(i) Reflecting telescope
(ii) No chromatic aberration, mirrors are relatively lighter and cheaper compared to the lens.
(iii) First hand experience, inculcating a scientific temper, team work, enthusiasm, finding curiosity.

(a) The light rays from the two (coherent) slits, reaching a point 'P' on the screen, have a path
difference (S₂P - S₁P). The point 'P' would, therefore be a 
(i) Point of maxima (bright fringe), If
S₂P - S1P = n\(\lambda\) .
(ii) Point of minima (dark fringle), If
S₂P- S₁P = (2n + 1) \(\lambda\)/2

We have (S₂P)²- (S₁P)²
\(=\left\{D^2-\left(x+{d\over2}\right)^2\right\}-\left\{D^2+\left(x-{d\over2}\right)^2\right\}\)
= 2 xd
\(S_2P-S_1P={2xd\over S_3P+S_1P}={2xd\over 2D}={xd\over D}\)
We have maxima at point, where
\({xd\over D}=n\lambda\)
and minima at points where
\({xd\over D}=\left(2n+1\over 2\right)\lambda\)
Now, fringe width \(\beta\)  = separation between two successive maxima (or two successaive minima)
= x_n - x_n-1
\(\beta={\lambda D\over d}\)
(b) We have \({I_{max}\over {I_{min}}}={(a_1+a_2)^2\over (a_1-a_2)^2}={25\over 9}\)
\(\therefore\ \ {a_1+a_2\over a_1-a_2}={5\over 3}\Rightarrow{a_1\over a_2}={4\over 1}\)
\(\therefore\ {W_1\over W_2}={I_1\over I_2}={(a_1)^2\over (a_2)^2}={16\over 1}\)  

(i)

From figure
(S₂P)²-(S₁P²) =
\(\left[D_2+\left(x+{d\over 2}\right)^2\right]-\left[D_2+\left(x-{d\over 2}^2\right)\right]\)
= 2xd
\(\Rightarrow\ S_2P-S_1P={2xd\over (S_2P+S_1P)}\)
Since S₂P \(\approx \) S₁P\(\approx \)D
For constructive interference,
S₂P- S₁P = n\(\lambda_1\), n = 0, 1, 2, 3
\(\Rightarrow{xd\over D}=n\lambda\)
\(x={n\lambda D\over d}\)
Position of the nth bright fringe on screen:
\(={n\lambda D\over d}\)
And position of the (n + 1)^th bright fringe on screen
\(={(n+1)\lambda D\over d}\)
Fringe Width
\(={(n+1)\lambda D\over d}-{n\lambda D\over d}\)
\(={\lambda D\over d}\)
(ii) On shifting principal source point '5' little upwards i.e., towards SI, the position of the central maximum on the screen will shift downwards on the screen, i.e., below its previous position, Hence whole interference pattern will get shifted little downwards but Fringe width will remain same as that of the initial arrangement.

(i) Here,
\(\lambda =600 \ nm=600\times { 10 }^{ -9 }m=6\times { 10 }^{ -7 }m\)
\(\theta ={ 0.1 }^{ \circ }=\frac { 0.1\pi }{ 180 } rad,d=?\)
From angular width,\(\theta =\frac { \lambda }{ d } \)
\(\Rightarrow \ d=\frac { \lambda }{ \theta } =\frac { 6\times { 10 }^{ -7 } }{ \frac { \pi }{ 180 } \times 0.1 } =3.44\times { 10 }^{ -4 }m\)
(ii) The frequency and wavelength of reflected wave will not change. The refracted wave will have same frequency. The velocity of light in water is given by \(v=f\lambda \)
where, v = velocity of light
f = frequency of light
​​​​​​​​​​​​\( ​​lambda =wavelength \ of \ light\)
If velocity will decrease, then wavelength (\(\lambda \)) will also decrease.

(i) Yes, plane and convex mirrors can produce the real image, if the rays incident on the plane or convex mirror are converging to a point behind the mirror. A plane or convex mirror can produce a real image, if object is virtual.
(ii) No, there is no contradiction because virtual image formed by the spherical mirror acts as virtual object for eye lens. Our eye lens is convergent and it forms a real image of virtual object on retina.
(iii) As, the fisherman is in air, the rays of light travels from rarer to denser medium, so they bend towards the normal. Therefore, the fisherman appears taller.

(iv) Yes, the apparent depth decreases, further when water tank is viewed obliquely as compared to the depth when seen normally.

(i) The values we observe in Kanchan are as follows:
(a) Concern about others
(b) Observation of traffic rules
(c) Resposible citizen
(ii) Convex mirror is used in rear viwe in scooties. The ray diagram for convex mirror is

Here, distance of the screen from the slit, D = 2m,
a = ?, x = 5 mm = \(5\times { 10 }^{ -3 }m\),
\(\lambda =5000\mathring { A } =5000\times { 10 }^{ -10 }m\)
\(For \ the \ first \ secondary \ minima,\)
\( sin\theta =\frac { \lambda }{ a } =\frac { x }{ D } \)
\(a=\frac { D\lambda }{ x } =\frac { 2\times 5000\times { 10 }^{ -10 } }{ 5\times { 10 }^{ -3 } } =2\times { 10 }^{ -4 }m\)

Here m =-50,  L = 102cm
As \(m=-{f_0\over f_e}=-50, f_0=50f_e\)
Also, L = f₀ + f_e= 102;
50f₀ + f_e= 102, f_e = 2 cm
f₀ = 50 f_e= 50 x 2= 100 cm
\(P_o={100\over f_0}={100\over100}\)= 1D,
\(P_e={100\over f_e}={100\over2}\)= 50D

m = -100, f₀ + f_e = 101 cm, f₀ = ?, f_e = ?
\(m=-{f_0\over f_e}=-100\ \ \therefore f_0=100 f_3\)
Now f₀ + f_e = 101
100 f_e + f_e = 101,
f_e = 1 cm,
f₀ = 100 f_e = 100 cm