Exercise Solutions (Based on Maharashtra Board Textbook)

Q1: Choose the correct option and rewrite the statements.

1. The angle of incidence is equal to the angle of reflection.
2. The image formed by a plane mirror is always virtual and erect.
3. A concave mirror is also known as a converging mirror.
4. The focal length of a spherical mirror is half of its radius of curvature.
5. Rear-view mirrors in vehicles are convex mirrors.

Q2: Match the pairs.

Q3: Give scientific reasons.

1. We can see objects due to reflection.
   Answer: We can see objects due to reflection because most objects do not produce their own light. Instead, they reflect light that falls on them from a light source (like the sun or a bulb). When this reflected light enters our eyes, our brain processes it, allowing us to perceive the object. Without reflection, objects would appear black or invisible in the absence of self-emitted light.
2. A plane mirror forms a virtual image.
   Answer: A plane mirror forms a virtual image because the light rays reflected from its surface do not actually meet at a point behind the mirror. Instead, they appear to diverge from a point behind the mirror. Since the image is formed by the apparent intersection of light rays and not the actual intersection, it cannot be projected onto a screen, making it a virtual image.
3. Convex mirrors are used as rear-view mirrors in vehicles.
   Answer: Convex mirrors are used as rear-view mirrors in vehicles because they always form a virtual, erect, and diminished image of the objects. This property allows them to provide a much wider field of view compared to plane mirrors, enabling the driver to see a larger area of traffic behind the vehicle, which is crucial for safe driving.
4. Concave mirrors are used as shaving mirrors.
   Answer: Concave mirrors are used as shaving mirrors because when an object (like the face) is placed between the pole (P) and the principal focus (F) of a concave mirror, it forms a virtual, erect, and enlarged image behind the mirror. This enlarged image helps in seeing a magnified view of the face, making shaving easier and more precise.

Q4: Write answers to the following questions in your own words.

1. Explain the laws of reflection with a neat diagram.
   Answer: The two laws of reflection explain how light behaves when it strikes a surface.
   **First Law of Reflection:** This law states that the incident ray (the light ray striking the surface), the reflected ray (the light ray bouncing off the surface), and the normal (an imaginary line perpendicular to the surface at the point where the incident ray strikes) all lie in the same plane. Imagine a flat sheet of paper; all three of these lines would lie perfectly flat on that paper.
   **Second Law of Reflection:** This law states that the angle of incidence ($\angle i$) is equal to the angle of reflection ($\angle r$). The angle of incidence is the angle between the incident ray and the normal, and the angle of reflection is the angle between the reflected ray and the normal. So, if light hits a mirror at a 30-degree angle to the normal, it will bounce off at a 30-degree angle.
   (Diagram: A plane mirror with a normal line. An incident ray striking the mirror, making angle 'i' with the normal. A reflected ray bouncing off, making angle 'r' with the normal, where i=r. All rays and normal shown in the same plane.)
2. Describe the image formation by a concave mirror when an object is placed at different positions.
   Answer: A concave mirror can form both real and virtual images, depending on the object's position.
   • Object at infinity: Image at F, real, inverted, highly diminished (point size).
   • Object beyond C: Image between F and C, real, inverted, diminished.
   • Object at C: Image at C, real, inverted, same size.
   • Object between C and F: Image beyond C, real, inverted, enlarged.
   • Object at F: Image at infinity, real, inverted, highly enlarged.
   • Object between F and P: Image behind the mirror, virtual, erect, enlarged. (This is the only case where a concave mirror forms a virtual image).
   (Note: In a full answer, one would draw ray diagrams for each case.)
3. Explain the uses of concave and convex mirrors in daily life.
   Answer:
   • Uses of Concave Mirrors:
     • Shaving mirrors/Dentist mirrors: They produce enlarged, erect virtual images when the object is close, making it easier to see fine details.
     • Torches, headlights of cars, searchlights: A bulb placed at the focus produces a strong parallel beam of light for illumination.
     • Solar furnaces: They concentrate parallel rays of sunlight to a single focal point, generating intense heat.
   • Uses of Convex Mirrors:
     • Rear-view mirrors in vehicles: They always form virtual, erect, and diminished images, providing a wide field of view, which is crucial for seeing traffic behind.
     • Shop security mirrors: Mounted in shops, they provide a wide view of the store, allowing staff to monitor a large area.
     • Street light reflectors: They diverge light over a large area for street illumination.

