(b) CB=-uCF = -fCB=-v 
The relation between u, v and f is given by the lens formula:
\(\frac { 1 }{ f } =\frac { 1 }{ v } -\frac { 1 }{ u } \)
As both u and v are negative the above equation will change to
\(\frac { 1 }{ f } =\frac { 1 }{ (-v) } -\frac { 1 }{ (-u) } \)
\(\frac { 1 }{ f } =\frac { -1 }{ v } +\frac { 1 }{ u } \)
\(\Rightarrow \frac { 1 }{ f } =\frac { 1 }{ u } -\frac { 1 }{ v } \)
We know that the focal length of a concave lens is negative, so the above equation will be changed to
\(\frac { 1 }{ -f } =\frac { 1 }{ u } -\frac { 1 }{ v } \)
\(\Rightarrow \frac { 1 }{ f } =\frac { 1 }{ v } -\frac { 1 }{ u } \)
(c) Magnification m=-1 
v = + 40 cm (real and inverted) 
Nature of the lens = ? 
Power of the lens, P =?
\(\frac { v }{ u } =m\Rightarrow \frac { +40 }{ u } =-1\)
40=-u ⇒ u = -40 cm 
According to the lens formula,
\(\frac { 1 }{ f } =\frac { 1 }{ v } -\frac { 1 }{ u } \)
\(\Rightarrow \frac { 1 }{ f } =\frac { 1 }{ 40 } -\frac { 1 }{ -40 } =\frac { 1 }{ 40 } +\frac { 1 }{ 40 } =\frac { 2 }{ 40 } =\frac { 1 }{ 20 } \)
f = +20 cm 
fis +ve thus the lens is convex
\(P=\frac { 1 }{ f(metres) } =\frac { 1\times 100 }{ 20 } =+5D\)
Since power of lens is positive, lens will be converging in nature.

(i)  A convex lens can form a magnified erect image when the object is placed between the optical centre and principal focus of the convex lens (i.e. between O and F).

(ii) A convex lens can form a magnified erect image when the object is placed between focus and the centre of curvature (i.e. between F' and 2F).

Object height, h = 4 cm; 
Object distance, u = -20 cm 
Nature of the lens = concave lens; 
Focal length = f = +10 cm 
Image distance, v = ? 
According to the lens formula,
\(\frac { 1 }{ f } =\frac { 1 }{ v } -\frac { 1 }{ u } \)
\(\Rightarrow \frac { 1 }{ v } =\frac { 1 }{ f } +\frac { 1 }{ u } \)
\(\frac { 1 }{ v } =\frac { 1 }{ +10 } +\frac { 1 }{ -20 } =\frac { 1 }{ 10 } -\frac { 1 }{ 20 } =\frac { 2-1 }{ 20 } =\frac { 1 }{ 20 } \)
∴ v=20cm
The image is formed at a distance of 20 cm from the lens on the opposite side.

(a) A convex lens should be used. This is because it can produce a real image of the candle flame on the wall as it is a converging lens where refracted rays actually meet. 
(b)&(c)
(i) A real magnified image is formed when the candle flame is placed between F and 2F from the convex lens on the other side of the wall.

(ii) A real diminished image is formed when the candle flame is placed beyond 2F from the convex lens on the other side of the wall.

(a) Laws of Refraction: 
(i) The first law of refraction of light states that the incident ray, the refracted ray and the normal at the point of incidence, all lie in the same plane. 
(ii) The second law of refraction of light is the Snell's Law of Refraction. It state that the ratio of sine of the angle of incidence to the sine of angle of refraction is a constant for a given pair of medium.
\(\frac { sin\quad i }{ sin\quad r } =Constant(n)\)
This constant (n) is called refractive index of the medium. 
(i) When the light is going from vacuum to another medium, then the value of refractive index is called the absolute refractive index. 
(ii) The ratio of speed of light in vacuum to the speed of light in a medium is called the absolute refractive index of that medium, 
i.e. 
Absolute refractive index (of a medium)
\(=\frac { Speed\quad of\quad light\quad in\quad vacuum(c) }{ Speed\quad of\quad light\quad in\quad medium(v) } \)
(b) \(_{ a }{ { n }_{ w } }=\frac { 4 }{ 3 } ,_{ a }{ { n }_{ g } }=\frac { 3 }{ 2 } \)
Speed of light in glass, v_g= 2 x 10⁸ m/s 
Speed of light in air, v_g= ? 
Speed of light in water, v_w =?
\(_{ a }{ { n }_{ w } }=\frac { { v }_{ a } }{ { v }_{ w } } \)
\(\Rightarrow \frac { 4 }{ 3 } =\frac { { v }_{ a } }{ { v }_{ w } } \)      ...(i)
\(_{ a }{ { n }_{ g } }=\frac { { v }_{ a } }{ { v }_{ g } } \)
\(\Rightarrow \frac { 3 }{ 2 } =\frac { { v }_{ a } }{ 2\times { 10 }^{ 8 } } \)

Putting the value of v_a in equation (i),
\(\frac { 4 }{ 2 } =\frac { 3\times { 10 }^{ 8 } }{ { v }_{ w } } \)
\(\Rightarrow { v }_{ w }=3\times { 10 }^{ 8 }\times \frac { 3 }{ 4 } =\frac { 9 }{ 4 } \times { 10 }^{ 8 }\)
=2.25x10⁸ m/s.

