Light - Reflection and Refraction

Phenomenon and Theories of Light

• Perception of Light:

  • Objects are visible when they reflect light that falls on them into the human eye.
  • Transparent media allow vision through them because light is transmitted through the medium.
  • Light exhibits straight-line propagation (rectilinear propagation), as evidenced by small light sources casting sharp shadows of opaque objects.
  • A straight-line path of light is conventionally represented as a ray of light.

• Diffraction of Light:

  • When an opaque object placed in the path of light becomes extremely small, light exhibits a tendency to bend around the object rather than proceeding in a straight line.
  • This phenomenon is defined as the diffraction of light.
  • In diffraction scenarios, ray optics (straight-line treatment using rays) fails to explain the observed behavior.

• Evolution of Light Theories:

  • Ray Theory: Treats light as traveling in straight rays; successfully explains shadow formation, reflection, and simple refraction.
  • Wave Theory: Treats light as a wave to explain phenomena such as diffraction and interference.
  • Particle Behavior: Discovered at the beginning of the 20th century; wave theory proved inadequate to explain light-matter interactions (such as the photoelectric effect), where light behaves as a stream of particles.
  • Modern Quantum Theory of Light: A unified framework emerging in the 20th century where light is reconciled as possessing dual natures — exhibiting both wave-like and particle-like properties depending on the interaction, rather than being strictly a classical wave or particle.

Reflection of Light and Reflection Properties

• Concept of Reflection:

  • A highly polished surface, such as a mirror, reflects the vast majority of light incident upon it.

• Laws of Reflection:

  • First Law: The angle of incidence ($i$) is equal to the angle of reflection ($r$):     $i = r$
  • Second Law: The incident ray, the normal to the reflecting surface at the point of incidence, and the reflected ray all lie within the same plane.
  • Scope of Application: The laws of reflection apply universally to all reflecting surfaces, including plane surfaces and curved/spherical surfaces.

• Image Formation by Plane Mirrors:

  • Nature of Image: Always virtual and erect.
  • Size of Image: Equal to the size of the object ($m = +1$).
  • Position of Image: Located as far behind the mirror as the object is situated in front of it.
  • Lateral Inversion: The left side of the object appears as the right side of the image, and vice versa.

• Observational Activity — Curved Reflecting Surfaces (Activity 9.1):

  • Viewing a face in the curved surface of a large, shining spoon demonstrates curved mirror behavior:
  • Inward Curved Surface: Functions as a concave mirror. Looking close shows an enlarged image; moving slowly away causes the image to invert and change size.
  • Outward Curved Surface (Bulged Out): Functions as a convex mirror. Shows an erect, diminished image regardless of distance.

Spherical Mirrors and Terminology

• Definition of Spherical Mirrors:

  • A mirror whose reflecting surface forms part of a sphere is defined as a spherical mirror.

• Types of Spherical Mirrors:

  • Concave Mirror: A spherical mirror whose reflecting surface is curved inwards (facing toward the centre of the sphere).
  • Convex Mirror: A spherical mirror whose reflecting surface is curved outwards (facing away from the centre of the sphere).

• Key Technical Parameters and Terminology:

  • Pole ($P$): The geometric centre of the reflecting surface of a spherical mirror. It lies directly on the mirror's surface.
  • Centre of Curvature ($C$): The centre of the sphere of which the mirror's reflecting surface forms a part.
  • Note: The centre of curvature is not a part of the mirror itself; it lies outside the reflecting surface.
  • For a concave mirror, $C$ lies in front of the reflecting surface.
  • For a convex mirror, $C$ lies behind the reflecting surface.
  • Radius of Curvature ($R$): The radius of the sphere of which the reflecting surface forms a part. The distance $PC$ equals $R$.
  • Principal Axis: An imaginary straight line passing through the pole ($P$) and centre of curvature ($C$) of the spherical mirror. It is normal to the mirror surface at $P$.
  • Principal Focus ($F$):
  • Concave Mirror: The point on the principal axis where rays parallel to the principal axis intersect after reflection.
  • Convex Mirror: The point on the principal axis from which rays parallel to the principal axis appear to diverge after reflection.
  • Focal Length ($f$): The distance along the principal axis between the pole ($P$) and the principal focus ($F$).
  • Aperture ($MN$): The effective diameter of the circular boundary outline of the reflecting surface.

