Chapter 9: Light – Reflection and Refraction
Exercise & In-Text Questions & Answers

Answer & Reference:
The principal focus of a concave mirror is the point on its principal axis where all rays of light parallel to the principal axis converge after reflection. (Page 136, para 2)

Answer & Reference:
For spherical mirrors, focal length \( f = \frac{R}{2} \).
Given \( R = 20 \, \text{cm} \), so \( f = \frac{20}{2} = 10 \, \text{cm} \). (Page 137, para 3)

Answer & Reference:
A concave mirror gives an erect and enlarged image when the object is placed between the pole (P) and the principal focus (F). (Page 138, Table 9.1)

Answer & Reference:
Convex mirrors always give an erect, diminished image and have a wider field of view, enabling the driver to see more traffic behind. (Page 142, para 2)

Answer & Reference:
\( f = \frac{R}{2} = \frac{32}{2} = 16 \, \text{cm} \). For a convex mirror, focal length is positive. (Page 137)

Answer & Reference:
Given: \( u = -10 \, \text{cm} \), magnification \( m = -3 \) (real image ⇒ negative).
Using \( m = -\frac{v}{u} \Rightarrow -3 = -\frac{v}{-10} \Rightarrow v = -30 \, \text{cm} \).
The image is located 30 cm in front of the mirror. (Page 143, Eq. 9.3)

Answer & Reference:
It bends towards the normal because water is optically denser than air, and light slows down in denser media. (Page 149, para 2)

Answer & Reference:
Refractive index \( n = \frac{c}{v} \Rightarrow v = \frac{c}{n} = \frac{3 \times 10^8}{1.50} = 2 \times 10^8 \, \text{m/s} \). (Page 148, Eq. 9.7)

Answer & Reference:
Highest optical density: Diamond (refractive index = 2.42).
Lowest optical density: Air (refractive index = 1.0003). (Page 149, Table 9.3)

Answer & Reference:
Speed of light is highest in the medium with lowest refractive index.
Refractive indices: Water = 1.33, Kerosene = 1.44, Turpentine = 1.47.
So, light travels fastest in water. (Page 149, Table 9.3)

Answer & Reference:
It means that the ratio of the speed of light in vacuum to the speed of light in diamond is 2.42. In other words, light travels 2.42 times slower in diamond than in vacuum. (Page 148–149)

Answer & Reference:
1 dioptre is the power of a lens whose focal length is 1 metre. Power \( P = \frac{1}{f} \) (with \( f \) in metres). (Page 158, para 2)

Answer & Reference:
For a convex lens, image = object size ⇒ object at 2F, image at 2F on other side.
So, \( v = +50 \, \text{cm} \Rightarrow 2f = 50 \Rightarrow f = 25 \, \text{cm} = 0.25 \, \text{m} \).
Object distance \( u = -50 \, \text{cm} \).
Power \( P = \frac{1}{f} = \frac{1}{0.25} = +4 \, \text{D} \). (Page 152, Table 9.4; Page 158)

Answer & Reference:
Focal length \( f = -2 \, \text{m} \) (concave ⇒ negative).
Power \( P = \frac{1}{f} = \frac{1}{-2} = -0.5 \, \text{D} \). (Page 158)

Answer & Reference:
(d) Clay — because it is not transparent. Lenses must be made of transparent material to refract light. (Page 150)

Answer & Reference:
(d) Between the pole of the mirror and its principal focus. (Page 138, Table 9.1)

Answer & Reference:
(b) At twice the focal length. (Page 152, Table 9.4)

Answer & Reference:
(a) both concave. — Negative focal length indicates concave mirror and concave (diverging) lens. (Page 142, 155)

Answer & Reference:
(d) either plane or convex. — Both always produce erect images. (Page 135, 141)

Answer & Reference:
(c) A convex lens of focal length 5 cm. — Shorter focal length gives higher magnification for reading. (Page 150, Activity 9.11)

Answer & Reference:
Object must be placed between pole (P) and focus (F), i.e., less than 15 cm from mirror.
Image is virtual, erect, and enlarged. (Page 138, Table 9.1)

Answer & Reference:
(a) Concave mirror — produces powerful parallel beam when source is at focus. (Page 140)
(b) Convex mirror — wide field of view, erect image. (Page 142)
(c) Concave mirror — concentrates sunlight at focus to produce heat. (Page 140)

Answer & Reference:
Yes, a complete image will be formed, but with reduced brightness, because light from all parts of the object still reaches the uncovered half. (Implied from ray diagram principles, Page 153–154)