Q5: Numericals (from typical textbook exercises).

1. An object is placed at a distance of 15 cm from a concave mirror of focal length 10 cm. Find the position and nature of the image.
   Solution: Given: Object distance $u = -15$ cm (Left of pole, according to sign convention) Focal length of concave mirror $f = -10$ cm (Concave mirror has negative focal length)
   Using Mirror Formula: $\frac{1}{v} + \frac{1}{u} = \frac{1}{f}$
   $\frac{1}{v} + \frac{1}{-15} = \frac{1}{-10}$
   $\frac{1}{v} = \frac{1}{-10} + \frac{1}{15}$
   $\frac{1}{v} = \frac{-3 + 2}{30}$
   $\frac{1}{v} = \frac{-1}{30}$
   $v = -30$ cm
   Position: The image is formed at 30 cm in front of the mirror.
   Nature: Since $v$ is negative, the image is real.
   To find nature (inverted/erect) and size (magnified/diminished), calculate magnification $M = -\frac{v}{u}$.
   $M = -\frac{-30}{-15} = -2$
   Since $M$ is negative, the image is inverted. Since $|M| > 1$, the image is enlarged.
   Final Answer: The image is formed at 30 cm in front of the mirror, and it is Real, Inverted, and Enlarged.
2. A convex mirror has a focal length of 20 cm. An object is placed 10 cm from the mirror. Find the position and nature of the image.
   Solution: Given: Focal length of convex mirror $f = +20$ cm (Convex mirror has positive focal length) Object distance $u = -10$ cm
   Using Mirror Formula: $\frac{1}{v} + \frac{1}{u} = \frac{1}{f}$
   $\frac{1}{v} + \frac{1}{-10} = \frac{1}{20}$
   $\frac{1}{v} = \frac{1}{20} + \frac{1}{10}$
   $\frac{1}{v} = \frac{1 + 2}{20}$
   $\frac{1}{v} = \frac{3}{20}$
   $v = \frac{20}{3} \approx +6.67$ cm
   Position: The image is formed at approximately 6.67 cm behind the mirror.
   Nature: Since $v$ is positive, the image is virtual.
   To find nature (erect/inverted) and size (magnified/diminished), calculate magnification $M = -\frac{v}{u}$.
   $M = -\frac{20/3}{-10} = \frac{20}{30} = +\frac{2}{3} \approx +0.67$
   Since $M$ is positive, the image is erect. Since $|M| < 1$, the image is diminished.
   Final Answer: The image is formed at approximately 6.67 cm behind the mirror, and it is Virtual, Erect, and Diminished.

Q6: Draw neat and labeled diagrams.

1. Reflection of light by a plane mirror showing laws of reflection.
   Sketch: (Imagine a diagram similar to the one described in Q4(a). A flat line representing the mirror, a dashed line perpendicular to it representing the normal. An incoming ray (incident ray) hitting the mirror, and an outgoing ray (reflected ray) bouncing off. Label the angles of incidence ($\angle i$) and reflection ($\angle r$). Show arrows on rays. Ensure $\angle i = \angle r$.)
2. Image formation by a concave mirror when the object is placed between F and C.
   Sketch: (Imagine a concave mirror, principal axis, P, F, C. Place an object (arrow) between F and C.
   - Draw a ray parallel to the principal axis from the top of the object, reflecting through F.
   - Draw a ray passing through F from the top of the object, reflecting parallel to the principal axis.
   - The intersection of the reflected rays will form the image (inverted arrow) beyond C. Label Object, Image, F, C, P.)
3. Image formation by a convex mirror when the object is placed anywhere.
   Sketch: (Imagine a convex mirror, principal axis, P, F, C (F and C are behind the mirror). Place an object (arrow) in front of the mirror.
   - Draw a ray parallel to the principal axis from the top of the object, reflecting such that it appears to come from F behind the mirror.
   - Draw a ray directed towards C from the top of the object, which will retrace its path.
   - The intersection of the *extended* reflected rays behind the mirror will form the image (erect, diminished arrow) between P and F. Label Object, Image, F, C, P.)