes.
2. They are used in cameras and in projectors to form real images.
3. They are used in spectacles to correct hypermetropia (far-sightedness).
4. They are used in the human eye as the eye lens.

Teacher's Note:
a) Mention at least four uses.
b) Link each use to the type of image formed.

27. With the help of a ray diagram show how an object gets magnified in a simple microscope.

Answer:
1. A simple microscope is a convex lens of short focal length.
2. The object is placed between the optical centre O and the focus F1, that is, at a distance less than the focal length.
3. A ray parallel to the axis passes through F2 after refraction, and a ray through O goes straight.
4. These two rays, when produced backwards, meet on the same side as the object, giving a virtual, erect and enlarged image, as shown in the figure.

Teacher's Note:
a) The object must be within one focal length.
b) The image formed is virtual, erect and magnified.

28. Define the power of a lens. What is its unit?

Answer:
1. The power of a lens is the reciprocal of its focal length (in metres): P = \( \frac{1}{f} \).
2. It measures the ability of a lens to converge or diverge a beam of light.
3. The SI unit of power is the dioptre (D). 1 D = 1 m⁻¹.

Teacher's Note:
a) A lens of focal length 1 m has a power of 1 D.
b) Do not use cm in the formula.

29. With the help of a diagram explain how light gets refracted when it passes through a rectangular glass slab.

Answer:
1. In the figure, a ray of light is incident on one face of the glass slab at an angle of incidence i.
2. On entering the glass (a denser medium), the ray bends towards the normal and the angle of refraction r is smaller than i.
3. At the opposite face, the ray goes from glass to air and bends away from the normal.
4. The emergent ray is parallel to the incident ray, because the angle of emergence e equals the angle of incidence i.
5. The emergent ray is displaced sideways by the lateral shift d.

Teacher's Note:
a) The two faces are parallel, so the net deviation is zero.
b) The lateral shift depends on the thickness of the slab and on the angle of incidence.

30. An object is placed at a distance of 50 cm from a concave lens of focal length 20 cm. Find the nature and position of the image.

Answer:
1. Given: u = -50 cm, f = -20 cm.
2. Here the lens is concave, so f = -20 cm, and the object is on the left, so u = -50 cm.
3. Lens formula: \( \frac{1}{v} - \frac{1}{u} = \frac{1}{f} \)
4. \( \frac{1}{v} = \frac{1}{f} + \frac{1}{u} = \frac{1}{-20} + \frac{1}{(-50)} = -\frac{7}{100} \) per cm
5. v = -14.29 cm
6. Magnification m = \( \frac{v}{u} = \frac{-14.29}{-50} \) = 0.29
7. The negative value of v shows that the image is formed on the same side as the object, 14.29 cm from the lens.
8. Hence, the image is virtual, erect and diminished (m = 0.29), formed at 14.29 cm from the lens on the object side.

Teacher's Note:
a) Use f = -20 cm for a concave lens.
b) A concave lens always forms a virtual, erect and diminished image.

31. An object is placed 50 cm from a lens which produces a virtual image at a distance of 10 cm in front of the lens. Draw a diagram to show the formation of image and calculate the focal length of the lens.

Answer:
1. Given: u = -50 cm and v = -10 cm (the image is on the same side as the object).
2. Lens formula: \( \frac{1}{f} = \frac{1}{v} - \frac{1}{u} \)
3. \( \frac{1}{f} = \frac{1}{-10} - \frac{1}{-50} = \frac{-5+1}{50} = \frac{-4}{50} \)
4. f = -12.5 cm.
5. The focal length is negative, so the lens is a concave lens. The diagram shows the formation of a virtual, erect and diminished image.

Teacher's Note:
a) A virtual image on the object side of a real object with a diminished size means a concave lens.
b) Substitute with signs: u and v are both negative.

32. An object of height 4 cm is placed at a distance of 10 cm from a convex lens of focal length 20 cm. Find the position, nature and size of the image.

Answer:
1. Given: u = -10 cm, f = 20 cm. Height of object h = 4 cm.
2. Lens formula: \( \frac{1}{v} - \frac{1}{u} = \frac{1}{f} \)
3. \( \frac{1}{v} = \frac{1}{f} + \frac{1}{u} = \frac{1}{20} + \frac{1}{(-10)} = -\frac{1}{20} \) per cm
4. v = -20 cm
5. Magnification m = \( \frac{v}{u} = \frac{-20}{-10} \) = 2
6. Height of image h' = m × h = 2 × 4 = 8 cm.
7. The image is formed at 20 cm from the lens on the same side as the object.
8. Hence, the image is virtual, erect and enlarged, of size 8 cm, at 20 cm from the lens on the object side.

Teacher's Note:
a) The object is inside the focal length, so a virtual enlarged image results.
b) The answer must give position, nature and size.

33. What is the power of a lens having a focal length of a) 50 cm b) -50cm

Answer:
1. Power P = \( \frac{1}{f} \) with f in metres.
2. a) f = 50 cm = 0.5 m, so P = \( \frac{1}{0.5} \) = +2 D (convex lens).
3. b) f = -50 cm = -0.5 m, so P = \( \frac{1}{-0.5} \) = -2 D (concave lens).

Teacher's Note:
a) Convert centimetres into metres first.
b) The sign of the power is the sign of the focal length.

34. Draw a ray diagram to show the position and nature of the image formed by a convex lens when the object is placed a) at 2F₁ b) between F₁ and 2F₁ c) beyond 2F₁

Answer:
1. a) Object at 2F1: the image is formed at 2F2. It is real, inverted and of the same size as the object (first figure).
2. b) Object between F1 and 2F1: the image is formed beyond 2F2. It is real, inverted and enlarged (second figure).
3. c) Object beyond 2F1: the image is formed between F2 and 2F2. It is real, inverted and diminished (third figure).

Teacher's Note:
a) Use two rays: one parallel to the axis through F2 and one through the optical centre.
b) Remember the nature of the image for each of the three positions.
c) Label F1, F2, 2F1, 2F2 and O in each diagram.

35. State and verify Snell's law.

Answer:
1. Statement: for light of a given colour and a given pair of media, the ratio of the sine of the angle of incidence to the sine of the angle of refraction is constant: \( \frac{\sin i}{\sin r} = n \).
2. Verification: place a rectangular glass slab on a drawing board and fix a sheet of paper.
3. Draw a normal NN' to the face of the slab and fix two pins P and Q on the incident ray. Looking through the other face, fix two pins R and S so that all four pins appear in a line.
4. Draw the incident ray and the refracted ray, and measure the angles i and r.
5. Repeat for different angles of incidence and find the ratio \( \frac{\sin i}{\sin r} \) each time.
6. The ratio is the same for every angle (for example about 1.5 for glass), which verifies Snell's law.

Teacher's Note:
a) Write the law, the apparatus and the observation.
b) A graph of sin i against sin r is a straight line through the origin.
c) Pins must appear exactly in one line when seen through the glass.

36. The bending of a beam of light when it passes obliquely from one medium to another is known as _______.
1. reflection
2. refraction
3. dispersion
4. deviation

Answer: 2. refraction

1. Bending of light on passing from one medium to another is refraction.
2. Reflection is the bouncing back of light, and dispersion is the splitting into colours.
3. Hence, the correct option is 2. refraction.

Teacher's Note:
a) Link the term to the cause: a change in the speed of light.
b) Dispersion needs a prism.

37. The part of the lens through which the ray of light passes without suffering deviation is called ________.
1. optical centre
2. focus
3. centre of curvature
4. pole

Answer: 1. optical centre

1. A ray passing through the optical centre of a thin lens goes straight.
2. The pole is the term used for a mirror.
3. Hence, the correct option is 1. optical centre.

Teacher's Note:
a) Pole is for mirrors and optical centre is for lenses.
b) The ray is not deviated at O.

