Modulation index ,mf = \(\frac { Maximum\ frequency\ deviation }{ minimum\ signal\ freqency } \)

(i) As the lens forms a real iamge, it must be a convex lens.
(ii) From the graph, when u = 20 cm , we have v = 20 cm.
For the convex lens forming a real iamge, u is negative and v and f are positive.
U = -20 cm v = +20cm
Using this lens formula,
1/f = 1/v – 1/u = 1/20 – 1/-20 = 1/10 or f = + 10 cm

Same as velocity of light

Anticlockwise

q \(\rightarrow\)_______Q
1/2 mv2 = kQq/r
Or, v2 a1/r
Or, r a 1/v²
Or, r' = r/4

Magnetic field at O due to current through one infinitely long wire is
\(B=\frac { \mu _{ 0 } }{ 4\pi } \times\frac { 2I }{ r } \hat{k}\)(i.e. along + Z-direction)
So, magnetic field at O due to all the wires is
\(B=\frac { \mu _{ 0 } }{ 4\pi } \times2\times2[\frac { 1 }{ 1 } -\frac { 1 }{ 2 } +\frac { 1 }{ 4 } -\frac { 1 }{ 8 } +\frac { 1 }{ 16 } +...] \hat{ k } \\ =\frac { { \mu }_{ 0 } }{ \pi } [(1+\frac { 1 }{ 4 } +\frac { 1 }{ 16 } +...)\hat{ k } -(\frac { 1 }{ 2 } +\frac { 1 }{ 8 } +...)\hat{ k } ]\\ =\frac { { \mu }_{ 0 } }{ \pi } [(\frac { 1 }{ 1-\frac { 1 }{ 4 } } )\hat{ k } \frac { 1 }{ 2 } (\frac { 1 }{ 1-\frac { 1 }{ 4 } } )\hat{ k } ]\\ \Rightarrow B=\frac { { \mu }_{ 0 } }{ \pi } \times\frac { 2 }{ 3 } \hat{ k } =\frac { { \mu }_{ 0 } }{ 4\pi } \times\frac { { { 8 }\hat{k} } }{ 3 } \)​​​​​​

Total energy of an electron E_e = mc²
Total energy of a photon, \(E_p=\frac{h c}{\lambda}\)
de-Broglie wavelength of electron of mass m moving with velocity V
\( \lambda=\frac{h}{m v} \\ \Rightarrow m=\frac{h}{\lambda v} \)
\(E_e=m c^2=\frac{h c^2}{\lambda v}\)
\(\therefore \frac{E_e}{E_p}=\frac{\frac{h_0^2}{\lambda v}}{\frac{h o}{\lambda}}=\frac{c}{v}\)
As c > > v, therefore the total energy of electron is more than the total energy of photon.

Since, the object distance is u. Let us consider the two ends of the object be at distance \({ u }_{ 1 }=u-L/2\) and \({ u }_{ 2 }=u+L/2\), respectively, so that \(|{ u }_{ 1 }-{ u }_{ 2 }|=L\). Let the image of the two ends be formed at \({ v }_{ 1 } \ and \ { v }_{ 2 }\), respectively so that the image length would be
 \({ L }^{ ' }=|{ v }_{ 1 }-{ v }_{ 2 }|\)
Applying mirror formula, we have
\(\frac { 1 }{ u } +\frac { 1 }{ v } =\frac { 1 }{ f } \ or\ v=\frac { fu }{ u-f } \)
On shoving, the positions of two images are given by
\({ v }_{ 1 }=\frac { f\left( u-L/2 \right) }{ u-f-L/2 } ,{ v }_{ 2 }=\frac { f\left( u+L/2 \right) }{ u-f+L/2 } \)
For length, substituting these values in Eq. (i), we have
 \({ L }^{ ' }=|{ v }_{ 1 }-{ v }_{ 2 }|=\frac { { f }^{ 2 }L }{ \left( u-f \right) ^{ 2 }-{ L }^{ 2 }/4 } \)
Since, the object is short and kept away from focus, we have \({ L }^{ 2 }/4<<\left( u-f \right) ^{ 2 }\)
Hence, finally, \({ L }^{ ' }=\frac { { f }^{ 2 } }{ \left( u-f \right) ^{ 2 } } L\)
This is the required expression of length of an image.

\(\lambda =\frac { c }{ v } =\frac { 3\times { 10 }^{ 8 } }{ 4\times { 10 }^{ 9 } } =0.075 \ m.\) This wavelength belongs to microwave region the microwaves are used 
(i) In microwave ovens 
(ii) in a radar system for the navigation aircraft

This cells are connected
(i) in series, to get maximum voltage,
(ii) in parallel, to get maximum current and
(iii) in mixed grouping, to get maximum power.

A constant current can be kept inside a conductor by maintaining a constant potential difference across the two ends of conductor.