(a) As we can see in the figure given, when the lower half of the convex lens is covered with a black paper, it still forms lens. However the intensity of the image is reduced when the convex lens is covered with black paper.

(b) (i) Converging lens (Convex lens): 
Height of the object, h₁= 5 cm 
Object distance, u = -25 cm 
Focal length, f= +10 cm

(ii) Image distance, v ? 
Image size, h₂ = ? 
According to lens formula:
\(\frac { 1 }{ v } -\frac { 1 }{ u } =\frac { 1 }{ f } \)
\(\Rightarrow \frac { 1 }{ v } -\frac { 1 }{ -25 } =\frac { 1 }{ 10 } \)
\(\Rightarrow \frac { 1 }{ v } +\frac { 1 }{ 25 } =\frac { 1 }{ 10 } \)
\(\frac { 1 }{ v } =\frac { 1 }{ 10 } -\frac { 1 }{ 25 } \)
\(\Rightarrow \frac { 1 }{ v } =\frac { 5-2 }{ 50 } =\frac { 3 }{ 50 } \)
\(\Rightarrow v=\frac { 50 }{ 3 } =+16.6cm\)
\(\frac { { h }_{ 2 } }{ { h }_{ 1 } } =\frac { v }{ u } \)
\(\Rightarrow \frac { { h }_{ 2 } }{ 5 } =\frac { 50 }{ 3\times -25 } \)

(iii) Nature of the image 
(a) The +ve of v shows that image is real. 
(b) The -ve sign of h₂ shows that image is inverted. 
∴ Size of image = 3.3 cm

(a) Optical centre of the lens. It is a point within the lens that lies on the principal axis through which away of light passes undeflected. 
= f=-20cm  h₁ = 4 cm 
v = -10 cm u=?
h₂=?
\(\frac { 1 }{ f } =\frac { 1 }{ v } -\frac { 1 }{ u } \)
\(\Rightarrow \frac { 1 }{ -20 } -\frac { 1 }{ 10 } -\frac { 1 }{ u } \)
\(\Rightarrow \frac { 1 }{ u } =\frac { 1 }{ 10 } +\frac { 1 }{ 20 } \)
\(\Rightarrow \frac { 1 }{ u } -\frac { 1 }{ 10 } +\frac { 1 }{ 20 } \)
\(\Rightarrow \frac { 1 }{ u } =\frac { -2+1 }{ 20 } \)
\(\Rightarrow \frac { 1 }{ u } =\frac { -1 }{ 20 } \)
Now,
\(\frac { { h }_{ 2 } }{ { h }_{ 1 } } =\frac { v }{ u } \)
\(\Rightarrow \frac { { h }_{ 2 } }{ 4 } =\frac { -10 }{ -20 } \)
\(\Rightarrow { h }_{ 2 }=\frac { 10 }{ 20 } \times 4=2cm\)
h₂=2cm

(a) The distance between optical centre and focus of the lens is called focal length of a spherical lens. 
(b) Diverging lens: Concave lens Focal -30 cm
Object distance, u =? 
Height of the object, h₁ = 5 cm 
Image distance, v = - 15 cm 
Image size, h₂ = ? 
According to the lens formula,
\(\frac { 1 }{ f } =\frac { 1 }{ v } +\frac { 1 }{ u } \)
\(\frac { 1 }{ -30 } =\frac { -1 }{ -15 } +\frac { 1 }{ u } \)
\(\Rightarrow \frac { 1 }{ u } =\frac { -1 }{ 15 } +\frac { 1 }{ 30 } =\frac { -2+1 }{ 30 } =\frac { -1 }{ 30 } \)
u=-30cm
\(\frac { { h }_{ 2 } }{ { h }_{ 1 } } =\frac { -v }{ u } \)
\(\Rightarrow \frac { { h }_{ 2 } }{ 5 } =\frac { -15 }{ -30 } \)
\(\Rightarrow { h }_{ 2 }=\frac { 5 }{ 2 } =2.5cm\)

(i) Yes.If a convex lens of focal length 10cm is covered one half with a black paper, it can produce an image of the complete object between F₂ and 2F₂.The rays of light coming from the object get refracted by the upper half of the lens. The image formed will be real, inverted and diminished.

(ii)Object height, h₁=4cm
Focal length, f=+20cm
Object distance, u=-15cm
Image distance, v=?
Image height, h₂=?
By lens formula,
\({1\over f}={1\over v}-{1\over u}\)
\(\Rightarrow\ {1\over v}={1\over f}+{1\over u}={1\over +20}+{1\over -15}={1\over 20}-{1\over 15}\)
\(\Rightarrow\ {1\over V}={3-4\over 60}={-1\over 60}\)
v=-60cm
Negative sign of v shows that the image is virtual.

Given m = -3, v = 80 cm, u = ?
Using the expression \(m=\frac { -v }{ u } \)
we have \(-3=\frac { 80 }{ u } \)  or \(u=\frac { -80 }{ 3 } =-26.67cm\)
The image is real and the lens is a convex lens.

Given m = 1/3, f = 20 cm, u = ?
Using the expression m = -v/u
We have 1/3 = -v/u or u = -3 V
Now using the expression
\(\frac{1}{f}=\frac{1}{u}+\frac{1}{v}\)
We have 
\(\frac{1}{20}=\frac{1}{u}+\frac{-3}{u}=\frac{-2}{u}\)
or u = -40 cm