38. Convex lens always gives a real image if the object is situated beyond _______.
1. optical centre
2. centre of curvature
3. Focus
4. radius of curvature

Answer: 3. Focus

1. A convex lens forms a real image when the object is beyond its focus (u greater than f).
2. Inside the focus, the image is virtual.
3. Hence, the correct option is 3. Focus.

Teacher's Note:
a) The boundary between real and virtual images is the focus.
b) Do not choose the centre of curvature; the image is real at positions between F and 2F as well.

39. Parallel rays of light entering a convex lens always converge at _______.
1. centre of curvature
2. the principal focus
3. optical centre
4. the focal plane

Answer: 2. the principal focus

1. A convex lens converges a parallel beam to a point on the principal axis.
2. This point is the principal focus.
3. Hence, the correct option is 2. the principal focus.

Teacher's Note:
a) This is the definition of the principal focus.
b) If the beam is not parallel to the axis, it meets in the focal plane.

40. Where should an object be placed so that a real and inverted image of the same size is obtained, using a convex lens?
1. Between O and F
2. At F
3. At 2 F
4. At infinity

Answer: 3. At 2 F

1. For a real, inverted image of the same size, u = v = 2f.
2. So the object must be at 2F.
3. Hence, the correct option is 3. At 2 F.

Teacher's Note:
a) Magnification is -1 at 2F.
b) At F the image forms at infinity.

41. SI unit of the power of a lens is ___________.
1. dioptre
2. cm
3. metre
4. watt

Answer: 1. dioptre

1. Power of a lens = 1/f, with f in metres.
2. Its SI unit is dioptre (D).
3. Hence, the correct option is 1. dioptre.

Teacher's Note:
a) 1 D = 1 m⁻¹.
b) Do not confuse it with the watt, which is the unit of electrical power.

42. 1 D is the power of the lens of focal length of ______ cm.
1. 100
2. 10
3. 1/100
4. 1/10

Answer: 1. 100

1. P = \( \frac{1}{f} \) with f in metres.
2. For P = 1 D, f = 1 m = 100 cm.
3. Hence, the correct option is 1. 100.

Teacher's Note:
a) Convert the metre into centimetres.
b) 1 m = 100 cm.

43. In a simple microscope lens used is __________.
1. biconvex
2. biconcave
3. plano convex
4. cylindrical

Answer: 1. biconvex

1. A simple microscope is a magnifying glass.
2. It uses a convex lens of short focal length, which is a biconvex lens.
3. Hence, the correct option is 1. biconvex.

Teacher's Note:
a) A concave lens cannot magnify.
b) A convex lens of short focal length gives a larger magnification.

44. Reciprocal of focal length in metres is known as the ______ of a lens.
1. focus
2. power
3. power of accommodation
4. far point

Answer: 2. power

1. Power P = \( \frac{1}{f} \) (f in metres).
2. So the reciprocal of the focal length is the power.
3. Hence, the correct option is 2. power.

Teacher's Note:
a) Do not choose "power of accommodation", which is a property of the eye.
b) Unit: dioptre.

45. A convex lens is called _________.
1. converging lens
2. diverging lens
3. both converging and diverging lens
4. refracting lens

Answer: 1. converging lens

1. A convex lens converges a parallel beam of light to a point.
2. So it is called a converging lens.
3. Hence, the correct option is 1. converging lens.

Teacher's Note:
a) Convex is converging and concave is diverging.
b) Every lens refracts, so "refracting lens" is not a specific name.

46. A positive magnification greater than unity indicates _____________________.
1. real image
2. virtual image
3. neither real not virtual image
4. distorted image

Answer: 2. virtual image

1. Magnification is positive for an erect image.
2. A real image formed by a single lens is always inverted (negative m).
3. A positive magnification greater than 1 means an erect, enlarged, virtual image.
4. Hence, the correct option is 2. virtual image.

Teacher's Note:
a) Positive m means erect, and an erect image of a single lens is virtual.
b) Real images have a negative magnification.

47. The power of a convex lens of focal length 50 cm is ______.
1. + 2D
2. - 2D
3. 50 D
4. - 5D

Answer: 1. + 2D

1. f = 50 cm = 0.5 m.
2. P = \( \frac{1}{0.5} \) = +2 D (convex lens, positive).
3. Hence, the correct option is 1. + 2D.

Teacher's Note:
a) Convert 50 cm into 0.5 m.
b) A convex lens has positive power.

48. The focal length of a lens whose power is -1.5 D is _______.
1. -66.66 cm
2. + 1.5 m
3. + 66.66 cm
4. -1.5 m

Answer: 1. -66.66 cm

1. f = \( \frac{1}{P} = \frac{1}{-1.5} \) m = -0.667 m.
2. f = -66.66 cm (a concave lens).
3. Hence, the correct option is 1. -66.66 cm.

Teacher's Note:
a) Negative power means negative focal length.
b) Convert metres into centimetres at the end.

49. Real images formed by single convex lenses are always ________________.
1. on the same side of the lens as the object
2. Inverted
3. Erect
4. smaller than the object

Answer: 2. Inverted

1. A real image formed by a convex lens is on the opposite side of the lens from the object.
2. It is always inverted, but it can be smaller, equal or larger.
3. Hence, the correct option is 2. Inverted.

Teacher's Note:
a) Real means inverted for a single lens.
b) Size depends on the object position.

50. An object is placed 12 cm from a convex lens whose focal length is 10 cm. The image must be.
1. virtual and enlarged
2. virtual and reduced in size
3. real and reduced in size
4. real and enlarged

Answer: 4. real and enlarged

1. u = -12 cm, f = +10 cm.
2. \( \frac{1}{v} = \frac{1}{10} - \frac{1}{12} = \frac{1}{60} \), so v = +60 cm.
3. m = \( \frac{v}{u} = \frac{60}{-12} \) = -5, so the image is real, inverted and 5 times larger.
4. Hence, the correct option is 4. real and enlarged.

Teacher's Note:
a) The object is between F and 2F, so the image is beyond 2F.
b) A real image has v positive.

51. When a person uses a convex lens as a simple magnifying glass, the object must be placed at a distance.
1. less than one focal length
2. more than one focal length
3. less than twice the focal length
4. more than twice the focal length

Answer: 1. less than one focal length

1. A convex lens gives a virtual, erect and enlarged image only when the object lies between O and F.
2. So the object distance must be less than one focal length.
3. Hence, the correct option is 1. less than one focal length.

Teacher's Note:
a) Inside the focus gives a magnified virtual image.
b) Between F and 2F the image is real.

52. The image produced by a concave lens is ________.
1. always virtual and enlarged
2. always virtual and reduced in size
3. always real
4. sometimes real, sometimes virtual

Answer: 2. always virtual and reduced in size

1. A concave lens always forms a virtual, erect and diminished image.
2. Hence, the correct option is 2. always virtual and reduced in size.

Teacher's Note:
a) A concave lens never forms a real image.
b) The image lies between O and F on the object side.

53. A virtual image is formed by _______.
1. a slide projector in a cinema hall
2. the ordinary camera
3. a simple microscope
4. Telescope

Answer: 3. a simple microscope

1. A slide projector and a camera form real images on a screen or film.
2. A simple microscope forms a virtual, erect and enlarged image.
3. Hence, the correct option is 3. a simple microscope.

Teacher's Note:
a) A virtual image cannot be taken on a screen.
b) The object is placed within one focal length in a simple microscope.

54. An object is placed 25 cm from a convex lens whose focal length is 10 cm. The image distance is ________ cm.
1. 50 cm
2. 16.66 cm
3. 6.66 cm
4. 10 cm

Answer: 2. 16.66 cm

1. u = -25 cm, f = +10 cm.
2. \( \frac{1}{v} = \frac{1}{f} + \frac{1}{u} = \frac{1}{10} - \frac{1}{25} = \frac{3}{50} \)
3. v = \( \frac{50}{3} \) = 16.66 cm.
4. Hence, the correct option is 2. 16.66 cm.

