Light: Reflection & Refraction
Overview: Reflection, Mirrors (Concave/Convex), Refraction, Lenses (Concave/Convex), Formulas.
1. Reflection of Light
Bouncing back of light.
Laws: 1. Angle of incidence = Angle of reflection (i = r). 2. Incident ray, reflected ray and normal lie in same plane.
Spherical Mirrors
Concave: Converging. Real (mostly) & Inverted images. Used in Torch, Dentist mirror.
Convex: Diverging. Virtual, Erect & Diminished images. Used as Rear-view mirror.
Mirror Formula: 1/v + 1/u = 1/f
Magnification (m) = h'/h = -v/u
2. Refraction of Light
Bending of light going from one medium to another.
Snell's Law: n = sin i / sin r
Refractive Index (n): n = c/v.
Spherical Lenses
Convex: Converging. Real (mostly). Magnifying glass.
Concave: Diverging. Virtual, Erect, Diminished.
Lens Formula: 1/v - 1/u = 1/f
Power (P) = 1/f (m). Unit: Dioptre (D).
NCERT In-Text Questions (Solved)
Set 1
Q1. Define the principal focus of a concave mirror.
The principal focus of a concave mirror is a point on its principal axis where all the light rays which are parallel to the principal axis converge after reflection from the mirror.
Q2. The radius of curvature of a spherical mirror is 20 cm. What is its focal length?
R = 20 cm
Focal length (f) = R/2 = 20/2 = 10 cm.
Focal length (f) = R/2 = 20/2 = 10 cm.
Q3. Name a mirror that can give an erect and enlarged image of an object.
Concave Mirror. (When the object is placed between the pole and the principal focus).
Q4. Why do we prefer a convex mirror as a rear-view mirror in vehicles?
1. It always produces an erect (though diminished) image.
2. It has a wider field of view as it is curved outwards, enabling the driver to view a much larger area.
2. It has a wider field of view as it is curved outwards, enabling the driver to view a much larger area.
Set 2
Q1. Find the focal length of a convex mirror whose radius of curvature is 32 cm.
R = +32 cm (Convex)
f = R/2 = 32/2 = +16 cm.
f = R/2 = 32/2 = +16 cm.
Q2. A concave mirror produces three times magnified (enlarged) real image of an object placed at 10 cm in front of it. Where is the image located?
Magnification m = -3 (Real image is inverted).
Object distance u = -10 cm.
m = -v/u
-3 = -v/(-10) => -3 = v/10
v = -30 cm.
The image is located at a distance of 30 cm from the mirror on the same side as the object (in front).
Object distance u = -10 cm.
m = -v/u
-3 = -v/(-10) => -3 = v/10
v = -30 cm.
The image is located at a distance of 30 cm from the mirror on the same side as the object (in front).
Set 3
Q1. A ray of light travelling in air enters obliquely into water. Does the light ray bend towards the normal or away from the normal? Why?
It bends towards the normal. Because water is optically denser than air, the speed of light decreases in water, causing it to bend towards the normal.
Q2. Light enters from air to glass having a refractive index of 1.50. What is the speed of light in the glass? (c = 3 x 10^8 m/s).
Refractive index (n) = c/v
1.50 = (3 x 108) / v
v = (3 x 108) / 1.5 = 2 x 108 m/s.
1.50 = (3 x 108) / v
v = (3 x 108) / 1.5 = 2 x 108 m/s.
Q3. Find out, from Table 10.3, the medium having highest optical density. Also find the medium with lowest optical density.
Highest Optical Density: Diamond (Refractive index = 2.42).
Lowest Optical Density: Air (Refractive index = 1.0003).
Lowest Optical Density: Air (Refractive index = 1.0003).
Q4. You are given kerosene, turpentine and water. In which of these does the light travel fastest?
Speeds is inversely proportional to refractive index (v ∝ 1/n).
nwater = 1.33
nkerosene = 1.44
nturpentine = 1.47
Water has the lowest refractive index, so light travels fastest in Water.
nwater = 1.33
nkerosene = 1.44
nturpentine = 1.47
Water has the lowest refractive index, so light travels fastest in Water.
