From the lens maker's formula,
\(\cfrac { 1 }{ f } =\left( \mu -1 \right) \left( \frac { 1 }{ { R }_{ 1 } } -\frac { 1 }{ { R }_{ 2 } } \right) \) 
Here, μ = 1.5; R₁ = R₂ and  =-R
\(\therefore \cfrac { 1 }{ f } =\left( 1.5-1 \right) \left( \cfrac { 1 }{ R } -\cfrac { 1 }{ -R } \right) =0.5\times \cfrac { 2 }{ R } =\cfrac { 1 }{ R } \) 
or f = R.

Formula:
From the condition of diffraction, 
a sin θ= nλ.. (for minima)
= \(\left( n+\cfrac { 1 }{ 2 } \right) \lambda \) 

Provided n = 1, 2, 3 ... and n = 0 for central
maximum
From the condition of minima
a sin θ = λ. (n = 1)
Since the value of λ. is nm, so
\(a.\theta =\lambda \Rightarrow a.\cfrac { y }{ D } =\lambda \left[ amgle=\cfrac { arc }{ radius } \right] \) 
= 2 x 10^-4 m.

Momentum, \(p=\frac { h }{ \lambda } =\frac { { 6.626\times 10 }^{ -34 } }{ 2\times { 10 }^{ -10 } } kg{ \quad ms }^{ -1 }\)

(i) Radar basically stands for Radio Detection and Ranging System.
(ii) It is one of the important applications of communication systems' and is mainly used to sense, detect, and locate distant objects like aircraft, ships, spacecraft, etc.
(iii) The angle, range, or velocity of the objects that are invisible to the human eye can be determined.
(iii) Radar uses electromagnetic waves for communication. The electromagnetic signal is initially radiated into space by an antenna in all directions.
(iv) When this signal strikes the targeted object, it gets reflected or reradiated in many directions.
(v) This reflected (echo) signal is received by the radar antenna which in turn is delivered to the receiver.
(vi) Then, it is processed and amplified to determine the geographical statistics of the object. The range is determined by calculating the time taken by the signal to travel from RADAR to the target and back.
Applications :
Radars find extensive applications in almost all fields.
(i) In military, it is used for locating and detecting the targets.
(ii) It is used in navigation systems such as ship borne surface search, air search and weapons guidance systems.
(iii) To measure precipitation .rate and wind speed in meteorological observations, Radars are used.
(iv) It is employed to locate and rescue people in emergency situations.

Chain reaction :

(i) When one \({ }_{92}^{235} \mathrm{U}\) Unucleus undergoes fission,the energy released might be small. But from each fission reaction, three neutrons are released.
(ii) These three neutrons can cause further fission in three other \({ }_{92}^{235} \mathrm{U}\) nuclei which in turn produce nine neutrons. These nine neutrons initiate fission in another 27 \({ }_{92}^{235} \mathrm{U}\) nuclei and so on.
(iii) This process is called a chain reaction and the number of neutrons goes on increasíng almost in geonetric progression. There are two kinds of chain reactions :
(a) uncontrolled chain rcaction               
(b) controlled chain reaction.
(a) uncontrolled chain rcaction :
The number of neutrons multiplies indefinitely and the entire amount of energy released in a fraction of second.
Example : Atom bomb.
(b) controlled chain reaction :
The average number of neutrons released in each stage is kept as one such that it is possible to store the released energy.
Example : Nuclear reactor.

E and B vectors must be in x and y directions.
Formula: We know \(\lambda =\frac { v }{ \gamma } =\frac { 3\times { 10 }^{ 8 } }{ 30\times { 10 }^{ 6 } } \)
λ = 10m.

The cyclotron's frequency v = 18 μHz
= 8 x 10⁶ Hz
The mass of the proton m = 1.67 x 10^-27 kg
The charge of the proton q = 1.6 x 10^-19C
Radius of the dees r = 50 cm = 50 x 10^-2 m
Magnetic field B = ?
k.E of the proton k.E = ?
Magnetic field B = \(\frac { 2\pi mv }{ q } \)
B = \(\frac { 2\times 3.14\times 1.67\times 10^{ -27 }\times 8\times 10^{ 6 } }{ 1.6\times 10^{ -19 } } \)
= \(\frac { 83.90\times 10^{ -21 } }{ 1.6\times 10^{ -19 } } \)
= 52.438 x 10^-2
B = 0.524T
k.E = \(\frac { 1 }{ 2 } \) mv²
v = rω = r x 2πv
= 0.5 x 2 x 3.14 x 8 x 10⁶
= 25.12 x 10⁶ m/s
k.E = \(\frac { 1 }{ 2 } \) x 1.67 x 10^-27 x (25.12 x 10⁶)²
=\(\frac { 41.95\times 10^{ -21 }\times 25.12\times 10^{ 6 } }{ 2 } \)
= 526.89 x 10^-15 J (or) 5.269 x 10^-13J
To convert J into MeV
= \(\frac { 526.89\times 10^{ -15 } }{ 1.6\times 10^{ -19 } } \)  = 3.29 x 10⁶ eV
k.E = 3.29 x 10⁶ MeV

Given:
We know that e = -\(\frac { d\Phi }{ dt } \)
e = \(\frac{d}{dt}\) (5t³ + 4t² + 2t)
e = 15t² + 8t + 2
for t = 2s, e = 15 x (2)² + 8 (2) + 2
e = 78V

(i) In series the circuit remains the same

\(\therefore I=neA_{ 1 }{ V }_{ d_{ 1 } }=n{ A }_{ 2 }{ ev }_{ { d }_{ 2 } }\)or \(\cfrac { { vd }_{ 1 } }{ { vd }_{ 2 } } =\cfrac { { A }_{ 2 } }{ { A }_{ 1 } } \)
(ii) In parallel the potential difference is the same but the circuits are different
\(V={ I }_{ 1 }{ R }_{ 1 }=n{ A }_{ 1 }{ ev }_{ d_{ 1 } }\times \cfrac { \rho l }{ { A }_{ 1 } } ={ n }_{ 1 }e\rho v_{ d_{ 1 } }l\)
\(V={ I }_{ 2 }{ R }_{ 2 }={ n }_{ 1 }e\rho { v }_{ { d }_{ 1 } }l\) \(\left[ \because { R }_{ 1 }=\cfrac { \rho l }{ { A }_{ 1 } } \right] \)
Now I₁R₁ =  I₂R₂  \(\therefore \cfrac { { V }_{ d_{ 1 } } }{ { V }_{ d_{ 2 } } } =1\)

\(V=\frac { q }{ 4\pi { \varepsilon }_{ 0 }r } \)
\(F=\frac { { q }_{ 1 }{ q }_{ 2 } }{ { r }^{ 2 } } \)
= 40.9kV