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.

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.

\(1.9eV,4125\mathring { A;no \ change } \)

\((i) \ 3.3\times { 10 }^{ -25 }kg{ ms }^{ -1 } \)
\((ii) \ 6.2\times { 10 }^{ -2 }eV\)
\((iii) \ 0.377 \ eV\)

There is no contradiction. Field lines inside the bar go away from S towards N. The next flux of B over any surface fully enclosing N or S must be identically zero. 

The first refracting ABC forms the image I₁ of the object O. The image I₁ acts as virtual object for the second refracting surface ADC, which forms the real image I as shown in the diagram
For refraction at ABC 
\(\frac { { n }_{ 2 } }{ { v }_{ 1 } } -\frac { { n }_{ 1 } }{ u } =\frac { { n }_{ 2 }-{ n }_{ 1 } }{ { R }_{ 1 } } \)
For refraction at ADC
\(\frac { { n }_{ 1 } }{ v } -\frac { { n }_{ 2 } }{ { v }_{ 1 } } =\frac { { n }_{ 1 }-{ n }_{ 2 } }{ { R }_{ 2 } } \)
Adding equation (i) and equation (ii)
 \(\frac { { n }_{ 1 } }{ v } -\frac { { n }_{ 2 } }{ u } =\left( { n }_{ 2 }-{ n }_{ 1 } \right) \left( \frac { 1 }{ { R }_{ 1 } } -\frac { 1 }{ { R }_{ 2 } } \right) \)
\(\frac { 1 }{ v } -\frac { 1 }{ u } =\left( \frac { { n }_{ 2 } }{ { n }_{ 1 } } -1 \right) \left( \frac { 1 }{ { R }_{ 1 } } -\frac { 1 }{ { R }_{ 2 } } \right) \)
We know, If  \(u=\infty ,v=f\)
\(\frac { 1 }{ v } -\frac { 1 }{ u } =\frac { 1 }{ f } \)
\(\frac { 1 }{ f } =\left( \frac { { n }_{ 2 } }{ { n }_{ 1 } } -1 \right) \frac { 1 }{ { R }_{ 1 } } -\frac { 1 }{ { R }_{ 2 } } \)
\(\frac { 1 }{ f } =\left( { \mu }_{ 21 }-1 \right) \left( \frac { 1 }{ { R }_{ 1 } } -\frac { 1 }{ { R }_{ 2 } } \right) \)
\(\frac { 1 }{ f } =\left( { \mu }_{ 21 }-1 \right) \left( \frac { 1 }{ { R }_{ 1 } } -\frac { 1 }{ { R }_{ 2 } } \right) \)
\(\frac { 1 }{ f } =\left( 1.55-1 \right) \left( \frac { 1 }{ R } -\frac { 1 }{ -R } \right) \)
\(=0.55\times \frac { 2 }{ R } \)
\(R=0.55\times 2\times 20=22 \ cm\) 

(i) The teacher displays the qualities of deep knowledge of the phenomenon and eagerness to explain it to the child.
(ii) The phenomenon involved in a thin film is interference. Different colours of light interfere at different points in space and hence, child is able to see different colours.