• Mathematical Relationship Between Radius of Curvature and Focal Length:

  • For spherical mirrors with small apertures relative to their radius of curvature, the radius of curvature ($R$) is twice the focal length ($f$):     $R = 2f$
  • The principal focus ($F$) lies exactly midway between the pole ($P$) and centre of curvature ($C$).

• Experimental Determination of Focal Length (Activity 9.2 & Activity 9.3):

  • Directing a concave mirror toward distant sunlight and focusing the reflected light onto a sheet of paper creates a sharp, bright spot.
  • The spot is a real, inverted, highly diminished image of the Sun formed at the principal focus ($F$).
  • Concentrated thermal energy at $F$ causes the paper to smoke and ignite.
  • The measured distance between the mirror and the paper sheet gives the approximate focal length ($f$).

Image Formation and Ray Tracing for Spherical Mirrors

• Standard Rays for Constructing Ray Diagrams:

  • Ray 1 (Parallel to Principal Axis):
  • Concave: After reflection, passes directly through principal focus $F$.
  • Convex: After reflection, appears to diverge from principal focus $F$.
  • Ray 2 (Passing Through / Directed Toward Focus $F$):
  • Concave: Passes through $F$, emerges parallel to principal axis after reflection.
  • Convex: Directed toward $F$, emerges parallel to principal axis after reflection.
  • Ray 3 (Passing Through / Directed Toward Centre of Curvature $C$):
  • Concave or Convex: Reflected back along its exact incident path because the ray hits the mirror surface along the normal ($i = 0^\circ$, $r = 0^\circ$).
  • Ray 4 (Incident Obliquely at Pole $P$):
  • Reflected obliquely such that the angle of reflection equals the angle of incidence ($i = r$) with respect to the principal axis.

• Summary of Image Formation by Concave Mirror:

  | Position of Object | Position of Image | Size of Image | Nature of Image |   | :--- | :--- | :--- | :--- |   | At infinity | At focus $F$ | Highly diminished, point-sized | Real and inverted |   | Beyond $C$ | Between $F$ and $C$ | Diminished | Real and inverted |   | At $C$ | At $C$ | Same size | Real and inverted |   | Between $C$ and $F$ | Beyond $C$ | Enlarged | Real and inverted |   | At $F$ | At infinity | Highly enlarged / at infinity | Real and inverted |   | Between $P$ and $F$ | Behind the mirror | Enlarged | Virtual and erect |

• Practical Applications and Uses of Concave Mirrors:

  • Torches, Search-lights, and Vehicle Headlights: The light source is placed at focus $F$ to obtain powerful parallel beams of light.
  • Shaving Mirrors: Face placed within focal length (between $P$ and $F$) to view an enlarged, virtual, erect image.
  • Dentist Mirrors: Held close to teeth to produce magnified virtual images of oral cavity.
  • Solar Furnaces: Large concave mirrors focus incident parallel sunlight onto a single focal spot to generate ultra-high temperatures.

• Summary of Image Formation by Convex Mirror:

  | Position of Object | Position of Image | Size of Image | Nature of Image |   | :--- | :--- | :--- | :--- |   | At infinity | At focus $F$, behind mirror | Highly diminished, point-sized | Virtual and erect |   | Between infinity and pole $P$ | Between $P$ and $F$, behind mirror | Diminished | Virtual and erect |

• Practical Applications and Uses of Convex Mirrors:
  • Rear-View (Wing) Mirrors in Vehicles:
  • Produce erect, diminished images of traffic behind.
  • Provide a significantly broader field of view compared to plane mirrors due to outward curvature.
  • Wide-Angle Surveillance:
  • Fitted on walls (e.g., terrace wall at Agra Fort viewing full image of Taj Mahal) or store corridors to observe large areas in a compact mirror.

New Cartesian Sign Convention

• Reference Frame Rules:

  • Origin is established at the Pole ($P$) of the spherical mirror.
  • Principal Axis is treated as the x-axis ($X'X$).

• Five Cartesian Rules:

  1. Object Location: Object is always placed to the left of the mirror, meaning incident light strikes the mirror from left to right.
  2. Measurement Reference: All distances parallel to the principal axis are measured starting from the pole ($P$).
  3. Horizontal Distance Signs: Distances measured to the right of origin ($+x\text{-axis}$) are positive ($+$); distances measured to the left of origin ($-x\text{-axis}$) are negative ($-$).
  4. Vertical Distance Above Axis: Distances measured perpendicular to and above the principal axis ($+y\text{-axis}$) are positive ($+$).
  5. Vertical Distance Below Axis: Distances measured perpendicular to and below the principal axis ($-y\text{-axis}$) are negative ($-$).