Answer & Reference:
Given: \( h = 5 \, \text{cm}, u = -25 \, \text{cm}, f = +10 \, \text{cm} \)
Using lens formula: \( \frac{1}{v} - \frac{1}{u} = \frac{1}{f} \Rightarrow \frac{1}{v} = \frac{1}{10} + \frac{1}{-25} = \frac{5 - 2}{50} = \frac{3}{50} \Rightarrow v = +16.67 \, \text{cm} \)
Magnification \( m = \frac{v}{u} = \frac{16.67}{-25} = -0.67 \)
Image height \( h' = m \cdot h = -3.33 \, \text{cm} \)
Image is real, inverted, diminished, located 16.7 cm on opposite side. (Page 156, Example 9.4)

Answer & Reference:
Given: \( f = -15 \, \text{cm}, v = -10 \, \text{cm} \)
Lens formula: \( \frac{1}{v} - \frac{1}{u} = \frac{1}{f} \Rightarrow \frac{1}{u} = \frac{1}{v} - \frac{1}{f} = \frac{-1}{10} - \frac{-1}{15} = \frac{-3 + 2}{30} = \frac{-1}{30} \)
So, \( u = -30 \, \text{cm} \). Object is 30 cm from lens. (Page 156, Example 9.3)

Answer & Reference:
\( u = -10 \, \text{cm}, f = +15 \, \text{cm} \)
Mirror formula: \( \frac{1}{v} + \frac{1}{u} = \frac{1}{f} \Rightarrow \frac{1}{v} = \frac{1}{15} - \frac{1}{-10} = \frac{1}{15} + \frac{1}{10} = \frac{5}{30} = \frac{1}{6} \)
\( v = +6 \, \text{cm} \). Image is virtual, erect, diminished, 6 cm behind mirror. (Page 141, Table 9.2)

Answer & Reference:
It means the image is virtual, erect, and of the same size as the object. (Page 135)

Answer & Reference:
\( R = +30 \, \text{cm} \Rightarrow f = +15 \, \text{cm}, u = -20 \, \text{cm}, h = 5 \, \text{cm} \)
\( \frac{1}{v} = \frac{1}{f} - \frac{1}{u} = \frac{1}{15} + \frac{1}{20} = \frac{7}{60} \Rightarrow v = +8.57 \, \text{cm} \)
\( m = -\frac{v}{u} = -\frac{8.57}{-20} = +0.43 \Rightarrow h' = 0.43 \times 5 = 2.15 \, \text{cm} \)
Image is virtual, erect, diminished. (Page 141, Table 9.2)

Answer & Reference:
\( u = -27 \, \text{cm}, f = -18 \, \text{cm}, h = 7 \, \text{cm} \)
\( \frac{1}{v} = \frac{1}{f} - \frac{1}{u} = -\frac{1}{18} + \frac{1}{27} = \frac{-3 + 2}{54} = -\frac{1}{54} \Rightarrow v = -54 \, \text{cm} \)
\( m = -\frac{v}{u} = -\frac{-54}{-27} = -2 \Rightarrow h' = -14 \, \text{cm} \)
Screen at 54 cm in front; image is real, inverted, enlarged. (Page 138, Table 9.1)

Answer & Reference:
\( P = -2.0 \, \text{D} \Rightarrow f = \frac{1}{P} = -0.5 \, \text{m} = -50 \, \text{cm} \).
It is a concave (diverging) lens. (Page 158)

Answer & Reference:
\( f = \frac{1}{1.5} = +0.67 \, \text{m} \).
Positive power ⇒ converging (convex) lens. (Page 158)

Position of Object	Position of Image	Size	Nature
At infinity	At 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	Image not formed	Image not formed
Between P and F	Behind mirror	Enlarged	Virtual and erect

(Page 138)

Position of Object	Position of Image	Size	Nature
At infinity	At F, behind mirror	Highly diminished, point-sized	Virtual and erect
Between ∞ and P	Between P and F, behind mirror	Diminished	Virtual and erect

(Page 141)

Object Position	Image Position	Size	Nature
At infinity	At F₂	Highly diminished	Real, inverted
Beyond 2F₁	Between F₂ and 2F₂	Diminished	Real, inverted
At 2F₁	At 2F₂	Same size	Real, inverted
Between F₁ and 2F₁	Beyond 2F₂	Enlarged	Real, inverted
At F₁	At infinity	Not formed	Not formed
Between F₁ and O	Same side as object	Enlarged	Virtual, erect

(Page 152)

Object Position	Image Position	Size	Nature
At infinity	At F₁	Highly diminished	Virtual, erect
Between ∞ and O	Between F₁ and O	Diminished	Virtual, erect

(Page 153)