Teacher's Note:
a) Use 1/v = 1/f + 1/u with u negative.
b) The image is real since v is positive.

55. The least distance of distinct vision is ______.
1. 25 cm
2. 25 m
3. 0.25 cm
4. 2.5 m

Answer: 1. 25 cm

1. The least distance of distinct vision is the nearest distance at which the eye sees clearly without strain.
2. For a normal eye it is 25 cm.
3. Hence, the correct option is 1. 25 cm.

Teacher's Note:
a) It is also the near point of the eye.
b) Do not mix up the units: 25 cm = 0.25 m.

56. A convex lens has a focal length of 20 cm. Its power in dioptres is ___________.
1. 2
2. 5
3. 0.5
4. 0.2

Answer: 2. 5

1. f = 20 cm = 0.2 m.
2. P = \( \frac{1}{0.2} \) = 5 D.
3. Hence, the correct option is 2. 5.

Teacher's Note:
a) Change cm into m before using P = 1/f.
b) A convex lens gives positive power.

57. An object is placed before a concave lens. The image formed _____________-.
1. is always erect
2. may be erect or inverted
3. is always inverted
4. is always real

Answer: 1. is always erect

1. A concave lens forms a virtual image for every position of the object.
2. This virtual image is always erect (and diminished).
3. Hence, the correct option is 1. is always erect.

Teacher's Note:
a) Virtual images from a single lens are always erect.
b) Real, inverted images are formed only by convex lenses.

58. A ray of light travels from a medium of refractive index n₁ to a medium of refractive index n₂. If angle of incidence is i and the angle of refraction is r, then \( \frac{\sin i}{\sin r} \) is equal to
1. n₁
2. n₂
3. n₂₁
4. n₁₂

Answer: 3. n₂₁

1. By Snell's law, \( \frac{\sin i}{\sin r} = \frac{n_2}{n_1} \).
2. The ratio n2/n1 is the refractive index of medium 2 with respect to medium 1, written n21.
3. Hence, the correct option is 3. n21.

Teacher's Note:
a) The first suffix is the second medium: n21 = n2 / n1.
b) Check the order of the suffixes carefully.

59. Two thin lenses of power +5 D and -2 D are placed in contact with each other. Focal length of the combination is
1. +3 m
2. -3 m
3. 0.33 m
4. -0.33 m

Answer: 3. 0.33 m

1. Power of the combination P = P1 + P2 = +5 - 2 = +3 D.
2. f = \( \frac{1}{P} = \frac{1}{3} \) m = 0.33 m.
3. Hence, the correct option is 3. 0.33 m.

Teacher's Note:
a) Powers add for lenses in contact.
b) The focal length is the reciprocal of the power, not the power itself.

60. The lens formula in cartesian frame is ______________.
1. \( \frac{1}{f} = \frac{1}{u} + \frac{1}{v} \)
2. \( \frac{1}{f} = \frac{1}{u} - \frac{1}{v} \)
3. \( \frac{1}{f} = \frac{1}{v} - \frac{1}{u} \)
4. \( \frac{1}{f} = \frac{1}{v} + \frac{1}{u} \)

Answer: 3. \( \frac{1}{f} = \frac{1}{v} - \frac{1}{u} \)

1. In the Cartesian sign convention, the lens formula is \( \frac{1}{f} = \frac{1}{v} - \frac{1}{u} \).
2. The formula with a plus sign (option 4) belongs to the mirror.
3. Hence, the correct option is 3.

Teacher's Note:
a) Lens: 1/v - 1/u = 1/f. Mirror: 1/v + 1/u = 1/f.
b) Check the sign of u before substitution.

61. An object is placed between two plane mirrors inclined at an angle of 45° to each other. How many images do you expect to see?

Answer:
1. Number of images n = \( \frac{360^{\circ}}{\theta} - 1 \), where θ is the angle between the mirrors.
2. Here θ = 45°, so n = \( \frac{360}{45} - 1 \) = 8 - 1 = 7.
3. Hence, 7 images are seen.

Teacher's Note:
a) The formula works when 360° divided by the angle is an even number.
b) Mirrors at 90° give 3 images.

62. How can an inverted image of an object be obtained with a plane mirror?

Answer:
1. Place the plane mirror horizontally (flat) and keep the object (for example an arrow standing upright) above it.
2. The image is formed behind the mirror, and it hangs downward, as far below the mirror as the object is above it.
3. So the image is inverted (upside down) with respect to the object.

Teacher's Note:
a) A vertical plane mirror gives only a laterally inverted, erect image.
b) To see an upside-down image, the mirror must be placed horizontally.

63. How many images of himself can an observer see in a room whose ceiling and two adjacent walls are mirrors?

Answer:
1. The ceiling and the two adjacent walls are three plane mirrors that are mutually perpendicular, like the corner of a cube.
2. Such an arrangement gives 7 images (three images from single reflection, three from double reflection and one from triple reflection).
3. Hence, the observer sees 7 images of himself.

Teacher's Note:
a) Three mutually perpendicular mirrors give 2³ - 1 = 7 images.
b) Count single, double and triple reflections: 3 + 3 + 1.

64. A convex mirror is held in water. What should be the change in its focal length?

Answer:
1. The focal length of a spherical mirror is f = \( \frac{R}{2} \), which depends only on the radius of curvature.
2. Reflection does not depend on the refractive index of the medium.
3. Hence, there is no change in the focal length of the mirror when it is held in water.

Teacher's Note:
a) A lens changes its focal length in water, but a mirror does not.
b) The formula f = R/2 has no refractive index.

65. If the light from the Sun subtend an angle Q at the pole of a concave mirror of focal length f, where will it converge? Also find the diameter of the image. Draw the necessary ray diagram.

Answer:
1. The Sun is very far away, so its rays are almost parallel and fall on the mirror at an angle Q with the axis.
2. The rays converge at the principal focus F, at a distance f from the pole of the mirror, as shown in the figure.
3. The image of the Sun is a small real, inverted image at F. Its diameter d subtends the same angle Q at the pole, so d = f × Q (Q in radians).
4. Hence, the light converges at the focus and the diameter of the image is f Q.

Teacher's Note:
a) Q must be in radians in d = f Q.
b) The image of a distant object forms in the focal plane.
c) Mark P, F and C in the ray diagram.

66. When does Snell’s law of refraction fail?

Answer:
1. Snell's law, \( \frac{\sin i}{\sin r} \) = constant, gives no angle of refraction when the ray is incident normally (i = 0), because then no bending occurs and the ratio becomes 0 divided by 0.
2. It also fails to describe light going from a denser to a rarer medium at an angle of incidence greater than the critical angle, when there is no refracted ray and total internal reflection takes place.

Teacher's Note:
a) The law applies to oblique incidence.
b) No refracted ray exists beyond the critical angle.

67. How does the frequency of a beam of ultraviolet light changes when it goes from air to glass?

Answer:
1. The frequency of light is fixed by its source.
2. It does not change when light goes from one medium to another.
3. Only the speed and the wavelength change.
4. Hence, the frequency of ultraviolet light remains the same.

Teacher's Note:
a) Speed and wavelength change on refraction, frequency does not.
b) Frequency belongs to the source, not to the medium.

68. State two conditions under which a ray of light suffering refraction from medium 1 to medium 2 does not undergo any change in direction

Answer:
1. Condition 1: the ray is incident normally on the surface (angle of incidence = 0°).
2. Condition 2: the two media have the same refractive index (n1 = n2), so there is no change in speed at the boundary.

