PHYSICAL SCIENCE CONTENT - CLASS x

 CHAPTER 1: REFLECTION OF LIGHT AT CURVED SURFACES

I. Reflections on concepts

1. Question:
Where will the image be formed when an object is placed between the focus (F) and centre of curvature (C) of a concave mirror?

Answer:
The image is formed beyond C, and it is real, inverted, and enlarged.

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2. Question:
State the differences between convex and concave mirrors.

Answer:

• Concave mirror: Converges light rays, forms real or virtual images.
• Convex mirror: Diverges light rays, forms only virtual, erect, and diminished images.

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3. Question:
Distinguish between real and virtual images.

Answer:

• Real image: Formed by actual intersection of rays, can be obtained on a screen, usually inverted.
• Virtual image: Formed by apparent intersection, cannot be obtained on a screen, always erect.

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4. Question:
How do you get a virtual image using a concave mirror?

Answer:
Place the object between the pole (P) and focus (F). The image formed is virtual, erect, and enlarged behind the mirror.

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5. Question:
What do you know about the following terms related to spherical mirrors?

Answer:

• Pole (P): Centre of the mirror surface
• Centre of curvature (C): Centre of the sphere of which mirror is a part
• Focus (F): Point where parallel rays meet or appear to meet
• Radius of curvature (R): Distance between P and C
• Focal length (f): Distance between P and F (f = R/2)
• Principal axis: Straight line passing through P and C
• Object distance (u): Distance of object from pole
• Image distance (v): Distance of image from pole
• Magnification (m): Ratio of image height to object height

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6. Question:
What do you infer from the experiment used to measure object distance and image distance?

Answer:
There is a relation between u, v, and f given by the mirror formula:

$\frac{1}{f} = \frac{1}{v} + \frac{1}{u}$

II. Application of concepts

1. Question:
Find the distance of the image when an object is placed on the principal axis at a distance of 10 cm in front of a concave mirror whose radius of curvature is 8 cm.

Answer:

Given:
Object distance (u) = −10 cm
Radius of curvature (R) = 8 cm → f = R/2 = −4 cm

Using mirror formula:
1/f = 1/v + 1/u

1/(-4) = 1/v + 1/(-10)

-1/4 = 1/v - 1/10

1/v = -1/4 + 1/10
1/v = (-5 + 2)/20 = -3/20

v = −20/3 ≈ −6.67 cm

👉 Image is formed 6.67 cm in front of the mirror (real image).

Image distance v = −6.67 cm, so the image is formed in front of the mirror (real image).

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2. Question:
The magnification produced by a mirror is +1. What does it mean?

Answer:
The image is same size as the object, erect, and virtual.

3. Question:
If the spherical mirrors were not known to human beings, given the consequences?

Answer:
We would lack many useful applications such as rear-view mirrors in vehicles, shaving/makeup mirrors, solar cookers, and reflecting telescopes. Daily life and scientific progress would be affected.

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4. Question:
Draw suitable rays by which we can guess the position of image formed by a concave mirror.

Answer:
Use two standard rays:

• A ray parallel to the principal axis reflects through the focus (F).
• A ray passing through the centre of curvature (C) reflects back along the same path.
  The intersection of reflected rays gives the image position.

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5. Question:
Show the formation of image with a ray diagram when an object is placed on the principal axis of a concave mirror away from the centre of curvature.

Answer:
When the object is beyond C, the image forms between F and C, and it is real, inverted, and diminished.

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6. Question:
Why do people prefer a convex mirror as a rear-view mirror in vehicles?

Answer:
Because it provides a wide field of view and forms erect, diminished, virtual images, helping drivers see more area behind them.

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III. Higher Order Thinking Questions

1. Question:
A convex mirror with a radius of curvature of 3 m is used as a rear-view mirror. If a bus is located 5 m from this mirror, find the position, nature, and size of the image.

Answer:
Given:
R = 3 m → f = +1.5 m
u = −5 m

Using mirror formula:
1/f = 1/v + 1/u

1/1.5 = 1/v − 1/5

1/v = 1/1.5 + 1/5 = 2/3 + 1/5 = 13/15

v = 15/13 ≈ +1.15 m

• Position: Behind the mirror
• Nature: Virtual and erect
• Size: Diminished

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2. Question:
To form the image on the object itself, how should we place the object in front of a concave mirror? Explain with a ray diagram.

Answer:
Place the object at the centre of curvature (C).
The image is formed at the same point (C), real, inverted, and same size.

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Multiple Choice Questions

1. Question:
If an object is placed at C on the principal axis in front of a concave mirror, the position of image is:
a) at infinity
b) between Focus and Centre of curvature
c) at Centre of curvature
d) beyond Centre of curvature

Answer:
c) at Centre of curvature

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2. Question:
We get a diminished image with a concave mirror when the object is placed:
a) at Focus
b) between pole and Focus
c) at Centre of curvature
d) beyond Centre of curvature

Answer:
d) beyond Centre of curvature

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3. Question:
We get a virtual image in a concave mirror when the object is placed:
a) at Focus
b) between the pole and Focus
c) at Centre of curvature
d) beyond Centre of curvature

Answer:
b) between the pole and Focus

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4. Question:
Which of the following represents magnification (m) for a spherical mirror?

Answer:
Correct expressions:

• $𝑚=\frac{{ℎ}_{\mathrm{𝑖/}}}{{ℎ}_{\mathrm{𝑜 }}}$
• $\frac{{\mathrm{m=-v/u}}_{}}{}$

Correct option: b) (ii) & (iii)

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5. Question:
Ray which seems to be travelling through the focus of a convex mirror, after reflection of an incident ray:
a) parallel to the axis
b) along the same path in opposite direction
c) through Focus
d) through Centre of curvature

Answer:
a) parallel to the axis