Position of minima due to diffraction at a single slit is given by
\( a\sin { \theta =n\lambda } \)
\( \therefore \ \sin { \theta } =\frac { n\lambda }{ a } =\frac { 1\times600\times{ 10 }^{ -9 } }{ 1.0\times{ 10 }^{ -3 } } \)
\( =6\times{ 10 }^{ -4 }m\)
\(For \ small\theta ,\ \sin { \theta } \simeq \ \theta =6\times{ 10 }^{ -4 } \ radian \)
Let y be the distance of first minima on either side of central maxima, then 
\( \theta =\frac { y }{ D } \)
\(or \ y=\theta D=6\times{ 10 }^{ -4 }\times2=12\times{ 10 }^{ -4 }m \)
So the distance between first dark fringes on either side of central maxima is
\(2y=2\times12\times{ 10 }^{ -4 }=24\times{ 10 }^{ -4 }m=2.4mm\)

Maximum amplitude, A_max = 10 V
Minimum amplitude, A_min = 2 V
Modulation index μ, is given by the relation:
\(\mu=\frac{A_{\max }-A_{\min }}{A_{\max }+A_{\min }}\)
\(=\frac{10-2}{10+2}=\frac{8}{12}=0.67\)
if A_"min" = 0
Then \(\mu=\frac{A_{\max }}{A_{\max }}=\frac{10}{10}=1\) 

Here,\(\beta =59,{ I }_{ e }=6.0\quad mA\)
\(\beta =\frac { { I }_{ c } }{ { I }_{ b } } =\frac { { I }_{ e }-{ I }_{ b } }{ { I }_{ b } } =\frac { { I }_{ e } }{ { I }_{ b } } =-1\)
\( \therefore \ 59=\frac { 6 }{ { I }_{ b } } -1 \ or \ \frac { 6 }{ { I }_{ b } } =60 \ or \ { I }_{ b }=\frac { 6 }{ 60 } =0.1 \ mA\)
\( { I }_{ c }={ I }_{ e }-{ I }_{ b }=6.0-0.1=5.9\ mA\)

Here, \(\beta \) = 50, \({ R }_{ 0 }=5\times10^{ 3 }\Omega \);
\(R_{ i }=10^{ 3 }\Omega ;V_{ 1 }=0.01 \ V\)
\({ A }_{ V }=\frac { { V }_{ 0 } }{ { V }_{ 1 } } =\beta .\frac { { R }_{ 0 } }{ { R }_{ i } } \)
or \({ V }_{ 0 }=\beta .\frac { R_{ 0 }\times{ V }_{ i } }{ { R }_{ i } } =50\times\frac { 50\times10^{ 3 }\times0.01 }{ 10^{ 3 } } =2.5 \ V\)

As, we know that
\(power=\frac { energy }{ time } =300\times { 10 }^{ 6 }W=3\times { 10 }^{ 8 }J/s\)
\( 170 \ MeV \ =170\times { 10 }^{ 6 }\times 1.6\times { 10 }^{ -19 }=27.2\times { 10 }^{ -12 }J\)
Number of atoms fissioned per second
\(=\frac { 3\times { 10 }^{ 8 } }{ 2.72\times { 10 }^{ -12 } } =\frac { 3\times { 10 }^{ 20 } }{ 27.2 } \)
Number of atoms fissioned per hour
\(=\frac { 3\times { 10 }^{ 20 }\times 3600 }{ 2.72 } =\frac { 3\times 36 }{ 27.2 } \times { 10 }^{ 22 }=4\times { 10 }^{ 22 }m\)

Force of attraction by the plate on the electron if it is at a distance X is given by
\(\frac { 1 }{ 4 } \frac { { q }^{ 2 } }{ 4\pi { e }_{ 0 }{ x }^{ 2 } } \)
Total W > D in taking the electron from x=d to x=\(\alpha \) will be
\(w=\int { F.dx } =\frac { 1 }{ 4 } .\frac { { q }^{ 2 } }{ 4\pi e_{ 0 } } \int _{ d }^{ \alpha }{ } \frac { 1 }{ { x }^{ 2 } } dx\)

\(=\frac { 9\times{ 10 }^{ 9 }(1.6\times{ 10 }^{ -19 })^{ 2 } }{ 4\times{ 10 }^{ -10 } } J\)
\(=\frac { 9\times{ 10 }^{ 9 }(1.6)^{ 2 }\times{ 10 }^{ -38 } }{ 4\times{ 10 }^{ 10 }\times1.6\times{ 10 }^{ -19 } } eV\)
\(=\frac { 1.6\times9 }{ 4 } =3.6\quad eV\)