Given, focal length of objective, f  = 1.25cm
Focal length of eyepiece, f_e = 5cm
Least distance of distinct vision, D = 25cm
Angular magnification of lens,m = 30
The magnification produced by eyepiece
m_e = \(1+\cfrac { D }{ { f }_{ e } } =1+\cfrac { 25 }{ 5 } =6\)      
The magnification produced by microscope,
m = m_o \(\times\) m_e 
30 = m_o \(\times\)6
Where, m_o is the magnification produced by objective lens.
m_o = 5
Again, we know that magnification of objective lens,
\(\Rightarrow \)  \(5=\cfrac { -{ v }_{ 0 } }{ { u }_{ 0 } } \)     
\(\Rightarrow \)   \({ v }_{ 0 }\ =-5{ u }_{ 0 }\)                ................(i)                              
Using lens formula for objective lens,
 \(\cfrac { 1 }{ { f }_{ 0 } } =\cfrac { 1 }{ { v }_{ 0 } } -\cfrac { 1 }{ { u }_{ 0 } } \)
\(\Rightarrow \) \(\cfrac { 1 }{ 1.25 } =\cfrac { 1 }{ -5{ u }_{ 0 } } -\cfrac { 1 }{ { u }_{ 0 } } =-\cfrac { 6 }{ 5{ u }_{ 0 } } \) [From Eq.(i)]
\({ u }_{ 0 }=-\cfrac { 6 }{ 5 } \times 1.25=-1.5cm\)
 \({ v }_{ 0 }=-5{ u }_{ 0 }\)
  \(=-5(-1.5)=7.5cm\)
Thus, the objective should be placed at a distance of 1.5cm from the objective lens to get the desired magnification.(1)
Now, using the lens formula for eyepiece,
\(\cfrac { 1 }{ { f }_{ e } } =\cfrac { 1 }{ { v }_{ e } } -\cfrac { 1 }{ { u }_{ e } } \)
\(\Rightarrow \)   \(\cfrac { 1 }{ { u }_{ e } } =\cfrac { 1 }{ { v }_{ e } } -\cfrac { 1 }{ { f }_{ e } } \)
\(=-\cfrac { 1 }{ 25 } -\cfrac { 1 }{ 5 } =-\cfrac { 6 }{ 25 } [\because \quad { v }_{ e }=-25cm]\)                       
\({ u }_{ e }=-4.17cm\)                                                                                                                         
The separation between objective and eyepiece
\(\left| { v }_{ 0 } \right| +\left| { u }_{ e } \right| =4.17+7.5=11.67cm\)
Thus, the microscope is settled as the distance between eyepiece and objective is 11.67cm.

(i) x-rays have the wavelength range,\({ 10 }^{ 13 } \ m \ to \ 3\times { 10 }^{ -8 }m\) which is smaller than that of ultraviolet rays of a wavelength range \(6\times { 10 }^{ -9 }m \ to \ 4\times { 10 }^{ -7 }m, \ i.e., \ { \lambda }_{ x }<{ \lambda }_{ uv }\)
since energy. \(E=\frac { hc }{ \lambda } \ or \ E\propto \frac { 1 }{ \lambda } ; \ so \ \frac { { E }_{ x } }{ { E }_{ uv } } =\frac { { \lambda }_{ uv } }{ \lambda _{ x } } >1 \ or \ { E }_{ x }>{ E }_{ uv }\) Thus x-rays are more energetic than ultraviolet rays
(ii) The wavelength of radio waves is of range. \(0.3m \ to \ 6\times { 10 }^{ 2 }m\) which is very large as compared to the size of molecules of our blood, flesh and bones etc. As the energy of radio waves is quite small so these radio waves can not penetrate the blood of our body. That is why we can not use the radio waves to detect the fracture in bones
(iii) Just as different em waves have different application depending on their wavelength, in the same way, every person has different qualities, which make him suitable for different purpose/fields.
The aim of a teacher or manager is to is identify the quality in different children/persons and put them to their best use accordingly.

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

Let the light be incident at the angle \(\theta \) at the plane at r and leave r+dr at an angle \(\theta +d\theta \) 
From snell's law
\(n(r)sin \ \theta \ = \ n(r+dr) \ sin(\theta +d\theta )\)
\(=\left[ n(r)+\frac { dn }{ dr } dr \right] \ \left[ sin\theta \ cos \ d\theta +cos\theta \ sind\theta \right] \)