Mirror Formula and Magnification

• Mirror Formula:

  • Relates object distance ($u$), image distance ($v$), and focal length ($f$):     $\frac{1}{v} + \frac{1}{u} = \frac{1}{f}$

• Linear Magnification ($m$):

  • Defined as the ratio of image height ($h'$) to object height ($h$):     $m = \frac{h'}{h}$
  • Expressed in terms of object and image distances:     $m = -\frac{v}{u}$
  • Combined formula:     $m = \frac{h'}{h} = -\frac{v}{u}$

• Sign Conventions for Magnification:

  • Object height ($h$) is always positive ($+$) as object is placed above principal axis.
  • Real images are formed below principal axis $\implies h'$ is negative ($-$) $\implies m$ is negative ($-$).
  • Virtual images are formed above principal axis $\implies h'$ is positive ($+$) $\implies m$ is positive ($+$).

• Worked Examples for Spherical Mirrors:

  • Example 9.1:
  • Given: Convex rear-view mirror, radius of curvature $R = +3.00\,\text{m}$, object distance $u = -5.00\,\text{m}$.
  • Focal Length:       $f = \frac{R}{2} = \frac{+3.00\,\text{m}}{2} = +1.50\,\text{m}$
  • Image Distance Calculation:       $\frac{1}{v} + \frac{1}{u} = \frac{1}{f} \implies \frac{1}{v} = \frac{1}{f} - \frac{1}{u}$$\frac{1}{v} = \frac{1}{1.50} - \frac{1}{-5.00} = \frac{1}{1.50} + \frac{1}{5.00} = \frac{5.00 + 1.50}{7.50} = \frac{6.50}{7.50}$$v = \frac{7.50}{6.50} = +1.15\,\text{m}$
  • Magnification Calculation:       $m = -\frac{v}{u} = -\frac{+1.15\,\text{m}}{-5.00\,\text{m}} = +0.23$
  • Conclusion: Image is located $1.15\,\text{m}$ behind the mirror, virtual, erect, and diminished by a factor of 0.23$.\n\n - **Example 9.2**:\n - **Given**: Object height h = +4.0\,\text{cm}$, object distance$u = -25.0\,\text{cm}$, concave mirror focal length$f = -15.0\,\text{cm}.\n - **Image Distance Calculation**:\n      \frac{1}{v} = \frac{1}{f} - \frac{1}{u} = \frac{1}{-15.0} - \frac{1}{-25.0} = -\frac{1}{15.0} + \frac{1}{25.0}\n      \frac{1}{v} = \frac{-5.0 + 3.0}{75.0} = -\frac{2.0}{75.0}\n      v = -37.5\,\text{cm}\n - **Image Size Calculation**:\n      m = \frac{h'}{h} = -\frac{v}{u} \implies h' = -h \times \left(\frac{v}{u}\right)\n      h' = -(+4.0\,\text{cm}) \times \left(\frac{-37.5\,\text{cm}}{-25.0\,\text{cm}}\right) = -6.0\,\text{cm}\n - **Conclusion**: Screen should be placed 37.5\,\text{cm}$in front of mirror. The image is real, inverted, and enlarged ($h' = -6.0\,\text{cm}).\n\n# Questions and Worked Solutions (Mirror Section)\n\n- **In-Text Questions (Page 9 / 142)**:\n - **Question 1**: Define the principal focus of a concave mirror.\n - **Answer**: Principal focus of a concave mirror is a point on its principal axis where rays parallel to principal axis converge after reflection.\n - **Question 2**: The radius of curvature of a spherical mirror is 20\,\text{cm}. What is its focal length?\n - **Answer**:\n      f = \frac{R}{2} = \frac{20\,\text{cm}}{2} = 10\,\text{cm}\n - **Question 3**: Name a mirror that can give an erect and enlarged image of an object.\n - **Answer**: A concave mirror (when object is placed between pole P$and principal focus$F).\n - **Question 4**: Why do we prefer a convex mirror as a rear-view mirror in vehicles?\n - **Answer**: Convex mirrors produce erect, diminished images and offer a much wider field of view than plane mirrors.\n\n- **In-Text Questions (Page 12 / 145)**:\n - **Question 1**: Find the focal length of a convex mirror whose radius of curvature is 32\,\text{cm}.\n - **Answer**:\n      f = \frac{R}{2} = \frac{+32\,\text{cm}}{2} = +16\,\text{cm}\n - **Question 2**: A concave mirror produces three times magnified (enlarged) real image of an object placed at 10\,\text{cm} in front of it. Where is the image located?\n - **Answer**:\n - Real image implies magnification m = -3$. Object distance$u = -10\,\text{cm}.