Real image is inverted. So, magnification is negative
Thus, m=-3=\(\frac{Image \quad size}{Object \quad size}\)
\(=\frac{h_{i}}{h_{o}}=-\frac{v}{u}\)
[where, v = image distance and u = object distance]
\(\Rightarrow -3=\frac{-(v)}{(-10)}\) [\(\because\) object is placed in front of mirror]
\(\Rightarrow -3=\frac{v}{10}\)
Image distance, v = -30 cm
Negative sign shows that image is real, so it will be formed in front of the mirror.

(a) (i) Optical centre: It is a point within the lens that lies on the principal axis through which a ray of light passes undeflected. 
(ii) Centre of curvature: The centre of curvature of the surface of a lens is the centre of the sphere of which it forms a part. A lens has two centres of curvature because it has two surfaces. 
(iii) Principal axis: It is a line through the centres of curvatures of the lens. 
(iv) Aperture: The diameter of the circular boundary of the lens is called the aperture of the lens. 
(v) Principal focus: A beam of light parallel to the principal axis either converges to a point or appears to diverge from a point on the principal axis after refraction through the lens, is called the principal focus. All lenses have two principal focuses. 
(vi) Focal length: The distance between the optical centre and the principal focus of the lens is called its focal length. 
(b)  A converging lens is a convex lens.
f = + 12 cm 
u=?
v = +48 cm (+ve as it is formed on other side of the object) 
According to the lens formula,
\(\frac { 1 }{ f } =\frac { 1 }{ v } -\frac { 1 }{ u } \)
\(\Rightarrow \frac { 1 }{ 12 } =\frac { 1 }{ 48 } -\frac { 1 }{ u } \)
\(\frac { 1 }{ u } =\frac { 1 }{ 48 } -\frac { 1 }{ 12 } =\frac { 1-4 }{ 48 } =\frac { -3 }{ 48 } =\frac { 1 }{ 16 } \)
∴ u=-16cm
∴ The object should be placed 16 cm away from the lens.

(a) Sign conventions
1. The object is always placed to the left of the mirror.
2. All the distances parallel to the principal axis are always measured from the pole of the spherical mirror.
3. All the distances measured along the direction of incident light (along +ve x-axis), are considered to be positive.
4. Those distances measured opposite to the direction of incidence light (i.e. along -ve x-axis), are taken as negative.
5. The distances measured in upward direction, i.e. perpendicular to and above the principal axis (along +ve y-axis), are taken as positive.
6. The distances measured in the downward direction, (along -ve y-axis), i.e. perpendicular to and below the principal axis are taken as negative.
(b) u=-16cm, m=-3 for real But \(m=-\frac { v }{ u } =-3\)
v = 3u = 3 (-16) = -48 cm. 
Using mirror formula
\(\frac { 1 }{ f } =\frac { 1 }{ v } +\frac { 1 }{ u } \)
We get, \(\frac { 1 }{ f } =\frac { 1 }{ -48 } +\frac { 1 }{ -16 } \)
\(=\frac { 1 }{ -48 } -\frac { 1 }{ 16 } =\frac { -1-3 }{ 48 } =\frac { -4 }{ 48 } =\frac { -1 }{ 12 } \)
f=-12cm
(c) Negative sign of focal length indicated that mirror is concave in nature.

(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) 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

(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.

The following are the laws of refraction of light.
(i)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.
(ii)The ratio of sine of angle of incidence to the sine of angle of refraction is a constant, for the light of a given pair of media. This law is also known as Snell's law of refraction. The ray diagram is as shown. As seen in the refracted ray are in the same plane.

Since the focal length = 15 cm, the range of object distance - 0cm to 15 cm
A concave mirror gives an erect image when an object is placed between its pole (P) and the principal focus (F). Hence, to obtain an erect image of an object from a concave mirror of focal length 15 cm, the object must be placed anywhere between the pole and the focus. The image formed will be virtual, erect, and magnified in nature, as shown in the given figure.

(a) Used in torches, search lights and vehicle headlights.
(b) Used as shaving mirror.
(c) Used by dentist.
(d) Used in solar furnance

A ray of light passing through the optical centre of the convex lens will continue to move along the same direction after refracting through the lens.

Real

The negative sign of magnification shows that the image is real and inverted. Hence, the object must be positioned between F and 2F. 

Image is real

The rough focal length of a convex lens is obtained by forming sharp image of a very distant object on a screen. The distance of the screen from the lens gives us the rough focal length of the lens.
This method is not applicable to a concave lens, as image formed by a concave lens is virtual and it cannot be taken on the screen.

Given, power of lens, P = + 1.5 D
It means the lens is convex
As power. \(P=\frac{1}{f(in \quad m)}\)
So, \(f=\frac{1}{p}=\frac{1}{1.5}=\frac{10}{15}=0.66 \quad m\)
\(\Rightarrow\) f = 0.66 \(\times\) 100 = 66 cm
[It is a converging lens because its focal length is positive.]