\(Here,\lambda =5000\overset { \circ }{ A } =5\times { 10 }^{ -7 }m;\)
Energy of one photon,
\(E=\frac { hc }{ \lambda } =\frac { \left( 6.63\times { 10 }^{ -34 } \right) \left( 3\times { 10 }^{ 8 } \right) }{ 5\times { 10 }^{ -7 } } =3.96\times { 10 }^{ -19 }J\)
A 200 W bulb supplies 200 J of energy per second.Energy emitted by lamp per second as visible light,
\({ E }_{ 1 }=200\times \frac { 10 }{ 100 } =20J{ s }^{ -1 }\)
Number of photons emitted per second as visible light.
 \(N=\frac { { E }_{ 1 } }{ E } =\frac { 20 }{ 3.96\times { 10 }^{ -19 } } =5.05\times { 10 }^{ 19 }\)

Here, v_c = 1200 kHz
v_m = 20 kHz
Lower side band frequency = v_c - v_m
 = 1200 - 20 = 1180 kHz
Upper side band frequency = v_c + v_m
 = 1200 + 20 = 1220 kHz
Unmodulated carrier amplitude
  \({ A }_{ C }=\frac { { A }_{ max }+{ A }_{ min } }{ 2 } =\frac { 110+90 }{ 2 } =100V\)
Modulation index (\(\mu\))
\(\mu =\frac { { A }_{ max }-{ A }_{ min } }{ { A }_{ max }+{ A }_{ min } } =\frac { 110-90 }{ 110+90 } =0.1\)
Amplitude of each side band
\(=\frac { \mu { A }_{ c } }{ 2 } =\frac { 0.1\times 100 }{ 2 } =5V\) 

Wavelength of light of a sodium line, λ = 589 nm = 589 x 10⁻⁹ m
Mass of an electron, m_e= 9.1 x 10⁻³¹ kg
Mass of a neutron, m_n= `1.66 x 10⁻²⁷ kg
Planck’s constant, h = 6.6 x 10⁻³⁴ Js
(a) For the kinetic energy K, of an electron accelerating with a velocity v, we have the relation:
\(K=\frac{1}{2} m_{e} v^{2}\)
We have the relation for de Broglie wavelength as:
\(\lambda=\frac{h}{m_{e} v}\)
\(\therefore v^{2}=\frac{h^{2}}{\lambda^{2} m_{e}^{2}}\)
Substituting equation (2) in equation (1), we get the relation:
\(K=\frac{1}{2} \frac{m_{e} h^{2}}{\lambda^{2} m e_{e}^{2}}=\frac{h^{2}}{2 \lambda^{2} m_{e}}\)
\(=\frac{\left(6.6 \times 10^{-34}\right)^{2}}{2 \times\left(589 \times x 10^{-9}\right)^{2} \times 9.1 \times 10^{-31}}\)
\(= 6.9 \times 10^{-25} J\)
\(=\frac{6.9 \times 10^{-25}}{1.6 \times 10^{-19}}=4.31 \times 10^{-6} \mathrm{eV}=4.31 \mu \mathrm{eV}\)
Hence, the kinetic energy of the electron is 6.9 x 10⁻²⁵ J or 4.31 μeV.
(b) Using equation (3), we can write the relation for the kinetic energy of the neutron as:
\(\frac{h^{2}}{2 \lambda^{2} m_{n}}\)
\(=\frac{\left(6.6 \times 10^{-34}\right)^{2}}{2 \times\left(589 \times 10^{-9}\right)^{2} \times 1.66 \times 10^{-27}}\)
\(=3.78 \times 10^{-28} J\)
\(=\frac{3.78 \times 10^{-28}}{1.6 \times 10^{-19}}=2.36 \times 10^{-9} e V=2.36 \neq V\)
Hence, the kinetic energy of the neutron is 3.78 x 10⁻²⁸ J or 2.36 neV.

\(Here, \ { \tau }_{ \alpha }=1620years,{ \tau }_{ \beta }=405years\)
\( t=?N={ N }_{ 0 }-\frac { 3 }{ 4 } { N }_{ 0 }=\frac { { N }_{ 0 } }{ 4 } \)
\( If \ { \lambda }_{ \alpha }and{ \lambda }_{ \beta }are \ decay \ constants \ for \ alpha \ and \ beta \ emission \ respectively,\)
\( then \ { \lambda }_{ \alpha }=\frac { 1 }{ { \tau }_{ \alpha } } =\frac { 1 }{ 1620 } { yr }^{ -1 }\)
\(and \ { \lambda }_{ \beta }=\frac { 1 }{ { \tau }_{ \beta } } =\frac { 1 }{ 405 } { yr }^{ -1 }\)
\(Total \ decay \ constant \ \lambda ={ \lambda }_{ \alpha }+{ \lambda }_{ \beta }\)
\( =\frac { 1 }{ 1620 } +\frac { 1 }{ 405 } =\frac { 1+4 }{ 1620 } =\frac { 1 }{ 324 } { yr }^{ -1 }\)
\( from \ N={ N }_{ 0 }{ e }^{ -\lambda t }; \ \frac { 1 }{ 4 } { N }_{ 0 }={ N }_{ 0 }{ e }^{ -\lambda t }\)
\(t=\frac { { log }_{ e }4 }{ \lambda } =\frac { 1.386 }{ sfrac { 1 }{ 324 } } yr=449.1yr\)