Teacher's Note:
a) No change in speed means no bending.
b) Normal incidence gives i = r = 0.

69. Name the factor on which lateral shift produced by glass slab depends?

Answer:
1. The lateral shift depends on the thickness of the glass slab.
2. It also depends on the angle of incidence and on the refractive index of the glass.
3. A thicker slab gives a larger lateral shift.

Teacher's Note:
a) The lateral shift increases with the thickness of the slab.
b) It is zero at normal incidence.

70. A glass lens is immersed in water. How is the power of the lens affected?

Answer:
1. The refractive index of glass relative to water is less than relative to air.
2. So the lens bends light less and its focal length increases.
3. Hence, the power of the glass lens decreases when it is immersed in water.

Teacher's Note:
a) Power is the reciprocal of the focal length.
b) The relative refractive index is n(glass) / n(water).

71. Why is the power of a lens is measured as reciprocal of focal length? Give reason.

Answer:
1. A lens of short focal length bends the rays through a large angle, while a lens of long focal length bends them only slightly.
2. So the ability of a lens to bend light is larger when its focal length is smaller.
3. Hence, power is defined as the reciprocal of the focal length: P = \( \frac{1}{f} \).

Teacher's Note:
a) Power measures the converging or diverging ability.
b) Short focal length means strong bending.

72. A convex lens is made of a material of refractive index m₁ when placed in a medium of refractive index m₂, behaves as a diverging lens. How are m₁ and m₂ related?

Answer:
1. A convex shaped lens converges light only when its material is optically denser than the surrounding medium.
2. If the lens behaves as a diverging lens, light bends away from the normal on entering the lens, so the lens is optically rarer than the medium.
3. Hence, m1 < m2 (the refractive index of the lens material is less than that of the medium).

Teacher's Note:
a) Compare the lens material with the medium, not only the shape.
b) If m1 = m2, the lens would have no effect.

73. An object is held at a distance of 60 cm from a convex mirror of focal length 20 cm. At what distance from the mirror, should a plane mirror be held so that images in the two mirrors coincide?(ans. 22.5 cm)

Answer:
1. For the convex mirror: u = -60 cm and f = +20 cm.
2. \( \frac{1}{v} = \frac{1}{f} - \frac{1}{u} = \frac{1}{20} + \frac{1}{60} = \frac{4}{60} \)
3. v = +15 cm, so the image is 15 cm behind the convex mirror.
4. Let the plane mirror be at a distance x from the convex mirror, in front of it. The object is (60 - x) in front of the plane mirror, so its image is (60 - x) behind the plane mirror, that is, at a distance x - (60 - x) = 2x - 60 in front of the convex mirror.
5. For the images to coincide, this image must be 15 cm behind the convex mirror: 2x - 60 = -15.
6. x = 22.5 cm.
7. Hence, the plane mirror should be held 22.5 cm from the convex mirror, which matches the printed answer.

Teacher's Note:
a) A plane mirror forms an image as far behind it as the object is in front.
b) Put both images at the same position and solve for x.
c) Use signs carefully: images behind a mirror have negative distance in front.

74. How do you find the rough focal length of a convex lens? Is the same method applicable to a concave lens?

Answer:
1. Hold the convex lens facing a distant object such as the Sun or a window, and place a screen behind it.
2. Move the lens until a sharp, bright image of the distant object forms on the screen.
3. The distance between the lens and the screen is the rough focal length, since rays from a distant object are almost parallel.
4. This method is not applicable to a concave lens, because a concave lens forms only a virtual image, which cannot be obtained on a screen.

Teacher's Note:
a) Never look at the Sun directly through the lens.
b) A concave lens is tested with a convex lens in contact.

75. Which factors determines the focal length of a lens?

Answer:
1. The refractive index of the material of the lens.
2. The radii of curvature of its two surfaces.
3. The refractive index of the medium in which the lens is placed.

Teacher's Note:
a) The lens maker's formula contains all these terms.
b) A thicker, more curved lens has a shorter focal length.

76. The formula for linear magnification of a spherical mirror is m=h’/h = -v/u. What determines the sign of m? What is the significance of this sign?

Answer:
1. The sign of m is decided by the sign of the image height h' (with the object height h taken as positive), which is equal to the sign of -v/u.
2. A positive m means h' is positive: the image is erect and virtual (formed behind the mirror).
3. A negative m means h' is negative: the image is inverted and real (formed in front of the mirror).
4. Hence, the sign of m tells us the nature of the image.

Teacher's Note:
a) Positive m: erect and virtual. Negative m: inverted and real.
b) The size of the image is given by |m|.

77. A convex lens made of material of refractive index n₂ is kept in a medium of refractive index n₁. A parallel beam of light is incident on the lens. Compare the path of rays of light emerging from the convex lens if (i)n₁<n₂ (ii)n₁=n₂ (iii) n₁>n₂

Answer:
1. (i) If n1 < n2, the lens is optically denser than the medium. The rays converge, and the lens behaves as a converging lens.
2. (ii) If n1 = n2, there is no change in speed at the surfaces. The rays go straight without bending, and the lens has no effect.
3. (iii) If n1 > n2, the lens is optically rarer than the medium. The rays diverge, and the convex lens behaves as a diverging lens.

Teacher's Note:
a) The behaviour depends on the relative refractive index of lens and medium.
b) Case (ii): the lens becomes invisible in the medium.

78. A concave lens of focal length 25 cm and a convex lens of focal length 20 cm are placed in contact with each other. What is the power of this combination? Also, calculate focal length of this combination. (ans. 1m)

Answer:
1. Concave lens: f1 = -25 cm = -0.25 m, so P1 = \( \frac{1}{-0.25} \) = -4 D.
2. Convex lens: f2 = +20 cm = +0.2 m, so P2 = \( \frac{1}{0.2} \) = +5 D.
3. Power of the combination P = P1 + P2 = -4 + 5 = +1 D.
4. Focal length of the combination f = \( \frac{1}{P} = \frac{1}{1} \) m = 1 m.
5. Hence, the power is +1 D and the focal length is 1 m (a converging combination).

Teacher's Note:
a) Powers of thin lenses in contact simply add.
b) Convert focal lengths to metres before finding the power.

79. A convergent lens of power 8D is combined with a divergent lens of power -10 D. Calculate focal length of the combination. (ans.-0.5 m)

Answer:
1. Power of the combination P = 8 + (-10) = -2 D.
2. Focal length f = \( \frac{1}{P} = \frac{1}{-2} \) m = -0.5 m.
3. The negative sign shows that the combination is diverging.

Teacher's Note:
a) P = P1 + P2.
b) Matches the printed answer of -0.5 m.

80. A concave lens has a focal length of 15 cm. At what distance should an object from the lens be placed so that it forms an image at 10 cm from the lens? Also find the magnification of the lens.

Answer:
1. A concave lens forms only a virtual image on the same side as the object, so v = -10 cm and f = -15 cm.
2. Lens formula: \( \frac{1}{u} = \frac{1}{v} - \frac{1}{f} \)
3. \( \frac{1}{u} = \frac{1}{-10} - \frac{1}{-15} = \frac{-3+2}{30} = \frac{-1}{30} \)
4. u = -30 cm, so the object must be placed 30 cm from the lens.
5. Magnification m = \( \frac{v}{u} = \frac{-10}{-30} \) = +0.33 (about \( \frac{1}{3} \)).
6. Hence, the object distance is 30 cm and the magnification is +0.33, which means the image is virtual, erect and diminished.

Teacher's Note:
a) In a concave lens the image is on the same side, so v is negative.
b) Use the lens formula in the form 1/u = 1/v - 1/f.

81. Under what condition in an arrangement of two plane mirrors, incident ray and reflected ray will always be parallel to each other, whatever be the angle of incidence? Show the same with the help of diagram.