\n - Using m = -\frac{v}{u}:\n        -3 = -\frac{v}{-10\,\text{cm}} \implies v = -30\,\text{cm}\n - The image is located 30\,\text{cm} in front of the concave mirror.\n\n# Phenomenon and Laws of Refraction\n\n- **Refraction Concept**:\n - Refraction is defined as the bending of light at the boundary when it travels obliquely from one transparent medium into another due to a change in the speed of light.\n\n- **Everyday Physical Manifestations**:\n - Bottom of a water-filled tank or pond appearing raised.\n - Printed text appearing raised when viewed through a thick glass slab.\n - Pencil partly submerged in water appearing bent or displaced at the air-water interface.\n - Lemons placed in water inside a glass tumbler appearing larger from the sides.\n\n- **Refraction Through Rectangular Glass Slab (Activity 9.10)**:\n - **First Boundary (Air to Glass at O$)**: Light travels from rarer to denser medium$\implies bends **towards the normal**.\n - **Second Boundary (Glass to Air at O'$)**: Light travels from denser to rarer medium$\implies bends **away from the normal**.\n - **Emergent Ray Behavior**: Bending at opposite parallel boundaries (AB$and$CD$) is equal and opposite. Emergent ray ($O'H$) is parallel to incident ray ($EO) but shifted laterally.\n - **Normal Incidence**: A ray incident normally (i = 0^\circ$) passes straight without deviation ($r = 0^\circ).\n\n- **Laws of Refraction**:\n - **First Law**: The incident ray, the refracted ray, and the normal to the interface of two transparent media at the point of incidence all lie in the same plane.\n - **Second Law (Snell's Law of Refraction)**: For a given pair of media and light of a given color/wavelength, the ratio of the sine of angle of incidence (i$) to the sine of angle of refraction ($r$) is constant ($0 < i < 90^\circ):\n    \frac{\sin(i)}{\sin(r)} = \text{constant}\n - This constant is defined as the **refractive index** of the second medium with respect to the first medium.\n\n# Refractive Index and Optical Density\n\n- **Refractive Index (n)**:\n - Links bending ability to the relative speed of light propagation in different media.\n - Speed of light in vacuum (c$):$3 \times 10^8\,\text{m s}^{-1}.\n\n- **Relative Refractive Index Equations**:\n - Refractive index of medium 2 with respect to medium 1 (n_{21}):\n    n_{21} = \frac{\text{Speed of light in medium 1}}{\text{Speed of light in medium 2}} = \frac{v_1}{v_2}\n - Refractive index of medium 1 with respect to medium 2 (n_{12}):\n    n_{12} = \frac{\text{Speed of light in medium 2}}{\text{Speed of light in medium 1}} = \frac{v_2}{v_1}\n\n- **Absolute Refractive Index (n_m)**:\n - Refractive index relative to vacuum (or air):\n    n_m = \frac{\text{Speed of light in air}}{\text{Speed of light in medium}} = \frac{c}{v}\n\n- **Absolute Refractive Indices Table (Table 9.3)**:\n - Air: 1.0003\n - Ice: 1.31\n - Water: 1.33$($n_w = 1.33)\n - Alcohol: 1.36\n - Kerosene: 1.44\n - Fused quartz: 1.46\n - Turpentine oil: 1.47\n - Benzene: 1.50\n - Crown glass: 1.52$($n_g = 1.52)\n - Canada Balsam: 1.53\n - Rock salt: 1.54\n - Carbon disulphide: 1.63\n - Dense flint glass: 1.65\n - Ruby: 1.71\n - Sapphire: 1.77\n - Diamond: 2.42\n\n- **Optical Density vs. Mass Density**:\n - Optical density measures a medium's capacity to refract light, defined by its refractive index (higher n \implies optically denser).\n - Optical density is distinct from mass density.\n - Example: Kerosene has a higher refractive index (1.44$) than water ($1.33), making kerosene **optically denser** than water, even though its mass density is lower (it floats on water).\n\n# Questions and Worked Solutions (Refraction Section)\n\n- **In-Text Questions (Page 17 / 150)**:\n - **Question 1**: A ray of light travelling in air enters obliquely into water. Does the light ray bend towards the normal or away from the normal? Why?