(i) Heavy doping makes the depletion region very thin. This makes the electric field of the junction.
(ii) When operated under reverse bias, the photodiode can detect changes in current with changes in light intensity more easily.
(iii) The photon energy, of visible light photons varies from about 1.8 eV. Hence for visible LED's, the semiconductor must have a band gap of 1.8 eV.

As,  
\( \beta =\frac { D\lambda }{ d } and \ { \beta }_{ 1 }=\frac { { D\lambda }_{ 1 } }{ d }\)
\( \because \ \frac { { \beta }_{ 1 } }{ \beta } =\frac { \frac { { D\lambda }_{ 1 } }{ d } }{ \frac { D\lambda }{ d } } =\frac { { \lambda }_{ 1 } }{ { \lambda } } =\frac { 1 }{ \mu } \)
or \({ \beta }_{ 1 }=\frac { \beta }{ \mu } =\frac { 2 }{ 1.22 } =1.5mm\)

For convex lens of focal length 10 cm,
f =+ 10cm, u = -3 cm
Using lens formula,
\( \frac { 1 }{ f } =\frac { 1 }{ v } -\frac { 1 }{ u }\)  
\(\Rightarrow \ \frac { 1 }{ 10 } =\frac { 1 }{ v } -\frac { 1 }{ \left( -30 \right) } \Rightarrow v=15cm\)
The image formed by first lens acts as a virtual object for plano-concave lens.
For plano-concave lens,
 \(u=+10cm,\quad f=-10cm,v=?\) 
Using lens formula,
 \(\frac { 1 }{ f } =\frac { 1 }{ v } -\frac { 1 }{ u } \) 
\(\Rightarrow \quad \frac { 1 }{ v } =0\)  
\( \Rightarrow \quad v=\infty \)
The refracted ray becomes parallel to principal axis for convex lens of focal length 30 cm.
 \(u=-\infty ,v=?,f=30cm\) 
 \(\frac { 1 }{ f } =\frac { 1 }{ v } -\frac { 1 }{ u } \) 
\(\Rightarrow \ \frac { 1 }{ 30 } =\frac { 1 }{ v } -\frac { 1 }{ \left( -\infty \right) } \ \Rightarrow \ v=30cm\) 
So, final image is formed at a distance of 30 cm from second convex lens on the other side of u.   

Given \(B=8\times 10^{ 6 } \ sin \ (2\times 10^{ 11 }t+300\pi x) \ T\) wave from equation of magnetic field.
\({ B }_{ y }={ B }_{ 0 } \ sin \ \left( \frac { 2\pi }{ T } t+\frac { 2\pi }{ \lambda } x \right) \) 
or  \(300\pi =\frac { 2\pi }{ \lambda } \) 
or  \(\lambda =\frac { 2\pi }{ 300\pi } =\frac { 1 }{ 150 } =6.3\times { 10 }^{ -3 } \ m\) 
(ii) Electric field is given as
\({ E }_{ Z }={ B }_{ Z } \ c=24\times 10^{ 14 }sin\left[ 2\times 10^{ 11 }t+300\pi x \right] Vm^{ -1 }\)

1.33

(i) The de-Broglie wavelength of a particle is given by
 \(\lambda =\frac { 12.27 }{ \sqrt { V_{ } } } \mathring { A } \)
[Where, V is the accelerating potential of the particle]
\(\because \quad \lambda _{ p }=\lambda _{ \alpha }\)                              [given]
\(=\frac { 12.27 }{ \sqrt { V_{ p } } } =\frac { 12.27 }{ \sqrt { V_{ \alpha } } }\)
\( \Rightarrow \frac { V_{ p } }{ V_{ \alpha } } =1\)
(ii) The de-Broglie wavelength of the particle is given by
\(\lambda =\frac { h }{ mv } \)
\( \lambda _{ p }=\frac { h }{ m_{ p }.v_{ p } } \ and \ \lambda _{ \alpha }=\frac { h }{ m_{ \alpha }v_{ \alpha } } \)
We know that, \(\\ m_{ \alpha }=4 m _{ p }\)
\( \because \lambda_{p}=\lambda_{\alpha} \)                            [given]
\( \therefore \frac{h}{m_{p} \cdot v_{p}}=\frac{h}{4 m_{p} \cdot v_{\alpha}} \Rightarrow \frac{v_{p}}{v_{\alpha}}=4 \)

(i) The fringe width interference pattern increase with the decrase in separation between S₁ and S₂ as  \(\beta \alpha \frac { 1 }{ d } \)
(ii) The fring width decreases as wavelength gets reduced, when interferences set up is taken from air to water as, \(\beta \alpha \lambda \)

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.

Here, I_D = 1 mA = 10^-3A ; C = \(2\mu F\) = 2 x 10^-6 F, \({ I }_{ D }=I=\frac { d }{ dt } \left( CV \right) =C\frac { dV }{ dt } \)
Therefore, \(\frac { dV }{ dt } =\frac { { I }_{ D } }{ C } =\frac { { 10 }^{ -3 } }{ 2\times { 10 }^{ -6 } } =500V/s\)
So, by applying a varying potential difference of 500 V/s, we would produce a displaccment current of desired value.

(ii) We know that, \(I={ I }_{ \circ }cos^{ 2 }\theta \)
Intensity at \({ O }_{ 1 },I={ I }_{ \circ }cos^{ 2 }\theta \)
Intensity at \({ O }_{ 2 }{ I }_{ 1 }={ I }cos^{ 2 }\theta _{ 1 }\)
\(={ I }_{ \circ }cos^{ 2 }\theta cos^{ 2 }{ 60 }^{ \circ } \ \left[ \therefore \theta _{ 1 }={ 60 }^{ \circ } \right]\)
\({ I }_{ 1 }=\frac { { I }cos^{ 2 }\theta _{ 1 } }{ 4 } \)
Intensity at \({ O }_{ 3 },{ I }_{ 2 }={ I }_{ 1 }cos^{ 2 }\theta _{ 2 }\)
\(=\frac { { I }_{ \circ }cos^{ 2 }\theta }{ 4 } \times cos^{ 2 }{ 90 }^{ \circ }\quad \left[ \therefore \theta _{ 2 }={ 90 }^{ \circ } \right]\)
\( =0\)

(i) \(\frac { { n }_{ 1 } }{ v } -\frac { { n }_{ 2 } }{ u } =\frac { { n }_{ 2 }-{ n }_{ 1 } }{ R } \)      ............(i)
(ii) Now, the image I^' acts as a virtual object for the second surface that will form a real at I. As, refraction takes place from denser to rarer medium,

\(\therefore \quad \frac { { -n }_{ 2 } }{ v } +\frac { { n }_{ 1 } }{ { v }^{ ' } } =\frac { { n }_{ 2 }-{ n }_{ 1 } }{ { R }^{ ' } } \)   ............(ii)
On adding Eqs. (i) and (ii), we get
\(\frac { 1 }{ f } =\left( { n }_{ 21 }-1 \right) \left( \frac { 1 }{ R } -\frac { 1 }{ { R }^{ ' } } \right) \ \left[ \therefore { n }_{ 21 }=\frac { { n }_{ 2 } }{ { n }_{ 1 } } ,\frac { 1 }{ f } =\frac { 1 }{ v } -\frac { 1 }{ u } \right] \)