(i) (a) The doping of pure silicon with boron or aluminium will give us \(p-\)type semiconductor 
(b) The doping of pure silicon with arsenic or phosphorous will give us  \(n-\)type semiconductor.
(ii) From the above study, we find that to get better output current ,pure semiconductor has to be adopted with suitable (impurity) atoms.Similarly,in day to day life,the appearence of right kind of persons(leaders) in society can improve their lot. 

Given, the equation of carrier wave,
 c(t) = 1 sin (8\(\pi \) t)        ..........(i)
(i) According to the figure,
Amplitude of modulating signal,
A_m = 1 V
Amplitude of carrier wave,
A_C =2
T_m = 1 s
From Eq.(i), we get
 \({ \omega }_{ m }=\frac { 2\pi }{ { T }_{ m } } =\frac { 2\pi }{ 1 } =2\pi \ rad/s\)    ....(ii)
 c(t) = 2 sin (8 \(\pi \) t)
So,  \({ \omega }_{ c }=4{ \omega }m_{ }\)
From Eq. (ii) , we get
So,   \({ \omega }_{ c }=4{ \omega }_{ m }\)
Amplitude of modulated wave,
A = A_m + A_c
 = 1 + 2= 3 V
The sketch of the amplitude modulated waveform is shown below:

For carrier signal, \({ \omega }\) = 8 \(\pi \) 
\(T=\frac { 2\pi }{ \omega } =\frac { 2\pi }{ 8\pi } =\frac { 1 }{ 4 } =0.25s\)
(ii) Modulation index, \(m=\frac{A_{m}}{A_{c}}=\frac{1}{2}=0.5\) 

\(N={ N }_{ 0 }{ e }^{ -\lambda t }\)
\(\frac { 3 }{ 4 } { N }_{ 0 }={ N }_{ 0 }{ e }^{ -(0.3465)^{ t }}\) =(\(​​​\because N\)75% of N₀)1/2  \(N=\frac { 3 }{ 4 } { N }_{ 0 }\)
\({ e }^{ (0.3465)t }=\frac { 4 }{ 3 } \)
\(0.3465\times t={ log }_{ e }(4/3)\)
\(=2.303[log4-log3]\)
\(=2.303[0.6020-0.4771]\)
\(=2.303\times 0.1249\)
\(t=\frac { 2.303\times 0.1249 }{ 0.3465 } \)
\( \therefore \ t=0.83days \ or \ 19.92\ hours\)

Balmer series lies in the visible region.

As   \(\frac { 1 }{ \lambda } \propto \left( \frac { 1 }{ { n }_{ 1 }^{ 2 } } -\frac { 1 }{ { n }_{ 2 }^{ 2 } } \right) \)
\(\frac { 1 }{ \lambda } \propto \frac { 1 }{ { n }_{ 1 }^{ 2 } } \ (\because { n }_{ 2 }=\infty )\)
For Lyman series \({ n }_{ 1 }=1\)
For Balmer series \({ n }_{ 1 }=2\)
For Paschen series\({ n }_{ 1 }=3\)
\(\therefore { \lambda }_{ L }<{ \lambda }_{ B }<{ \lambda }_{ P }.\)

It is a p-n junction device which converts solar energy into electrical energy.

Light waves reflected from outer and inner surface of soap bubble interfere. For different wavelengths, conditions for constructive interference are satisfied at different positions. That is why beautiful colours are seen.

Range of frequency used for TV transmission is 54 MHz to 890 MHz (VHF and UHF).These waves and light waves are electromagnetic waves.The ionosphere is unable to reflect back these waves to earth.

\(\lambda =\frac { h }{ mv }\)
\(=\frac { 6.63\times { 10 }^{ -34 } }{ 9.1\times { 10 }^{ -31 }\times 2.2\times { 10 }^{ 8 } }\)
\(=3.31\times { 10 }^{ -12 }m\)

Working:
During one-half of the input AC, the diode is forward biased and a current flows through RL. During the other half of the input AC, the diode is reverse biased and no current flows through the load RL. Hence, the given AC input is rectified.