As \(d\theta \) is small, so \(cos \ d\theta = \ 1\)  and  \(sind\theta =d\theta \) and neglecting the product of differentiates, we get
\( n(r)sin \ \theta \ = \ n(r) \ sin\theta +n(r) \ cos \ d\theta +\frac { dn }{ dr } dr \ sin\theta \)
\( -\frac { dn }{ dr } dr \ sin\theta \ = \ n(r)cos\theta \ d\theta\)
\(-\frac { dn }{ dr } dr \ tan\theta \ = \ n(r)\frac { d\theta }{ dr } \)
\( As n(r) \ =\ 1+\frac { 2GM }{ { rc }^{ 2 } } \)
\(\\ \therefore \frac { dn }{ dr } \ = \ \frac { -2GM }{ { r }^{ 2 }{ c }^{ 2 } } \)
Putting in eq, we get
\( \frac { -2GM }{ { r }^{ 2 }{ c }^{ 2 } } tan\theta \ = \ \left( 1+\frac { 2GM }{ { rc }^{ 2 } } \right) \frac { d\theta }{ dr } =\frac { d\theta }{ dr } \)
\( or d\theta \ =\ \frac { 2GM }{ { c }^{ 2 } } \frac { tan\theta }{ { r }^{ 2 } } dr\)
Integrating both sides, we get
\(\int _{ 0 }^{ { \theta }_{ 0 } }{ d\theta } \ = \ \frac { 2GM }{ { c }^{ 2 } } \int _{ -\infty }^{ \infty }{ \frac { tan\theta \ dr }{ { r }^{ 2 } } }\)
\(=\frac { 2GM }{ { c }^{ 2 } } \int _{ -\infty }^{ \infty }{ \frac { tan\theta \ dr }{ { r }^{ 3 } } } .r\)
\(As \ { r }^{ 2 }={ x }^{ 2 }+{ R }^{ 2 } \ and \ tan\theta =\frac { R }{ X } ,\ we \ get\)
\(2r \ dr=2xdx\)
\(\therefore \int _{ 0 }^{ { \theta }_{ 0 } }{ d\theta } =\frac { 2GM }{ { c }^{ 2 } } \int _{ -\infty }^{ \infty }{ \frac { R }{ x } \frac { x\quad dx }{ ({ { X }^{ 2 }+{ R }^{ 2 } })^{ 3/2 } } } \) 
\( let\ x=R \ tan \ \phi \)
\(\therefore dx=R\quad { sec }^{ 2 }\phi \ d\phi \)
\( \therefore{ \theta }_{ 0 }=\frac { 2GMR }{ { c }^{ 2 } } \int _{ -\pi /2 }^{ \pi /2 }{ \frac { R{ sec }^{ 2 }\phi \ d\phi }{ { R }^{ 3 }{ sec }^{ 3 }\phi } }\)
\(=\frac { 2GMR }{ { c }^{ 2 } } \int _{ -\pi /2 }^{ \pi /2 }{ cos\phi \quad d\phi } =\frac { 4GM }{ { Rc }^{ 2 } } \)

K = Energy of the photon \(\frac { hc }{ \lambda } \)
de Broglie wavelength,
\({ \lambda }_{ B }=\frac { h }{ p } =\frac { h }{ \sqrt { 2mk } } \)
\({ \lambda }_{ B }=\frac { h }{ \sqrt { 2mc\frac { hc }{ { \lambda } } } } \)
= \(\sqrt { \frac { h\lambda }{ 2mc } }\)
= \({ \left[ \frac { 6.63\times { 10 }^{ -34 }\times 5460\times { 10 }^{ -9 } }{ 2\times 9.1\times { 10 }^{ -31 }\times 3\times { 10 }^{ 8 } } \right] }^{ \frac { 1 }{ 2 } }\)
= 25.75\(\times \)\({ 10 }^{ -10 }\)m

(a) The electromagnetic waves are produced by the accelerated charge. The electric and magnetic field produced by the accelerated charge change with time. Hence it radiates electromagnetic waves.
(b) The EM waves are produced by the accelerated charge. The electron jumping from outer to inner orbit of the electron radiates EM waves. EM waves are propagated as electric & magnetic fields oscillation in mutually perpendicular directions which is why cause of momentum & energy
(c) 

\(\frac { { E }^{ 0 } }{ { B }^{ 0 } } =c\)

de-Broglie wavelength of accelerating charged particle is given by
\(\lambda =\frac { h }{ \sqrt { 2mqV } } \Rightarrow \lambda \sqrt { V } =\frac { h }{ \sqrt { 2mq } } =constant\)
(i) The slope of the lines represent \(\frac { h }{ \sqrt { 2mq } } .\)
where, h is Planck's constant, q is the charge and m is the mass of charged particle.
(ii) ₁H² and ₁H³ carry same charge (as they have same atomic number).
\(\therefore \ \lambda \sqrt { V } \propto \frac { 1 }{ \sqrt { m } } \)
The lighter mass i.e ₁H²  is represented by line of greater slope i.e A and similarly ₁H³ by line B.

