(i) Convex mirror is used as rear view in vehicles because
(a) It always produces an erect image of the objects.
(b) The image formed in a convex mirror is highly diminished thus it gives a wide field of view.
(ii) Concave mirrors are used as shaving mirrors because 
(a) When the face is held within the focus of a concave mirror, then an enlarged image of the face is seen in the concave mirror. This helps in making a smooth shave.

(a) Image is of same size as the object.
(b) Image is virtual and erect.

Along the normal to the surface or at an incident angle of i = 0^o .

\(m=\frac { Image\quad size }{ Object\quad size } \\ +1=\frac { Image\quad size }{ 2 } =2cm\)
Image size = +2

Since,\(n=\frac { c }{ v } \)
we get, \(n=\frac { c }{ 0.6c } =\frac { 1 }{ 0.6 } \approx 1.67\)

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.

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.

Refractive index of any medium w.r.t. another indicates the extent to which light bends when it enters from first medium to the given medium. The given value of refractive index also states that speed of light in diamond is 1/2.42 times to the speed of light in vacuum.

Concave mirror can give erect and enlarged image of an object.

(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

(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.

(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.