\n - **Answer**: Bends **towards the normal** because water is optically denser than air (n_\text{water} = 1.33 > n_\text{air} = 1.0003), causing light to slow down.\n - **Question 2**: Light enters from air to glass having refractive index 1.50$. What is the speed of light in the glass? (Speed of light in vacuum$c = 3 \times 10^8\,\text{m s}^{-1}).\n - **Answer**:\n      n_g = \frac{c}{v_g} \implies v_g = \frac{c}{n_g} = \frac{3 \times 10^8\,\text{m s}^{-1}}{1.50} = 2 \times 10^8\,\text{m s}^{-1}\n - **Question 3**: Find out the medium having highest optical density and lowest optical density from Table 9.3.\n - **Answer**: Highest optical density is **Diamond** (n = 2.42$); lowest optical density is **Air** ($n = 1.0003).\n - **Question 4**: Given kerosene (1.44$), turpentine ($1.47$), and water ($1.33), in which does light travel fastest?\n - **Answer**: Light travels fastest in **water** because it has the lowest refractive index (n = 1.33) among the three.\n - **Question 5**: The refractive index of diamond is 2.42. What is the meaning of this statement?\n - **Answer**: The speed of light in vacuum is 2.42$times the speed of light in diamond, or speed of light in diamond is$\frac{1}{2.42} times its speed in vacuum.\n\n# Refraction by Spherical Lenses\n\n- **Definition of Spherical Lens**:\n - A transparent material bound by two surfaces, where at least one surface is spherical.\n\n- **Types of Lenses**:\n - **Convex Lens (Double Convex)**: Bulges outwards, thicker at centre than edges. Converges parallel rays to a point \implies **Converging Lens**.\n - **Concave Lens (Double Concave)**: Curved inwards, thicker at edges than centre. Diverges parallel rays \implies **Diverging Lens**.\n\n- **Anatomical Parameters of Lenses**:\n - **Centres of Curvature (C_1, C_2)**: Centres of the two spheres forming the lens surfaces.\n - **Principal Axis**: Imaginary straight line passing through both centres of curvature (C_1$and$C_2).\n - **Optical Centre (O$)**: Central point of the lens. Light rays passing through$O emerge without suffering any net deviation.\n - **Aperture**: Effective diameter of circular outline of spherical lens.\n - **Thin Lens**: Lens whose aperture is much smaller than its radii of curvature.\n - **Principal Foci (F_1, F_2)**: Points on principal axis where parallel incident rays converge (convex) or appear to diverge from (concave).\n - **Focal Length (f$)**: Distance from optical centre$O to principal focus.\n\n# Image Formation by Lenses and Ray Tracing Rules\n\n- **Three Standard Rays for Lens Ray Diagrams**:\n - **Ray 1**: Ray parallel to principal axis passes through F_2$(convex) or appears to diverge from$F_1 (concave).\n - **Ray 2**: Ray passing through F_1$(convex) or directed toward$F_2 (concave) emerges parallel to principal axis.\n - **Ray 3**: Ray passing through optical centre O continues straight without deviation.\n\n- **Summary of Image Formation by Convex Lens**:\n\n  | Position of Object | Position of Image | Relative Size | Nature of Image |\n  | :--- | :--- | :--- | :--- |\n  | At infinity | At focus F_2 | Highly diminished, point-sized | Real and inverted |\n  | Beyond 2F_1$| Between$F_2$and$2F_2 | Diminished | Real and inverted |\n  | At 2F_1$| At$2F_2 | Same size | Real and inverted |\n  | Between F_1$and$2F_1$| Beyond$2F_2 | Enlarged | Real and inverted |\n  | At focus F_1 | At infinity | Highly enlarged / at infinity | Real and inverted |\n  | Between F_1$and$O | Same side as object | Enlarged | Virtual and erect |\n\n- **Summary of Image Formation by Concave Lens**:\n\n  | Position of Object | Position of Image | Relative Size | Nature of Image |\n  | :--- | :--- | :--- | :--- |\n  | At infinity | At focus F_1 | Highly diminished, point-sized | Virtual and erect |\n  | Between infinity and optical centre O$| Between$F_1$and$O | Diminished | Virtual and erect |\n\n# Lens Formula, Magnification, and Lens Power\n\n- **Sign Convention for Lenses**:\n - Measurements taken from Optical Centre (O).