Answer:
1. The incident ray and the reflected ray are always parallel if the two plane mirrors are placed at 90° to each other (perpendicular mirrors).
2. In the figure, the ray is incident on mirror M2 at some angle, reflects to mirror M1, and then reflects again.
3. For perpendicular mirrors, the angles of incidence at the two mirrors add up to 90°, so the total deviation is 180°.
4. Hence, the final reflected ray is parallel to the incident ray (travelling in the opposite direction) for any angle of incidence.

Teacher's Note:
a) Perpendicular mirrors act as a corner reflector.
b) The two normals are also perpendicular to each other.

82. How much time will light take to cross 2 mm thick glass pane if refractive index of glass is 3/2? (ans. 10^-11s)

Answer:
1. Speed of light in glass v = \( \frac{c}{n} = \frac{3 \times 10^{8}}{3/2} \) = 2 × 10⁸ m/s.
2. Thickness d = 2 mm = 2 × 10⁻³ m.
3. Time t = \( \frac{d}{v} = \frac{2 \times 10^{-3}}{2 \times 10^{8}} \) = 10⁻¹¹ s.
4. Hence, light takes 10⁻¹¹ s, which matches the printed answer.

Teacher's Note:
a) Convert mm to metres.
b) Use v = c / n for the speed of light in the glass.

83. Light travel from rarer medium 1 to a denser medium 2. The angle of incidence and refraction are respectively 45° and 30°. Calculate the refractive index of second medium with respect to the first medium.

Answer:
1. Refractive index of medium 2 with respect to medium 1: n21 = \( \frac{\sin i}{\sin r} = \frac{\sin 45^{\circ}}{\sin 30^{\circ}} \)
2. sin 45° = \( \frac{1}{\sqrt{2}} \) and sin 30° = \( \frac{1}{2} \).
3. n21 = \( \frac{1}{\sqrt{2}} \times 2 = \sqrt{2} \) = 1.414.
4. Hence, the refractive index of the second medium with respect to the first medium is 1.414.

Teacher's Note:
a) Use sin 45° = 0.707 and sin 30° = 0.5.
b) The answer is greater than 1, as the second medium is denser.

84. A pond of depth 20 cm is filled with water of refractive index 4/3. Calculate apparent depth of the tank when viewed normally.

Answer:
1. Real depth = 20 cm and refractive index n = \( \frac{4}{3} \).
2. Apparent depth = \( \frac{\text{real depth}}{n} = \frac{20}{\frac{4}{3}} \) = 20 × \( \frac{3}{4} \) = 15 cm.
3. Hence, the apparent depth of the pond is 15 cm.

Teacher's Note:
a) The pond always appears shallower than it really is.
b) Apparent depth = real depth divided by n.

85. What is the real depth of a swimming pool when its bottom appears to be realized by 1m?given refractive index of water is 4/3

Answer:
1. The bottom appears to be raised by 1 m, so real depth - apparent depth = 1 m.
2. Apparent depth = \( \frac{\text{real depth}}{n} \) = \( \frac{3}{4} \) × real depth.
3. Let the real depth be d: d - \( \frac{3}{4} \)d = 1, so \( \frac{d}{4} \) = 1 and d = 4 m.
4. Hence, the real depth of the swimming pool is 4 m.

Teacher's Note:
a) The printed word "realized" is read as "raised" here.
b) The raised distance is real depth minus apparent depth.

86. An object is placed in front of a concave mirror of radius of curvature 15cm at a distance of (a) 10cm. and (b) 5cm. Find the position, nature and magnification of the image in each case.

Answer:
1. Radius of curvature R = -15 cm for a concave mirror, so f = \( \frac{R}{2} \) = -7.5 cm.
2. (a) u = -10 cm: \( \frac{1}{v} = \frac{1}{f} - \frac{1}{u} = \frac{1}{-7.5} + \frac{1}{10} = \frac{-4+3}{30} = \frac{-1}{30} \)
3. v = -30 cm, m = \( -\frac{v}{u} = -\frac{-30}{-10} \) = -3.
4. So in case (a) the image is at 30 cm in front of the mirror, real, inverted and enlarged (3 times).
5. (b) u = -5 cm: \( \frac{1}{v} = \frac{1}{-7.5} + \frac{1}{5} = \frac{-2+3}{15} = \frac{1}{15} \)
6. v = +15 cm, m = \( -\frac{v}{u} = -\frac{15}{-5} \) = +3.
7. So in case (b) the image is at 15 cm behind the mirror, virtual, erect and enlarged (3 times).

Teacher's Note:
a) Inside the focus (u = 5 cm) a concave mirror gives a virtual, erect, enlarged image.
b) Beyond the focus it gives a real, inverted image.
c) Use f = R/2 with the sign of R negative for a concave mirror.

87. An object is placed 15cm from a concave mirror of radius of curvature 60 cm. Find the position of image and its magnification?

Answer:
1. Radius of curvature R = -60 cm, so f = \( \frac{R}{2} \) = -30 cm. Object distance u = -15 cm.
2. Mirror formula: \( \frac{1}{v} + \frac{1}{u} = \frac{1}{f} \)
3. \( \frac{1}{v} = \frac{1}{-30} + \frac{1}{15} = \frac{-1+2}{30} = \frac{1}{30} \)
4. v = +30 cm.
5. Magnification m = \( -\frac{v}{u} = -\frac{30}{-15} \) = +2.
6. Hence, the image is formed 30 cm behind the mirror; it is virtual, erect and twice the size of the object.

Teacher's Note:
a) The object is between P and F, so the image is virtual and enlarged.
b) Positive m means an erect image.

88. An object is kept at a distance of 5cm in front of a convex mirror of focal length 10cm. Give the position, magnification and the nature of the image formed.

Answer:
1. For a convex mirror f = +10 cm and u = -5 cm.
2. Mirror formula: \( \frac{1}{v} = \frac{1}{f} - \frac{1}{u} = \frac{1}{10} + \frac{1}{5} = \frac{3}{10} \)
3. v = \( \frac{10}{3} \) = +3.33 cm.
4. Magnification m = \( -\frac{v}{u} = -\frac{3.33}{-5} \) = +0.67.
5. Hence, the image is formed 3.33 cm behind the mirror; it is virtual, erect and diminished (m = 0.67).

Teacher's Note:
a) A convex mirror always forms a virtual, erect and diminished image.
b) Its focal length is positive.

89. An object is placed at a distance of 50cmfrom a concave lens of focal length 20cm. Find the nature and position of the image.

Answer:
1. For a concave lens f = -20 cm and u = -50 cm.
2. Lens formula: \( \frac{1}{v} - \frac{1}{u} = \frac{1}{f} \)
3. \( \frac{1}{v} = \frac{1}{f} + \frac{1}{u} = \frac{1}{-20} + \frac{1}{(-50)} = -\frac{7}{100} \) per cm
4. v = -14.29 cm
5. Magnification m = \( \frac{v}{u} = \frac{-14.29}{-50} \) = 0.29
6. Hence, the image is virtual, erect and diminished (m = 0.29), formed 14.29 cm from the lens on the same side as the object.

Teacher's Note:
a) Concave lens: f is negative.
b) The image lies between the lens and its focus.

90. The power of a lens is 2.5 dioptre. What is the focal length and the type of lens?

Answer:
1. Focal length f = \( \frac{1}{P} = \frac{1}{2.5} \) m = 0.4 m = 40 cm.
2. The power is positive, so the focal length is positive, and the lens is a convex lens.
3. Hence, f = +40 cm and the lens is convex.

Teacher's Note:
a) Positive power means convex and negative power means concave.
b) Give the focal length in cm or m with the unit.

91. What is the power of a concave lens of focal length 50cm?

Answer:
1. For a concave lens f = -50 cm = -0.5 m.
2. Power P = \( \frac{1}{f} = \frac{1}{-0.5} \) = -2 D.

