The principal focus of a concave mirror is a point on the principal axis at which a light ray parallel to the principal axis converges after reflection.
Light rays that are parallel to the principal axis of a concave mirror converge at a specific point on its principal axis after reflecting from the mirror. This point is known as the principal focus of the concave mirror.
It is denoted by F.
 

The mirror formula is \(\frac { 1 }{ v } +\frac { 1 }{ u } =\frac { 1 }{ f } \), where the symbols have standard meaning. When u is changed, v changes, but f remains constant. This is because focal length of mirror depends only on radius of curvature of the mirror.

Yes, a convex lens of focal length 20 cm can produce a magnified virtual as well as real image. The statement is correct. For magnified virtual image, the object must be held at a distance less than 20 cm from the lens. For magnified real image, the object must be held at a distance between 20 cm and 40 cm from the lens.

It is a measure of the degree of convergence or divergence of light rays falling on it. It can also be defined as the reciprocal of its focal length in metres.

Convex mirror. This is because, the convex mirror shows wider field of view.

Since, \(n=\frac { c }{ v } \)
We get, \(v=\frac { c }{ v } =\frac { { 3\times 10 }^{ 8 } }{ 1.5 } =2\times { 10 }^{ 8 }m/s.\)

\( \therefore Power=\frac{1}{Focal length}=-\frac{1}{2}=-0.50 D\)
Negative sign arises due to the divergent nature of concave lens.

No bending of light ray occurs i.e., the light rays goes straight from one medium to another.

Concave mirror.
When a solar furnace is placed at the focus of a large concave mirror, it focusses a parallel beam of light on the furnace, as a parallel beam of light on the furnace, as a result a temperature is achieved after some time.

(i) Concave mirror, to get powerful beam of light.
(ii) Convex mirror to get larger field of view and erect image.
(iii) Concave mirror, as it can converge light rays of the sun in a small area.

Given, object size, h_o = 7 cm
Object distance, u = - 27 cm
Focal length, f = - 18 cm
By mirror formula, \(\frac{1}{v}=\frac{1}{f}-\frac{1}{u}=\frac{1}{-18}-\frac{1}{(-27)}=\frac{-1}{54}\)
Screen should be placed at 54 cm in front of the mirror.
Now, \(\frac{h_{i}}{h_{o}}=-\frac{v}{u}\Rightarrow \frac{h_{i}}{7}=-\frac{(-54)}{(-27)}or h_{i}=-14 cm\)
So, image is double the size of object. Also, image is real and inverted (since, h_i is negative).

Given 
(a) Position of candle = 12.0 cm 
Position of convex lens = 50.0 cm 
Position of the screen = 88.0 cm 
Therefore u = 50.0 - 12.0 = 38.0 cm 
v = 88.0-50.0 = 38.0 cm 
Therefore u = v 
Hence both lie at 2F, 
therefore F = 38/2 = 19 cm. 
(b) u = 50.0-31.0 = -19 cm 
f = +19 cm, v = ?
Using \(\frac { 1 }{ f } =\frac { 1 }{ u } +\frac { 1 }{ v } \)
We have \(\frac { 1 }{ 19 } =\frac { -1 }{ 19 } +\frac { 1 }{ v } \)
or v=infinity.
(c) If he further shifts the candle towards the lens the image will become virtual and erect. 
(d) The ray diagram is as shown

Height of the object , h₁ = 5cm
Focal length of the concave lens, f = - 10 cm
Position of the object, v =?
Size of the object, h₂ =?
Object distance, u = -20cm
According to the lens formula:
   \(\frac { 1 }{ v } -\frac { 1 }{ u } =\frac { 1 }{ f } \)
\(\Longrightarrow \frac { 1 }{ v } -\frac { 1 }{ -20 } =\frac { 1 }{ -10 } \)
\(\Longrightarrow \frac { 1 }{ v } +\frac { 1 }{ 20 } =\frac { -1 }{ 10 } \)
\(\Longrightarrow \frac { 1 }{ v } =\frac { -1 }{ 10 } -\frac { 1 }{ 20 } =\frac { -2-1 }{ 20 } =\frac { -3 }{ 20 } \) 
\(\therefore =\frac { -20 }{ 3 } = -6.67\)
\(\therefore \)  Image distance = - 6.67 cm
The negative (-) sign for image distance shows that image is formed on the left side of the concave lens. So the image is virtual.
Magnification, \(m=\frac { v }{ u } ,m=\frac { { h }_{ 2 } }{ { h }_{ 1 } } \)
\(\therefore m=\frac { \frac { -20 }{ 3 } }{ -10 } =\frac { 20 }{ 10\times 3 } =\frac { +2 }{ 3 } =0.67\)  
 \(\frac { { h }_{ 2 } }{ { h }_{ 1 } } =\frac { v }{ u } \)
\(\therefore { h }_{ 2 }=\frac { { h }_{ 1\quad }\times v }{ u } =\frac { 5\times -20 }{ 3\times -20 } =\frac { 5 }{ 3 } =+1.66 \ cm\)
Since h₂ < h₁ therefore image is diminished. The positive (+) sign for the magnification shows the image is erect and virtual.

Object distance u=-30 cm, 
Image distance, v = +60 cm 
(+ve sign is due to the image formed on the screen, hence it is real) 
height of the object, h₁= 2.4 cm, 
height of the image, h₂ = ? 
According to lens formula:
\(\frac { 1 }{ f } =\frac { 1 }{ v } -\frac { 1 }{ u } \)
\(\Rightarrow \frac { 1 }{ f } =\frac { 1 }{ 60 } -\frac { 1 }{ -30 } \)

⇒ f=+20cm
The positive (+ve) sign off shows that the lens is convex having focal length 20cm.
Now Magnification,
Formula:\(m=\frac { { h }_{ 2 } }{ { h }_{ 1 } } \Rightarrow \frac { { h }_{ 2 } }{ 2.4 } =-2\)
∴ h₂=-2x2.4=-4.8cm
The negative (-ve) sign of h₂ shows that the image is inverted.
Thus an inverted, magnified, 4.8 cm long image is formed on the screen.

Nature of image formed by spherical mirror = Real, inverted 
Magnification, m = -2 
Object distance, u=?
f=?
Image distance, v = -30 cm (-ve sign is for real image) 
\(m=\frac { -v }{ u } \)
\(\Rightarrow \frac { -v }{ u } =-2\)
\(\Rightarrow \frac { -30 }{ u } =2\)
2u=-30
∴ u=-15 cm
According to the mirror formula,
\(\frac { 1 }{ f } =\frac { 1 }{ v } +\frac { 1 }{ u } \)
\(\frac { 1 }{ f } =\frac { 1 }{ -30 } +\frac { 1 }{ -15 } \)
\(\Rightarrow \frac { 1 }{ f } =\frac { 1 }{ -30 } -\frac { 1 }{ 15 } =\frac { -1-2 }{ 30 } =\frac { -3 }{ 30 } \)
\(\frac { 1 }{ f } =\frac { -1 }{ 10 } \)
∴ f=-10cm
If the object is shifted 10 cm towards the mirror 
then, u = 15-10-5 cm 
In this case the object is placed between pole and focus of the spherical mirror. Therefore, the image formed is virtual, erect, magnified and behind the mirror.

(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.