During photelectric emission, the momentum of incident photon is transferred to the metal. At microscopic level, atoms of a metal absorb the photon and its momentum is transferred mainly to the nucleus and electrons.
The excited electron is emitted. Therefore, the conservation of momentum is to be considerd as the momentum of incident photon transferred to the nucleus and electrons.

When one slit is closed, amplitude becomes \(\frac { 1 }{ 2 } \) and hence, intensity becomes\(\frac { 1 }{ 4 } \)^th and there is no interference.

As we know that, fringe width \(\beta \)
 \(\beta =\frac { \lambda D }{ d } \)     
Here  \({ D }^{ ' }=2D\)
So  \({ \beta }^{ 1 }=\frac { 2\lambda D }{ D } \)      
\(\Rightarrow \)  \({ \beta }^{ 1 }=2\beta \) 

Let q be the charge on capacitor plates at any instant \(t\) and \(\sigma \) be the surface density of charge. Then electric field between the plates of capacitor will be \(E=\frac { \sigma }{ { \epsilon }_{ 0 } } =\frac { q }{ { \epsilon }_{ 0 }A } \)
Electric flux through area A/3 will be
\({ \phi }_{ E }=E\frac { A }{ 3 } =\frac { q }{ { \epsilon }_{ 0 }A } \times \frac { A }{ 3 } =\frac { 1 }{ { 3\epsilon }_{ 0 } } \)
The displacement current will be
\(I={ \epsilon }_{ 0 }\frac { { d\phi }_{ E } }{ dt } ={ \epsilon }_{ 0 }\frac { { d }_{ E } }{ dt } \left( \frac { q }{ 3{ \epsilon }_{ 0 } } \right) =\frac { { \epsilon }_{ 0 } }{ 3{ \epsilon }_{ 0 } } \frac { dq }{ dt } =\frac { 1 }{ 3 } \) 

During charging or discharging of a capacitor. Increasing or decreasing current in a circuit with time flows due to conduction current in wire and displacement current between the plates of a capacitor. when capacitor get fully charged both conduction and displacement current becomes zero. That is why galvanometer shows a momentary. deflection at the time of charging or discharging.
The expression to explain this observation is \(\oint { \overset { \rightarrow }{ B } .\overset { \rightarrow }{ dt } } ={ \mu }_{ 0 }(I+{ I }_{ D })\)

(i) Zero
(ii) \(2\pi\) radian
(iii) \(2\pi\) radian

(i) \(2.21\times { 10 }^{ -12 }\quad J{ m }^{ -3 }\)
(ii) \(2.21\times { 10 }^{ -12 }\quad J{ m }^{ -3 }\)
(iii) \(4.42\times { 10 }^{ -12 }\quad J{ m }^{ -3 }\)
(iv) \(3.3\times { 10 }^{ -12 }\quad J{ m }^{ -3 }\)

Ultraviolet rays

The basic source of an electromagnetic wave is the time varying electric field produces magnetic field and vice versa

\(v=\frac{1}{\sqrt{\mu_0 \mu_r \varepsilon_0 \varepsilon_r}}=\frac{c}{\sqrt{\mu_r \varepsilon_r}}\)
\(\left[\therefore c=\frac{1}{\sqrt{\mu_0 \epsilon_0}}\right]\)

\(M a x . K . E .=K_{\max }=e V_0=e \times 1.5 \mathrm{~V}=1.5 \mathrm{eV}\)

Photoelectric current is not affected by decreasing the wavelength of incident radiation provided its intensity remains unchanged. 

d = 25c.Yes

30^o as the reflected ray turns through twice the angle through which mirror is turned.