Teacher's Note:
a) A concave lens has negative power.
b) Convert cm into metres before using P = 1/f.

92. Find the power of a concave lens of focal length 2m.

Answer:
1. For a concave lens f = -2 m.
2. Power P = \( \frac{1}{f} = \frac{1}{-2} \) = -0.5 D.

Teacher's Note:
a) Concave lens: take f negative.
b) No conversion is needed here, because f is already in metres.

93. Two lens of power +3.5D and -2.5D are placed in contact. find the power and focal length of the lens combination.

Answer:
1. Power of the combination P = P1 + P2 = +3.5 + (-2.5) = +1 D.
2. Focal length f = \( \frac{1}{P} = \frac{1}{1} \) m = 1 m = 100 cm.
3. Hence, the power is +1 D and the focal length is 1 m (the combination is a converging lens).

Teacher's Note:
a) Powers add for lenses in contact.
b) The combined power is positive, so the combination is convex.

94. A convex lens has a focal length of 20 cm. Calculate at what distance from the lens should an object be placed so that it forms an image at a distance of 40cm on the other side of the lens. State the nature of the image formed?

Answer:
1. f = +20 cm and the image is on the other side of the lens, so v = +40 cm.
2. Lens formula: \( \frac{1}{u} = \frac{1}{v} - \frac{1}{f} \)
3. \( \frac{1}{u} = \frac{1}{40} - \frac{1}{20} = \frac{1-2}{40} = \frac{-1}{40} \)
4. u = -40 cm, so the object must be placed 40 cm from the lens (at 2F).
5. Magnification m = \( \frac{v}{u} = \frac{40}{-40} \) = -1.
6. Hence, the image is real, inverted and of the same size as the object.

Teacher's Note:
a) The object is at 2f, so the image is at 2f on the other side.
b) A real image is formed on the other side of the lens.

95. A 10cm tall object is placed perpendicular to the principal axis of a convex lens of focal length 30cm. The distance of the object from the line is 20cm.find the i)position ii)nature and iii) size of the image formed.

Answer:
1. Given: u = -20 cm, f = 30 cm. Height of object h = 10 cm.
2. Lens formula: \( \frac{1}{v} - \frac{1}{u} = \frac{1}{f} \)
3. \( \frac{1}{v} = \frac{1}{f} + \frac{1}{u} = \frac{1}{30} + \frac{1}{(-20)} = -\frac{1}{60} \) per cm
4. v = -60 cm
5. Magnification m = \( \frac{v}{u} = \frac{-60}{-20} \) = 3
6. Height of image h' = m × h = 3 × 10 = 30 cm.
7. The negative value of v shows that the image is on the same side as the object.
8. Hence, the image is at 60 cm from the lens on the object side; it is virtual and erect, and its size is 30 cm.

Teacher's Note:
a) The object is within the focal length, so the image is virtual and enlarged.
b) The printed words "from the line" mean "from the lens".

96. Find the focal length of a line power is given as +2.0D.

Answer:
1. The question means a lens of power +2.0 D.
2. Focal length f = \( \frac{1}{P} = \frac{1}{2.0} \) m = 0.5 m = 50 cm.
3. The power is positive, so it is a convex lens.

Teacher's Note:
a) The printed "line" is a misprint of "lens".
b) Positive power means a converging (convex) lens.

97. With respect to air the refractive index of ice and rock salt benzene are 1.31 and 1.54 respectively. Calculate the refractive index of rock salt with respect to ice.

Answer:
1. Refractive index of ice with respect to air: n(ice) = 1.31 and of rock salt with respect to air: n(rock salt) = 1.54.
2. Refractive index of rock salt with respect to ice = \( \frac{n_{\text{rock salt}}}{n_{\text{ice}}} = \frac{1.54}{1.31} \) = 1.18.
3. Hence, the refractive index of rock salt with respect to ice is 1.18.

Teacher's Note:
a) The printed words "rock salt benzene" refer to rock salt only.
b) n21 = n2 / n1, with both values taken with respect to air.

98. An object 5cm in length is placed at a distance of 20 cm in front of a convex mirror of radius of curvature 30cm. Find the position of the image, its nature and size.

Answer:
1. Radius of curvature R = 30 cm, so f = \( \frac{R}{2} \) = +15 cm (convex mirror).
2. Mirror formula: \( \frac{1}{v} + \frac{1}{u} = \frac{1}{f} \)
3. \( \frac{1}{v} = \frac{1}{f} - \frac{1}{u} = \frac{1}{(15)} - \frac{1}{(-20)} = \frac{7}{60} \) per cm
4. v = 8.57 cm
5. Magnification m = \( -\frac{v}{u} = -\frac{8.57}{(-20)} \) = 0.43
6. Height of image h' = m × h = 0.43 × 5 = 2.14 cm.
7. Hence, the image is formed 8.57 cm behind the mirror; it is virtual, erect and diminished, and its size is 2.14 cm.

Teacher's Note:
a) A convex mirror always gives a virtual, erect and diminished image.
b) Its focal length is positive.

99. The far point of a myopic person is 150cm in front of the eye. Calculate the focal length and the power of the lens required to enable him to see distant objects clearly.

Answer:
1. The lens must form the image of a distant object at the far point of the eye, so v = -150 cm and u = ∞.
2. Lens formula: \( \frac{1}{f} = \frac{1}{v} - \frac{1}{u} = \frac{1}{-150} - 0 \)
3. f = -150 cm = -1.5 m.
4. Power P = \( \frac{1}{f} = \frac{1}{-1.5} \) = -0.67 D.
5. Hence, a concave lens of focal length 150 cm and power -0.67 D is required.

Teacher's Note:
a) Myopia is corrected by a concave lens.
b) The focal length of the lens is equal to the far point distance.

100. To obtain magnified erect image on a concave mirror, object should be held
a) At pole
b) At focus
c) Between p and f
d) Beyond 2f

Answer: c) Between p and f

1. A concave mirror forms a virtual, erect and enlarged image only when the object is between the pole P and the focus F.
2. Hence, the correct option is c) Between p and f.

Teacher's Note:
a) At the focus the image forms at infinity.
b) Beyond F the image is real and inverted.

101. If f is the focal length and R is the radius of curvature of a spherical mirror, then
a) R=f
b) R=2f
c) R=3f
d) R=4f

Answer: b) R=2f

1. The focal length of a spherical mirror is half its radius of curvature: f = \( \frac{R}{2} \).
2. So R = 2f.
3. Hence, the correct option is b) R=2f.

Teacher's Note:
a) This is true for both concave and convex mirrors.
b) Use R = 2f to find the centre of curvature.

102. The image formed by a concave mirror is real, inverted and of same size as that of the object. The position of the object is
a) At C
b) At F
c) Between C and F
d) Beyond C

Answer: a) At C

1. A real, inverted image of the same size is obtained when the object is at the centre of curvature C.
2. The image is also formed at C.
3. Hence, the correct option is a) At C.

Teacher's Note:
a) At C the magnification is -1.
b) Here u = v = 2f.

103. The image formed by a concave mirror is virtual, erect and magnified. The position of the object is
a) At F
b) At c
c) At infinity
d) Between P and F

Answer: d) Between P and F

1. A concave mirror forms a virtual, erect and enlarged image when the object is between the pole P and the focus F.
2. Hence, the correct option is d) Between P and F.

Teacher's Note:
a) A virtual image forms only inside the focus.
b) At F the image is at infinity.

104. The image formed by a spherical mirror is virtual, erect and smaller in size. Whatever be the position of the object. The mirror is
a) Convex
b) Concave
c) Either convex or concave
d) Cannot say

Answer: a) Convex

1. A convex mirror always forms a virtual, erect and diminished image, for every position of the object.
2. A concave mirror forms such an image never.
3. Hence, the correct option is a) Convex.