\n - Focal length of convex lens is **positive** (+).\n - Focal length of concave lens is **negative** (-).\n\n- **Lens Formula**:\n - Relates u$,$v$, and$f:\n    \frac{1}{v} - \frac{1}{u} = \frac{1}{f}\n\n- **Magnification Produced by Lens (m)**:\n - Ratio of image height (h'$) to object height ($h):\n    m = \frac{h'}{h}\n - Ratio of image distance (v$) to object distance ($u):\n    m = \frac{v}{u}\n - Combined formula:\n    m = \frac{h'}{h} = \frac{v}{u}\n\n- **Worked Examples for Lenses**:\n - **Example 9.3**:\n - **Given**: Concave lens, f = -15\,\text{cm}$, image distance$v = -10\,\text{cm}.\n - **Object Distance Calculation**:\n      \frac{1}{v} - \frac{1}{u} = \frac{1}{f} \implies \frac{1}{u} = \frac{1}{v} - \frac{1}{f}\n      \frac{1}{u} = \frac{1}{-10} - \frac{1}{-15} = -\frac{1}{10} + \frac{1}{15} = \frac{-3 + 2}{30} = -\frac{1}{30}\n      u = -30\,\text{cm}\n - **Magnification Calculation**:\n      m = \frac{v}{u} = \frac{-10\,\text{cm}}{-30\,\text{cm}} = +\frac{1}{3} \approx +0.33\n - **Conclusion**: Object placed 30\,\text{cm} from lens. Image is virtual, erect, and one-third object size.\n\n - **Example 9.4**:\n - **Given**: Convex lens, h = +2.0\,\text{cm}$,$f = +10\,\text{cm}$,$u = -15\,\text{cm}.\n - **Image Distance Calculation**:\n      \frac{1}{v} = \frac{1}{f} + \frac{1}{u} = \frac{1}{10} + \frac{1}{-15} = \frac{3 - 2}{30} = \frac{1}{30}\n      v = +30\,\text{cm}\n - **Image Size Calculation**:\n      m = \frac{v}{u} = \frac{+30\,\text{cm}}{-15\,\text{cm}} = -2.0\n      h' = m \times h = (-2.0) \times (+2.0\,\text{cm}) = -4.0\,\text{cm}\n - **Conclusion**: Real, inverted image formed 30\,\text{cm}$on opposite side of lens, height$4.0\,\text{cm} below axis.\n\n- **Power of a Lens (P)**:\n - Defined as the reciprocal of focal length f measured in metres:\n    P = \frac{1}{f}\n - **SI Unit**: Dioptre (\text{D}$), where$1\,\text{D} = 1\,\text{m}^{-1}.\n - Power of convex lens is **positive** (+$); power of concave lens is **negative** ($-).\n - Examples:\n - Prescribed power +2.0\,\text{D} \implies$Convex lens,$f = +\frac{1}{2.0} = +0.50\,\text{m}.\n - Prescribed power -2.5\,\text{D} \implies$Concave lens,$f = \frac{1}{-2.5} = -0.40\,\text{m}.\n - **Combination of Lenses in Contact**:\n - Net power P of lenses placed in contact is algebraic sum of individual powers:\n      P = P_1 + P_2 + P_3 + \dots\n - Example: Combination of +2.0\,\text{D}$and$+0.25\,\text{D} \implies P = +2.25\,\text{D}.\n\n# Questions and Solutions (Lens Power Section)\n\n- **In-Text Questions (Page 25 / 158)**:\n - **Question 1**: Define 1 dioptre of power of a lens.\n - **Answer**: 1 dioptre is the power of a lens whose focal length is 1 metre (1\,\text{D} = 1\,\text{m}^{-1}).\n - **Question 2**: A convex lens forms a real and inverted image of a needle at a distance of 50\,\text{cm} from it. Where is the needle placed in front of the convex lens if the image is equal to the size of the object? Also, find the power of the lens.\n - **Answer**:\n - Real image equal to object size \implies$Object is at$2F_1$and image is at$2F_2$.
    • Image distance $v = +50\,\text{cm} \implies u = -50\,\text{cm}$. Needle is placed $50\,\text{cm}$ in front of lens.
    • Focal length:         $\frac{1}{f} = \frac{1}{v} - \frac{1}{u} = \frac{1}{50} - \frac{1}{-50} = \frac{2}{50} = \frac{1}{25\,\text{cm}} \implies f = +25\,\text{cm} = +0.25\,\text{m}$
    • Power:         $P = \frac{1}{f} = \frac{1}{+0.25\,\text{m}} = +4.0\,\text{D}$
  • Question 3: Find the power of a concave lens of focal length $2\,\text{m}$.
  • Answer:
    • For concave lens, $f = -2\,\text{m}$.
    • Power $P = \frac{1}{-2\,\text{m}} = -0.5\,\text{D}$.

