The phenomenon observed is the phenomenon of diffraction
(i) At the cental maxima The contributions due to the secondary wavelets, frotn all parts of the wave front (at the slit), arrive in phase at the central maxima \(\theta\) = a
At the secondary maxima
It is only the contributions from (nearly) 1/3 (or 1/5, or 117,...) of the secondary maxima. These occur at points for which
\(\theta\cong \left(n+{1\over2}\right){\lambda\over a}(n=0,1,2,3,...)\)
At the minima
The contribution, from 'corresponding pairs', of the sub-parts of the incident wavefront, cancel each other and the net contribution, at the location of the minima, is zero. The minima occur at points for which \(\theta=n{\lambda\over a}(n=1,2,3,...)\)
(ii) There is a significant fall in intensity at the secondary maxima because the intensity there, is only due to the contribution of (nearly)(1/3 or 1/5 or 1/7,......) of the incident wavefronts.
(iii) The size of the central maxima would get halved when width of the slit is doubled.

(i) Here, the direction of magnetic field is perpendicularly inwards to the plane of paper. If a wire of irregular shape turns into a circular shape, then its area increases (\(\because \) the circular loop has greater area than the loop of irregular shape) so that, the magnetic flux linked also increases. Now, the induced current is produced in a direction such that it decreases the magnetic field [i.e. the current will flow in such a direction, so that the wire forming the loop is pulled inward in all directions (to decrease the area)], i.e current is in anti-clockwise direction, ie. along adcba.  
(ii) When a circular loop deforms into a narrow straight wire, the magnetic flux linked with it also decreases. The current induced due to change in flux will flow in such a direction that it will oppose the decrease in magnetic flux, so it will flow anti-clockwise, i.e along adcba due to which the magnetic field produced will be out of the plane of paper. 

Let d be the diameter of the disc. The sopt shall be invisible, if the incident rays OA and OB suffer total internal reflection.
Let i be the angle of incidence

Using relationship between refractive index and critical angle, then
\(\sin { i=\frac { 1 }{ \mu } }\)
Using geometry and trigonometry,
Now,
\(\frac { d/2 }{ h } =\tan { i } \Rightarrow \frac { d }{ 2 } =h\tan { i } =h[\sqrt { { \mu }^{ 2 }-1{ ] }^{ -1 } }\)
\( [\because \ From \ the \ figure,\ tani=\frac { 1 }{ \sqrt { { \mu }^{ 2 } } -1 } ]\)
\( d=\frac { 2h }{ \sqrt { { \mu }^{ 2 } } -1 } \)
This is required expression of d.

Since the North pole of the bar magnet is approachng the loop, therefore the induced current wall flow in such a way that when loop viewed from left side. it will behave like a north pole and when viewed from right side, it will behave like a South pole. So, the flow of induced current in the loop will be clockwise. Hence, A acquires positive polarity and B acquires negative polarity.

(i) For statement of Faraday's law of electromagnetic induction
(ii) v = 1800 km/h = \(1800\times \frac { 5 }{ 18 } \)  m/s = 500 m/s, e = ?, l = 25m,
R = 5.0 \(\times\)10^-4 T and \(\delta ={ 30 }^{ \circ }\)
e = Blv = Vlv , where V is vertical component of earth's field
= (R sin \(\delta \))lv = 5.0\(\times\)10^-4 \(\times\) \(sin\ { 30 }^{ \circ }\) \(\times\) 25 \(\times\) 500 = 3.1 volt

\(First\ case,\)
\(angle \ of \ incidence \ i={ 60 }^{ \circ }\)
\(angle \ of \ refraction \ r={ 35 }^{ \circ }\)
\({ \alpha }_{ { \mu }_{ g } }= \ \frac { 1 }{ { g }_{ { \mu }_{ \alpha } } } =\frac { sin\ i }{ sin \ r } \)
\(Second \ case,\)
\( { \alpha }_{ { \mu }_{ \omega } }=\frac { sin\ { 60 }^{ \circ } }{ sin\ { 47 }^{ \circ } } =1.18\)
\( { \omega }_{ { \mu }_{ g } }= { \alpha }_{ { \mu }_{ g } }\times { \omega }_{ { \mu }_{ \alpha } }\)
\(=\frac { { \alpha }_{ { \mu }_{ g } } }{ { \alpha }_{ { \mu }_{ \omega } } } =\frac { 1.51 }{ 1.18 } =1.28\)
\(Third \ case,\)
\( angle \ of \ incidence \ i={ 45 }^{ \circ }\)
\( angle \ of \ refraction \ r=?\)
\( { \omega }_{ { \mu }_{ g } }=\frac { sin\ i }{ sin\ r } \)
\(\frac { sin \ i }{ sin\ r } =1.28\)
\(sin \ r=\frac { sin \ { 45 }^{ \circ } }{ 1.28 } =0.5525\)
\( sin \ r= \ sin{ 33 }^{ \circ }54'\)
\(r= \ { 33 }^{ \circ }54'\)

Resolving power of a microscope is defined as the reciprocal of its limit of resolution (d) i.e.
RP of microscope = 1/d
where, limit of resolution is equal to the smallest distance between two closest objects whose vivid or clean image can be seen through the
microscope and given by d = \(\frac { \lambda }{ 2\mu sin\theta } \)
Resolving power of microscope = \(\frac { 2\mu sin\theta }{ \lambda } \)
where, \(\lambda \) = wavelength of light used,
\(\theta =\) semivertical angle of the cone formed by object
at objective and \(\mu \)  = refractive index of molecule between object and lens.
(a) Resolving power increases with the increase of \(\mu \)
(b) Resolving power decreases as resolving power \(\propto 1/\lambda \)

According to Brewster's law, when a light ray fall on a surface of transparent medium in such a way that reflected ray is perpendicular to the refracted ray, the reflected ray is a totally polarised ray. The angle of incidence, in this case, is 'called Brewster's angle i_B.