Teacher's Note:
a) A convex mirror is used as a rear-view mirror because of this property.
b) The image lies behind the mirror between P and F.

105. When light goes from one medium to another, the characteristics that remain unaffected is
a) Speed
b) Direction
c) Wave length
d) Frequency

Answer: d) Frequency

1. When light changes the medium, its speed, wavelength and (generally) direction change.
2. The frequency is fixed by the source and does not change.
3. Hence, the correct option is d) Frequency.

Teacher's Note:
a) Frequency belongs to the source.
b) Speed and wavelength change by the same factor, n.

106. Reflective index of glass w.r.t air is 3/2. What is the refractive index of air w.r.t glass?
a) 2/3
b) 1
c) Zero
d) (3/2)2

Answer: a) 2/3

1. Refractive index of glass with respect to air = \( \frac{3}{2} \).
2. The refractive index of air with respect to glass is the reciprocal: n(air w.r.t. glass) = \( \frac{1}{3/2} = \frac{2}{3} \).
3. Hence, the correct option is a) 2/3.

Teacher's Note:
a) Interchanging the two media gives the reciprocal.
b) The printed "Reflective" means refractive.

107. A convex lens of focal length 15 cm is used to form an image of the size of the object. Where from the lens should be the object is placed.
a) 15 cm
b) 30 cm
c) 60 cm
d) 10 cm

Answer: b) 30 cm

1. An image of the same size is formed when the object is at 2f.
2. 2f = 2 × 15 = 30 cm.
3. Hence, the correct option is b) 30 cm.

Teacher's Note:
a) Same size means u = 2f.
b) The image is real and inverted.

108. To form an image twice the size of the object, using a convex lens of focal length 20 cm, the object distance must be
a) <20 cm
b) >20 cm
c) <20 cm and between 20 cm and 40 cm
d) Cannot say

Answer: c) <20 cm and between 20 cm and 40 cm

1. For m = +2 (virtual image): \( 2 = \frac{f}{f+u} \) gives u = -10 cm, which is less than 20 cm.
2. For m = -2 (real image): \( -2 = \frac{f}{f+u} \) gives u = -30 cm, which lies between 20 cm and 40 cm.
3. So the object can be at less than 20 cm, or between 20 cm and 40 cm.
4. Hence, the correct option is c).

Teacher's Note:
a) Both a virtual and a real image can be twice the size.
b) Option c) is the printed way of saying "less than 20 cm, or between 20 cm and 40 cm".

109. Bending of a ray of light due to change in velocity with medium is called
a) Reflection
b) Refraction
c) Diffraction
d) Dispersion

Answer: b) Refraction

1. The bending of light as it goes from one medium to another because its speed changes is called refraction.
2. Hence, the correct option is b) Refraction.

Teacher's Note:
a) Reflection is the bouncing back of light.
b) Dispersion is splitting of white light into colours.

110. If correct value of refractive index of a medium(μ) in terms of velocity of light in vacuum (C) and velocity of light in medium (v) is
a) n = vC
b) n = 1 / ve
c) n= C / v
d) n = v / C.

Answer: c) n= C / v

1. Refractive index n = \( \frac{\text{speed of light in vacuum}}{\text{speed of light in the medium}} = \frac{C}{v} \).
2. Hence, the correct option is c) n= C / v.

Teacher's Note:
a) Refractive index is always greater than or equal to 1, so C is in the numerator.
b) Option d) is the inverse and is incorrect.

111. Total internal reflection takes place when light travels
a) From denser to rarer medium
b) From rarer to denser medium
c) In same denser medium from one side to other.
d) In same rarer medium from one side to other.

Answer: a) From denser to rarer medium

1. Total internal reflection needs light going from a denser to a rarer medium.
2. The angle of incidence must also be larger than the critical angle.
3. Hence, the correct option is a) From denser to rarer medium.

Teacher's Note:
a) Both conditions must be met: denser to rarer, and i greater than critical angle.
b) Light going from rarer to denser always refracts.

112. For no bending of a ray of light through a glass slab, angle of incidence must be
a) 0°
b) 30°
c) 60°
d) 90°

Answer: a) 0°

1. A ray that falls normally on the slab, with angle of incidence 0°, goes straight through without bending.
2. Hence, the correct option is a) 0°.

Teacher's Note:
a) Normal incidence means no deviation and no lateral shift.
b) At 90° the ray grazes the surface.

113. A convex lens is
a) Thin in the middle, thick at the room.
b) Thick in the middle, thin at the rim.
c) Thick through out.
d) Thin throughout.

Answer: b) Thick in the middle, thin at the rim.

1. A convex lens is thicker in the middle and thinner at the rim (edges).
2. Hence, the correct option is b) Thick in the middle, thin at the rim.

Teacher's Note:
a) The printed "room" is a misprint of "rim".
b) Concave lenses are thin in the middle.

114. Power of convex lens is
a) Zero.
b) Infinite.
c) Positive
d) Negative.

Answer: c) Positive

1. A convex lens has a positive focal length, so its power P = \( \frac{1}{f} \) is positive.
2. Hence, the correct option is c) Positive.

Teacher's Note:
a) Concave lens: negative power.
b) The unit is the dioptre.

115. An object is placed in front of a concave mirror of radius of curvature 15cm at a distance of (a) 10cm. and (b) 5cm. Find the position, nature and magnification of the image in each case.

Answer:
1. Radius of curvature R = -15 cm for a concave mirror, so f = \( \frac{R}{2} \) = -7.5 cm.
2. (a) u = -10 cm: \( \frac{1}{v} = \frac{1}{f} - \frac{1}{u} = \frac{1}{-7.5} + \frac{1}{10} = \frac{-4+3}{30} = \frac{-1}{30} \)
3. v = -30 cm, m = \( -\frac{v}{u} = -\frac{-30}{-10} \) = -3.
4. So in case (a) the image is at 30 cm in front of the mirror, real, inverted and enlarged (3 times).
5. (b) u = -5 cm: \( \frac{1}{v} = \frac{1}{-7.5} + \frac{1}{5} = \frac{-2+3}{15} = \frac{1}{15} \)
6. v = +15 cm, m = \( -\frac{v}{u} = -\frac{15}{-5} \) = +3.
7. So in case (b) the image is at 15 cm behind the mirror, virtual, erect and enlarged (3 times).

Teacher's Note:
a) Inside the focus (u = 5 cm) a concave mirror gives a virtual, erect, enlarged image.
b) Beyond the focus it gives a real, inverted image.
c) Use f = R/2 with the sign of R negative for a concave mirror.

116. An object is placed 15cm from a concave mirror of radius of curvature 60 cm. Find the position of image and its magnification?

Answer:
1. Radius of curvature R = -60 cm, so f = \( \frac{R}{2} \) = -30 cm. Object distance u = -15 cm.
2. Mirror formula: \( \frac{1}{v} + \frac{1}{u} = \frac{1}{f} \)
3. \( \frac{1}{v} = \frac{1}{-30} + \frac{1}{15} = \frac{-1+2}{30} = \frac{1}{30} \)
4. v = +30 cm.
5. Magnification m = \( -\frac{v}{u} = -\frac{30}{-15} \) = +2.
6. Hence, the image is formed 30 cm behind the mirror; it is virtual, erect and twice the size of the object.

Teacher's Note:
a) The object is between P and F, so the image is virtual and enlarged.
b) Positive m means an erect image.

117. An object is kept at a distance of 5cm in front of a convex mirror of focal length 10cm. Give the position, magnification and the nature of the image formed.