Chapter End Exercises Solutions

• Exercise 1: Which one of the following materials cannot be used to make a lens?

  • Options: (a) Water (b) Glass (c) Plastic (d) Clay
  • Answer: (d) Clay (because clay is opaque and does not transmit light).

• Exercise 2: 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?

  • Options: (a) Between principal focus and centre of curvature (b) At centre of curvature (c) Beyond centre of curvature (d) Between pole of mirror and principal focus.
  • Answer: (d) Between the pole of the mirror and its principal focus.

• Exercise 3: Where should an object be placed in front of a convex lens to get a real image of the size of the object?

  • Options: (a) At principal focus (b) At twice the focal length (c) At infinity (d) Between optical centre and principal focus.
  • Answer: (b) At twice the focal length ($2F_1$).

• Exercise 4: A spherical mirror and a thin spherical lens have each a focal length of $-15\,\text{cm}$. The mirror and lens are likely to be:

  • Options: (a) Both concave (b) Both convex (c) Mirror concave, lens convex (d) Mirror convex, lens concave.
  • Answer: (a) Both concave (by sign convention, focal length is negative for both concave mirrors and concave lenses).

• Exercise 5: No matter how far you stand from a mirror, your image appears erect. The mirror is likely to be:

  • Options: (a) Only plane (b) Only concave (c) Only convex (d) Either plane or convex.
  • Answer: (d) Either plane or convex (both always produce erect images for all object distances).

• Exercise 6: Which lens would you prefer to use while reading small letters found in a dictionary?

  • Options: (a) Convex lens of focal length $50\,\text{cm}$ (b) Concave lens of focal length $50\,\text{cm}$ (c) Convex lens of focal length $5\,\text{cm}$ (d) Concave lens of focal length $5\,\text{cm}$.
  • Answer: (c) A convex lens of focal length $5\,\text{cm}$ (convex lens acts as magnifying glass when object is within focus; shorter focal length gives higher magnification and power).

• Exercise 7: Obtain erect image using concave mirror of $f = 15\,\text{cm}$. Range of object distance, nature, size, and ray diagram description.

  • Answer:
  • Range of distance: Object must be placed between pole and focus, i.e., $0\,\text{cm} < u < 15\,\text{cm}$.
  • Nature of image: Virtual and erect.
  • Size of image: Larger than object (enlarged).

• Exercise 8: Mirror type for:

  • (a) Headlights of a car: Concave mirror. Light bulb placed at focus produces powerful parallel beam.
  • (b) Side/rear-view mirror of a vehicle: Convex mirror. Forms erect, diminished images with wide field of view.
  • (c) Solar furnace: Concave mirror. Concentrates parallel sunlight onto focus to generate extreme heat.

• Exercise 9: One-half of a convex lens is covered with black paper. Will it produce a complete image?

  • Answer: Yes, it produces a complete image because light rays from every point on object pass through the uncovered half of the lens. However, the brightness / intensity of the image is reduced because fewer light rays contribute to image formation.

• Exercise 10: Object $h = 5\,\text{cm}$ held $25\,\text{cm}$ away from converging lens of $f = 10\,\text{cm}$. Position, size, nature.

  • Answer:
  • Given: $h = +5\,\text{cm}$, $u = -25\,\text{cm}$, $f = +10\,\text{cm}$.
  • Calculation:       $\frac{1}{v} = \frac{1}{f} + \frac{1}{u} = \frac{1}{10} + \frac{1}{-25} = \frac{5 - 2}{50} = \frac{3}{50}$$v = +\frac{50}{3}\,\text{cm} \approx +16.67\,\text{cm}$
  • Size and Magnification:       $m = \frac{v}{u} = \frac{+50/3}{-25} = -\frac{2}{3} \approx -0.67$$h' = m \times h = \left(-\frac{2}{3}\right) \times 5\,\text{cm} = -\frac{10}{3}\,\text{cm} \approx -3.33\,\text{cm}$
  • Nature: Real, inverted, diminished image formed $16.67\,\text{cm}$ on opposite side of lens.