The tangent of the polarising angle of incidence of a transparent medium is equal to its refractive index, i.e., \({ \mu }_{ 1 }=\tan { \left( { i }_{ B } \right) } \)
\(\therefore\) Brewster's angle \({ i }_{ B }=\tan ^{ -1 }{ { \mu } } \)
As we have seen, the value of i_B depends on the refractive index of the medium and the refractive index of a medium depends on the wavelength of incident light. Thus, the value of i_B will different for different colours of light (i.e. wavelength of light) 

Short radio waves \(\lambda>01 \mathrm{~m} \text { or } \mathrm{v}<3 \times 10^9 \mathrm{~Hz}\) are used in satellite communication.

(i) From given figure, A = r₁ + r2
As ray QR is parallel to the base BC, then
Therefore, 2r₁(or 2r₂) = A(ii) D = x + y,  r₁ = r₂ = A/2
 D = (i - r₁) + (e - r₂)
 D = (i + e) - (r₁ + r₂)
 D = 2i - A

We know that  \(\frac{W_{1}}{W_{2}}=\frac{I_{1}}{I_{2}}=\frac{a^{2}}{b^{2}}=\frac{4}{1} \Rightarrow \frac{a}{b}=2: 1\)

Radius of curvature of objectrive mirror, 
R₁ = 220 mm
\({ f }_{ 1 }=\frac { { R }_{ 1 } }{ 2 } =\frac { 220 }{ 2 } =110 \ mm\) 
Radius of curvature of secondary mirrors, R₂ = 140 mm
\({ f }_{ 2 }=\frac { { R }_{ 2 } }{ 2 } =\frac { 140 }{ 2 } =70 \ mm\) 
Distance between two mirrors, d = 20 mm from objective mirror.
Now, for secondary mirror, u = f₁ - d = 110 - 20
= 90 mm
From mirror formula,
\(\frac { 1 }{ v } +\frac { 1 }{ u } =\frac { 1 }{ { f }_{ 2 } } \Longrightarrow \frac { 1 }{ v } =\frac { 1 }{ { f }_{ 2 } } -\frac { 1 }{ u } \) 
\( =\frac { 1 }{ 70 } -\frac { 1 }{ 90 } \Longrightarrow v=\frac { 630 }{ 2 } =315\ mm\)
i.i., final image will be at 31.5 cm to the right of secondary mirror.

Given, primary voltage, V_p = 2300 V
Secondary voltage, V_s = 230V
Primary turns, N_p = 4000 turns
Her, we assume that the transformer is ideal. No power loss in the form of heat, etc.,
Using the formula, \(\frac { { V }_{ s } }{ { V }_{ p } } =\frac { N_{ s } }{ N_{ p } } or\frac { 230 }{ 2300 } =\frac { { N }_{ s } }{ 4000 } \)
\({ N }_{ s }=400\)

(a) For water purification, ultraviolet rays are used
(b) For remote sensing, microwaves are used 
(c) For treatment of cancer, Gamma rays are used order of their penetrating power of these radiations microwaves.

As current through element X leads the alternating voltage applied by \(\pi /2\) radian, therefore X is a pure capacitance. Its reactance is \(X_C=\frac{1}{\omega C}=\frac{1}{2 \pi v C} .\)

\(\text { (i) } \oint_s \vec{E} \cdot \overrightarrow{d s}=\frac{q}{\epsilon_0}\)
\(\text { (ii) } \oint \vec{E} \cdot \overrightarrow{d t}=0\)

(a) The ray diagram, showing image formation by a compound microscope, is given below:

Linear magnification due to the objective = \(=\frac { h^{ ' } }{ h } =\frac { L }{ f_{ 0 } } \)
\(\left( \therefore tan\beta =\frac { h }{ f_{ 0 } } =\frac { h^{ ' } }{ L } \right) \)
Here, L = tube length = distance between the second focal point of the objective and the first focal point of the eyepiece.
When the final image is formed at infinity, the angular magnification due to the eyepiece equals
\(\frac { D }{ f_{ e } } \) (D = least distance of distinct vision)
\(\therefore \) Total magnification when the final image is formed
at infinity = \(\left( \frac { L }{ f_{ 0 } } ,\frac { D }{ f_{ e } } \right) \)
(b) Resolving power increases when the focal length of the objective is decreased.
(i) This is because the minimum separation \(d_{ min }\left( =\frac { 1.22f\lambda }{ D } \right) \) decreases when f is decreased
(ii) Resolving power decreases when the wavelength of light is increased.
This is because the minimum separation,  \(d_{ min }\left( =\frac { 1.22f\lambda }{ d } \right) \) increases when \(\lambda \) is increased.

(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 \({I_{max}\over {I_{min}}}={(a_1+a_2)^2\over (a_1-a_2)^2}={25\over 9}\)
\(\therefore\ \ {a_1+a_2\over a_1-a_2}={5\over 3}\Rightarrow{a_1\over a_2}={4\over 1}\)
\(\therefore\ {W_1\over W_2}={I_1\over I_2}={(a_1)^2\over (a_2)^2}={16\over 1}\)