Q5. The refractive index of diamond is 2.42. What is the meaning of this statement?
It means that the ratio of the speed of light in air (or vacuum) to the speed of light in diamond is 2.42. Speed of light in diamond is 1/2.42 times speed of light in vacuum.
Set 4
Q1. Define 1 dioptre of power of a lens.
1 Dioptre is the power of a lens whose focal length is 1 meter. (1 D = 1 m-1).
Q2. A convex lens forms a real and inverted image of a needle at a distance of 50 cm from it. Where is the needle placed... if image is equal to size of object? Power?
Since image is real, inverted and same size, the object must be at 2F.
Image distance v = +50 cm. So 2f = 50 cm => f = 25 cm.
Since object is at 2F, object distance u = -50 cm.
Power P = 1/f(in meters) = 1/0.25 = +4.0 D.
Image distance v = +50 cm. So 2f = 50 cm => f = 25 cm.
Since object is at 2F, object distance u = -50 cm.
Power P = 1/f(in meters) = 1/0.25 = +4.0 D.
Q3. Find the power of a concave lens of focal length 2 m.
Focal length of concave lens is negative. f = -2 m.
P = 1/f = 1/(-2) = -0.5 D.
P = 1/f = 1/(-2) = -0.5 D.
NCERT Exercise Questions
Complete solutions for Chapter 9 exercises.
Q1. Which one of the following materials cannot be used to make a lens?
(a) Water (b) Glass (c) Plastic (d) Clay
Answer: (d) Clay. Because it is opaque.
Answer: (d) Clay. Because it is opaque.
Q2. The image formed by a concave mirror is observed to be virtual, erect and
larger than the object. Where should be the position of the object?
(d) Between the pole of the mirror and its principal focus.
Q3. Where should an object be placed in front of a convex lens to get a real
image of the size of the object?
(b) At twice the focal length (At 2F).
Q4. A spherical mirror and a thin spherical lens have each a focal length of -15
cm. The mirror and the lens are likely to be:
(a) both concave. (By convention, focal length is negative for concave mirror and
concave lens).
Q5. No matter how far you stand from a mirror, your image appears erect. The
mirror is likely to be:
(d) either plane or convex. (Plane mirror always gives erect image; Convex mirror
always gives erect diminished image. Concave gives inverted especially at distance).
Q6. Which of the following lenses would you prefer to use while reading small
letters found in a dictionary?
To read small letters, we need a magnifying glass (Convex lens) of high power (short focal length).
(c) A convex lens of focal length 5 cm.
(c) A convex lens of focal length 5 cm.
Q7. We wish to obtain an erect image of an object, using a concave mirror of
focal length 15 cm. Range of distance? Nature? Size? Ray Diagram.
Range: Between 0 and 15 cm (Between Pole and Focus).
Nature: Virtual and Erect.
Size: Larger than object (Magnified).
[Ray Diagram: Object placed between P and F. Rays diverge after reflection and appear to meet behind the mirror].
Nature: Virtual and Erect.
Size: Larger than object (Magnified).
[Ray Diagram: Object placed between P and F. Rays diverge after reflection and appear to meet behind the mirror].
Q8. Name the type of mirror used in: (a) Headlights (b) Side/rear-view (c) Solar
furnace.
(a) Concave Mirror: Source placed at Focus gives powerful parallel beam.
(b) Convex Mirror: Gives wider field of view and erect image.
(c) Concave Mirror: Converges sun's heat rays to a single point (Focus) to produce high temperature.
(b) Convex Mirror: Gives wider field of view and erect image.
(c) Concave Mirror: Converges sun's heat rays to a single point (Focus) to produce high temperature.
Q9. One-half of a convex lens is covered with a black paper. Will this lens
produce a complete image?