• Exercise 11: Concave lens $f = 15\,\text{cm}$ forms image $10\,\text{cm}$ from lens. Find object distance.

  • Answer:
  • Given: $f = -15\,\text{cm}$, $v = -10\,\text{cm}$.
  • Calculation:       $\frac{1}{u} = \frac{1}{v} - \frac{1}{f} = \frac{1}{-10} - \frac{1}{-15} = -\frac{1}{10} + \frac{1}{15} = \frac{-3 + 2}{30} = -\frac{1}{30}$$u = -30\,\text{cm}$
  • Object is placed $30\,\text{cm}$ in front of the concave lens.

• Exercise 12: Object at $10\,\text{cm}$ from convex mirror $f = 15\,\text{cm}$. Position and nature of image.

  • Answer:
  • Given: $u = -10\,\text{cm}$, $f = +15\,\text{cm}$.
  • Calculation:       $\frac{1}{v} = \frac{1}{f} - \frac{1}{u} = \frac{1}{15} - \frac{1}{-10} = \frac{1}{15} + \frac{1}{10} = \frac{2 + 3}{30} = \frac{5}{30} = \frac{1}{6}$$v = +6\,\text{cm}$
  • Nature: Virtual and erect image formed $6\,\text{cm}$ behind the mirror.

• Exercise 13: Magnification produced by plane mirror is $+1$. What does this mean?

  • Answer:
  • Positive sign ($+$) indicates that the image is virtual and erect.
  • Magnitude $1$ indicates that the image size is exactly equal to object size ($h' = h$).

• Exercise 14: Object $h = 5.0\,\text{cm}$ placed at $u = -20\,\text{cm}$ in front of convex mirror $R = +30\,\text{cm}$. Position, nature, size.

  • Answer:
  • Given: $h = +5.0\,\text{cm}$, $u = -20\,\text{cm}$, $f = \frac{R}{2} = \frac{+30\,\text{cm}}{2} = +15\,\text{cm}$.
  • Calculation:       $\frac{1}{v} = \frac{1}{15} - \frac{1}{-20} = \frac{4 + 3}{60} = \frac{7}{60} \implies v = +\frac{60}{7}\,\text{cm} \approx +8.57\,\text{cm}$
  • Size:       $m = -\frac{v}{u} = -\frac{60/7}{-20} = +\frac{3}{7} \approx +0.43$$h' = m \times h = \left(+\frac{3}{7}\right) \times 5.0\,\text{cm} = +\frac{15}{7}\,\text{cm} \approx +2.14\,\text{cm}$
  • Nature: Virtual, erect image of height $2.14\,\text{cm}$ formed $8.57\,\text{cm}$ behind the mirror.

• Exercise 15: Object $h = 7.0\,\text{cm}$ placed at $u = -27\,\text{cm}$ in front of concave mirror $f = -18\,\text{cm}$. Distance of screen, size, nature.

  • Answer:
  • Given: $h = +7.0\,\text{cm}$, $u = -27\,\text{cm}$, $f = -18\,\text{cm}$.
  • Calculation:       $\frac{1}{v} = \frac{1}{-18} - \frac{1}{-27} = -\frac{1}{18} + \frac{1}{27} = \frac{-3 + 2}{54} = -\frac{1}{54}$$v = -54\,\text{cm}$
  • Screen position: Screen should be placed $54\,\text{cm}$ in front of mirror.
  • Size:       $m = -\frac{v}{u} = -\frac{-54}{-27} = -2.0$$h' = m \times h = (-2.0) \times 7.0\,\text{cm} = -14.0\,\text{cm}$
  • Nature: Real, inverted, enlarged image of height $14.0\,\text{cm}$.

• Exercise 16: Focal length of lens of power $-2.0\,\text{D}$. Type of lens.

  • Answer:
  • $f = \frac{1}{P} = \frac{1}{-2.0\,\text{D}} = -0.5\,\text{m} = -50\,\text{cm}$.
  • Negative focal length indicates a concave (diverging) lens.

• Exercise 17: Doctor prescribed lens of power $+1.5\,\text{D}$. Focal length and type.

  • Answer:
  • $f = \frac{1}{P} = \frac{1}{+1.5\,\text{D}} = +\frac{2}{3}\,\text{m} \approx +0.67\,\text{m} = +66.7\,\text{cm}$.
  • Positive focal length indicates a convex (converging) lens.