Answer:
1. For a convex mirror f = +10 cm and u = -5 cm.
2. Mirror formula: \( \frac{1}{v} = \frac{1}{f} - \frac{1}{u} = \frac{1}{10} + \frac{1}{5} = \frac{3}{10} \)
3. v = \( \frac{10}{3} \) = +3.33 cm.
4. Magnification m = \( -\frac{v}{u} = -\frac{3.33}{-5} \) = +0.67.
5. Hence, the image is formed 3.33 cm behind the mirror; it is virtual, erect and diminished (m = 0.67).

Teacher's Note:
a) A convex mirror always forms a virtual, erect and diminished image.
b) Its focal length is positive.

118. An object is placed at a distance of 50cmfrom a concave lens of focal length 20cm. Find the nature and position of the image.

Answer:
1. For a concave lens f = -20 cm and u = -50 cm.
2. Lens formula: \( \frac{1}{v} - \frac{1}{u} = \frac{1}{f} \)
3. \( \frac{1}{v} = \frac{1}{f} + \frac{1}{u} = \frac{1}{-20} + \frac{1}{(-50)} = -\frac{7}{100} \) per cm
4. v = -14.29 cm
5. Magnification m = \( \frac{v}{u} = \frac{-14.29}{-50} \) = 0.29
6. Hence, the image is virtual, erect and diminished (m = 0.29), formed 14.29 cm from the lens on the same side as the object.

Teacher's Note:
a) Concave lens: f is negative.
b) The image lies between the lens and its focus.

119. The power of a lens is 2.5 dioptre. What is the focal length and the type of lens?

Answer:
1. Focal length f = \( \frac{1}{P} = \frac{1}{2.5} \) m = 0.4 m = 40 cm.
2. The power is positive, so the focal length is positive, and the lens is a convex lens.
3. Hence, f = +40 cm and the lens is convex.

Teacher's Note:
a) Positive power means convex and negative power means concave.
b) Give the focal length in cm or m with the unit.

120. What is the power of a concave lens of focal length 50cm?

Answer:
1. For a concave lens f = -50 cm = -0.5 m.
2. Power P = \( \frac{1}{f} = \frac{1}{-0.5} \) = -2 D.

Teacher's Note:
a) A concave lens has negative power.
b) Convert cm into metres before using P = 1/f.

121. Find the power of a concave lens of focal length 2m.

Answer:
1. For a concave lens f = -2 m.
2. Power P = \( \frac{1}{f} = \frac{1}{-2} \) = -0.5 D.

Teacher's Note:
a) Concave lens: take f negative.
b) No conversion is needed here, because f is already in metres.

122. Two lens of power +3.5D and -2.5D are placed in contact. find the power and focal length of the lens combination.

Answer:
1. Power of the combination P = P1 + P2 = +3.5 + (-2.5) = +1 D.
2. Focal length f = \( \frac{1}{P} = \frac{1}{1} \) m = 1 m = 100 cm.
3. Hence, the power is +1 D and the focal length is 1 m (the combination is a converging lens).

Teacher's Note:
a) Powers add for lenses in contact.
b) The combined power is positive, so the combination is convex.

123. A convex lens has a focal length of 20 cm. Calculate at what distance from the lens should an object be placed so that it forms an image at a distance of 40cm on the other side of the lens. State the nature of the image formed?

Answer:
1. f = +20 cm and the image is on the other side of the lens, so v = +40 cm.
2. Lens formula: \( \frac{1}{u} = \frac{1}{v} - \frac{1}{f} \)
3. \( \frac{1}{u} = \frac{1}{40} - \frac{1}{20} = \frac{1-2}{40} = \frac{-1}{40} \)
4. u = -40 cm, so the object must be placed 40 cm from the lens (at 2F).
5. Magnification m = \( \frac{v}{u} = \frac{40}{-40} \) = -1.
6. Hence, the image is real, inverted and of the same size as the object.

Teacher's Note:
a) The object is at 2f, so the image is at 2f on the other side.
b) A real image is formed on the other side of the lens.

124. A 10cm tall object is placed perpendicular to the principal axis of a convex lens of focal length 30cm. The distance of the object from the line is 20cm.find the i)position ii)nature and iii) size of the image formed.

Answer:
1. Given: u = -20 cm, f = 30 cm. Height of object h = 10 cm.
2. Lens formula: \( \frac{1}{v} - \frac{1}{u} = \frac{1}{f} \)
3. \( \frac{1}{v} = \frac{1}{f} + \frac{1}{u} = \frac{1}{30} + \frac{1}{(-20)} = -\frac{1}{60} \) per cm
4. v = -60 cm
5. Magnification m = \( \frac{v}{u} = \frac{-60}{-20} \) = 3
6. Height of image h' = m × h = 3 × 10 = 30 cm.
7. The negative value of v shows that the image is on the same side as the object.
8. Hence, the image is at 60 cm from the lens on the object side; it is virtual and erect, and its size is 30 cm.

Teacher's Note:
a) The object is within the focal length, so the image is virtual and enlarged.
b) The printed words "from the line" mean "from the lens".

125. Find the focal length of a line power is given as +2.0D.

Answer:
1. The question means a lens of power +2.0 D.
2. Focal length f = \( \frac{1}{P} = \frac{1}{2.0} \) m = 0.5 m = 50 cm.
3. The power is positive, so it is a convex lens.

Teacher's Note:
a) The printed "line" is a misprint of "lens".
b) Positive power means a converging (convex) lens.

126. With respect to air the refractive index of ice and rock salt benzene are 1.31 and 1.54 respectively. Calculate the refractive index of rock salt with respect to ice.

Answer:
1. Refractive index of ice with respect to air: n(ice) = 1.31 and of rock salt with respect to air: n(rock salt) = 1.54.
2. Refractive index of rock salt with respect to ice = \( \frac{n_{\text{rock salt}}}{n_{\text{ice}}} = \frac{1.54}{1.31} \) = 1.18.
3. Hence, the refractive index of rock salt with respect to ice is 1.18.

Teacher's Note:
a) The printed words "rock salt benzene" refer to rock salt only.
b) n21 = n2 / n1, with both values taken with respect to air.

127. An object 5cm in length is placed at a distance of 20 cm in front of a convex mirror of radius of curvature 30cm. Find the position of the image, its nature and size.

Answer:
1. Radius of curvature R = 30 cm, so f = \( \frac{R}{2} \) = +15 cm (convex mirror).
2. Mirror formula: \( \frac{1}{v} + \frac{1}{u} = \frac{1}{f} \)
3. \( \frac{1}{v} = \frac{1}{f} - \frac{1}{u} = \frac{1}{(15)} - \frac{1}{(-20)} = \frac{7}{60} \) per cm
4. v = 8.57 cm
5. Magnification m = \( -\frac{v}{u} = -\frac{8.57}{(-20)} \) = 0.43
6. Height of image h' = m × h = 0.43 × 5 = 2.14 cm.
7. Hence, the image is formed 8.57 cm behind the mirror; it is virtual, erect and diminished, and its size is 2.14 cm.

Teacher's Note:
a) A convex mirror always gives a virtual, erect and diminished image.
b) Its focal length is positive.

128. The far point of a myopic person is 150cm in front of the eye. Calculate the focal length and the power of the lens required to enable him to see distant objects clearly.

Answer:
1. The lens must form the image of a distant object at the far point of the eye, so v = -150 cm and u = ∞.
2. Lens formula: \( \frac{1}{f} = \frac{1}{v} - \frac{1}{u} = \frac{1}{-150} - 0 \)
3. f = -150 cm = -1.5 m.
4. Power P = \( \frac{1}{f} = \frac{1}{-1.5} \) = -0.67 D.
5. Hence, a concave lens of focal length 150 cm and power -0.67 D is required.

Teacher's Note:
a) Myopia is corrected by a concave lens.
b) The focal length of the lens is equal to the far point distance.