Yes, it will produce a complete image. The rays from the object can still pass through
the uncovered half to form the image. However, the intensity (brightness) of the image
will be reduced (it will appear dimmer) because less light contributes to it.
Q10. An object 5 cm in length is held 25 cm away from a converging lens of focal
length 10 cm. Find position, size and nature.
u = -25 cm, f = +10 cm, h = 5 cm.
1/v - 1/u = 1/f => 1/v = 1/10 + 1/(-25) = 1/10 - 1/25 = 3/50.
v = 50/3 = +16.67 cm. (Real image on other side).
m = v/u = (50/3) / -25 = -2/3 = -0.66.
h' = m * h = -0.66 * 5 = -3.33 cm.
Nature: Real, Inverted and Diminished.
1/v - 1/u = 1/f => 1/v = 1/10 + 1/(-25) = 1/10 - 1/25 = 3/50.
v = 50/3 = +16.67 cm. (Real image on other side).
m = v/u = (50/3) / -25 = -2/3 = -0.66.
h' = m * h = -0.66 * 5 = -3.33 cm.
Nature: Real, Inverted and Diminished.
Q11. A concave lens of focal length 15 cm forms an image 10 cm from the lens. How
far is object?
f = -15 cm, v = -10 cm (Virtual).
1/v - 1/u = 1/f => 1/u = 1/v - 1/f = 1/(-10) - 1/(-15) = -3/30 + 2/30 = -1/30.
u = -30 cm. Object is 30 cm from the lens.
1/v - 1/u = 1/f => 1/u = 1/v - 1/f = 1/(-10) - 1/(-15) = -3/30 + 2/30 = -1/30.
u = -30 cm. Object is 30 cm from the lens.
Q12. An object is placed at a distance of 10 cm from a convex mirror of focal
length 15 cm. Find position and nature.
u = -10 cm, f = +15 cm.
1/v + 1/u = 1/f => 1/v = 1/15 + 1/10 = 5/30 = 1/6.
v = +6 cm. (Behind mirror).
m = -v/u = -6/(-10) = +0.6.
Nature: Virtual, Erect, Diminished.
1/v + 1/u = 1/f => 1/v = 1/15 + 1/10 = 5/30 = 1/6.
v = +6 cm. (Behind mirror).
m = -v/u = -6/(-10) = +0.6.
Nature: Virtual, Erect, Diminished.
Q13. The magnification produced by a plane mirror is +1. What does this mean?
1. + sign means image is Erect and Virtual.
2. 1 means size of image is equal to size of object.
So, plane mirror forms virtual, erect image of same size.
2. 1 means size of image is equal to size of object.
So, plane mirror forms virtual, erect image of same size.
Q14. An object 5.0 cm in length is placed at a distance of 20 cm in front of a
convex mirror of radius 30 cm. Find position, nature and size.
u = -20 cm, R = +30 cm => f = +15 cm, h = 5 cm.
1/v + 1/u = 1/f => 1/v = 1/15 + 1/20 = 7/60.
v = 60/7 = +8.57 cm (Behind mirror).
m = -v/u = - (60/7) / -20 = +3/7 = 0.428.
h' = m * h = 0.428 * 5 = 2.14 cm.
Nature: Virtual, Erect, Diminished.
1/v + 1/u = 1/f => 1/v = 1/15 + 1/20 = 7/60.
v = 60/7 = +8.57 cm (Behind mirror).
m = -v/u = - (60/7) / -20 = +3/7 = 0.428.
h' = m * h = 0.428 * 5 = 2.14 cm.
Nature: Virtual, Erect, Diminished.
Q15. An object of size 7.0 cm is placed at 27 cm in front of a concave mirror of
focal length 18 cm. Position, Size, Nature?
u = -27 cm, f = -18 cm, h = 7 cm.
1/v + 1/u = 1/f => 1/v = 1/(-18) - 1/(-27) = -3/54 + 2/54 = -1/54.
v = -54 cm. (Screen should be placed here).
m = -v/u = -(-54)/(-27) = -2.
h' = -2 * 7 = -14 cm.