(i) \(1.7\times {{10}^{-5}}V\)
(ii) \(2.72\times {{10}^{-24}}J\)

Refractive index of glass, μ
Focal length of the double-convex lens, f = 20 cm
Radius of curvature of one face of the lens = R₁
Radius of curvature of the other face of the lens = R₂
Radius of curvature of the double-convex lens = R The value of R can be calculated as:
∴ R₁ = R and R₂ = -R
The value of R can be calculated as:
\(\frac { 1 }{ f } =(\mu -1)\left[ \frac { 1 }{ { R }_{ 1 } } -\frac { 1 }{ { R }_{ 2 } } \right] \)
\(\frac { 1 }{ 20 } =(1.55)\left[ \frac { 1 }{ R } +\frac { 1 }{ R } \right] \)
\(\frac { 1 }{ 20 } =0.55\times \frac { 2 }{ R } \)
∴ R = 0.55 x 2 x 20 = 22 cm
Hence, the radius of curvature of the double-convex lens is 22 cm.

(i) Fot region II,
\(E_{II} = \frac{1}{2\pi\epsilon_0}(\sigma_1-\sigma_2)\)
towards tight side/from sheet A to sheet B
(ii) For region III,
\(E_{III} = \frac{1}{2\pi\epsilon_0}(\sigma_1+\sigma_2)\)
towards tight side/away from the two sheets

Since, it is given that the current flows in the clockwise direction for an observer on the left side of the solenoid.
It means that the left face of the solenoid acts as south pole and right face acts as North pole. Inside a bar, the magnetic field lines are directed from South to North. \(\left( { 1\frac { 1 }{ 2 } } \right) \)
Therefore, the magnetic field lines are directed from left to right in the solenoid.
Magnetic moment of a single current carrying loop is given by, \({ m }^{ ' }={ IA }\)
So, magnetic moment of the whole solenoid is given by
 \({ m }={ N{ m }^{ ' } }=N\left( IA \right) \)

Given \(v=40\times { 10 }^{ 6 }Hz\) 
\(\therefore \)\(T=\frac { 1 }{ v } =\frac { 1 }{ 40\times 10^{ -6 } } =0.25\times 10^{ -6 }/s\) 
Magnetic field
\({ B }_{ 0 }=\frac { E_{ 0 } }{ c } =\frac { 750 }{ 3\times 10^{ 18 } } =2.5\times 10^{ -6 }T \ along \ (z \ direction)\) 
And angular frequency 
\(\omega =2\pi v=2\times \pi \times 40\times { 10 }^{ 6 }=8\pi \times { 10 }^{ 7 } \ Hz\)​​​​​

(i) Power = +2 dioptre.
(ii) Here, we have f = +12 cm, R₁ = +10 cm, R₂ = -15 cm.
Refractive index of air is taken as unity.
We use the lens formula. The sign convention has to be applied for f, R₁ and R₂.
Substituting the values, we have
\(\frac { 1 }{ 12 } =(n-1)\left( \frac { 1 }{ 10 } -\frac { 1 }{ 15 } \right) \)
This gives n = 1.5.
(iii) For a glass lens in air, n₂ = 1.5, n₁ = 1, f = +20 cm. Hence, the lens formula gives
\(\frac { 1 }{ 20 } =0.5\left( \frac { 1 }{ { R }_{ 1 } } -\frac { 1 }{ { R }_{ 2 } } \right) \)
For the same glass lens in water, n₂ = 1.5, n₁ = 1.33. Therefore \(\frac { 1.33 }{ f } =(1.5-1.33)\left[ \frac { 1 }{ { R }_{ 1 } } -\frac { 1 }{ { R }_{ 2 } } \right] \)
Combining these two equations, we find f = + 78.2 cm.

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  \(u=\infty ,v=f\)
\(\frac { 1 }{ v } -\frac { 1 }{ u } =\frac { 1 }{ f } \)
\(\frac { 1 }{ f } =\left( \frac { { n }_{ 2 } }{ { n }_{ 1 } } -1 \right) \frac { 1 }{ { R }_{ 1 } } -\frac { 1 }{ { R }_{ 2 } } \)
\(\frac { 1 }{ f } =\left( { \mu }_{ 21 }-1 \right) \left( \frac { 1 }{ { R }_{ 1 } } -\frac { 1 }{ { R }_{ 2 } } \right) \)
\(\frac { 1 }{ f } =\left( { \mu }_{ 21 }-1 \right) \left( \frac { 1 }{ { R }_{ 1 } } -\frac { 1 }{ { R }_{ 2 } } \right) \)
\(\frac { 1 }{ f } =\left( 1.55-1 \right) \left( \frac { 1 }{ R } -\frac { 1 }{ -R } \right) \)
\(=0.55\times \frac { 2 }{ R } \)
\(R=0.55\times 2\times 20=22 \ cm\) 

Here, l = 40cm = 0.4m,
A = 20cm² = 20\(\times\)10^-4m², N = 800
dI/dt = 0.2 A/s, e = ?
e = \(L\frac { dI }{ dt } =\frac { { \mu }_{ 0 }{ N }^{ 2 }A }{ l } .\frac { dI }{ dt } \)
\(=\frac { 4\pi \times { 10 }^{ -7 }\times \left( 800 \right)\ ^{ 2 }\times 20\times { 10 }^{ -4 } }{ 0.4 } \times 0.2\ =\ 8.03\times { 10 }^{ -4 }\ V\)