Nature: Real, Inverted, Enlarged.
1/v + 1/u = 1/f => 1/v = 1/(-18) - 1/(-27) = -3/54 + 2/54 = -1/54.
v = -54 cm. (Screen should be placed here).
m = -v/u = -(-54)/(-27) = -2.
h' = -2 * 7 = -14 cm.
Nature: Real, Inverted, Enlarged.
Q16. Find the focal length of a lens of power -2.0 D. What type of lens is this?
P = -2.0 D.
f = 1/P = 1/(-2) = -0.5 m = -50 cm.
Since power/focal length is negative, it is a Concave Lens (Diverging).
f = 1/P = 1/(-2) = -0.5 m = -50 cm.
Since power/focal length is negative, it is a Concave Lens (Diverging).
Q17. A doctor has prescribed a corrective lens of power +1.5 D. Find focal
length. Diverging or Converging?
P = +1.5 D.
f = 1/P = 1/1.5 = 2/3 m = +0.67 m (approx 66.7 cm).
Since power is positive, it is a Converging Lens (Convex).
f = 1/P = 1/1.5 = 2/3 m = +0.67 m (approx 66.7 cm).
Since power is positive, it is a Converging Lens (Convex).
Key Formulas & Facts
Formulas for Mirrors, Lenses and Refraction.
Important Formulas
Mirror Formula
1/v + 1/u = 1/f
Lens Formula
1/v - 1/u = 1/f
Magnification (Mirror)
m = -v/u
Magnification (Lens)
m = v/u
Power of Lens
P = 1/f(m)
Refractive Index
n = c/v
Snell's Law
sin i / sin r = constant
50 Important Facts
1. Light travels in straight line (Rectilinear
propagation).
2. Speed of light in vacuum: 3 x 108 m/s.
3. Reflection: Bouncing back of light.
4. Angle i = Angle r.
5. Plane mirror forms virtual, erect, same size
image.
6. Focal length (f) = Radius of curvature (R) / 2.
7. Concave Mirror: Converging mirror.
8. Convex Mirror: Diverging mirror.
9. Real image: Can be obtained on screen (Inverted).
10. Virtual image: Cannot be obtained on screen
(Erect).
11. Dentists use Concave mirror.
12. Shaving mirrors are concave (Enlarged image).
13. Rear-view mirrors are convex (Wide view).
14. Concave mirror forms real image mostly.
15. Convex mirror always forms virtual image.
16. Object distance (u) is always negative.
17. Focal length of Concave mirror is negative.
18. Focal length of Convex mirror is positive.
19. Refraction: Bending of light.
20. Rarer to Denser: Towards normal.
21. Denser to Rarer: Away from normal.
22. Refractive index of air approx 1.0003.
23. Refractive index of water 1.33.
24. Refractive index of glass 1.5.
25. Refractive index of diamond 2.42 (Highest).
26. Convex Lens: Converging lens.
27. Concave Lens: Diverging lens.
28. Convex lens forms real image mostly.
29. Concave lens always forms virtual image.
30. Focal length of Convex lens is positive.
31. Focal length of Concave lens is negative.
32. Power of lens P = 1/f (m).
33. Unit of Power is Dioptre (D).
34. Power of convex lens is positive.
35. Power of concave lens is negative.
36. Concave lens correction for Myopia.
37. Convex lens correction for Hypermetropia.
38. Magnification m > 1 means enlarged.
39. Magnification m < 1 means diminished.
40. Negative m means Real and Inverted.
41. Positive m means Virtual and Erect.
42. Absolute refractive index is w.r.t
vacuum.
43. Optical density is different from mass
density.
44. Kerosene is optically denser than water.
45. Twinkling of stars is due to atmospheric
refraction.
46. Advance sunrise/delay sunset due to
diffraction.
47. Combination of power P = P1 + P2.
48. Optical centre: No deviation of light.
49. Principal axis: Line joining centers of
curvature.
50. Aperture: Diameter of reflecting
surface.
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