(i) When the final image is at infinity,
\(m=-\frac{f_{0}}{f_{e}}=\frac{-20}{5}\)
⇒ m = -4
(ii) When the final image is at 25 cm from the eye,
i.e. D = 25 cm
\(m=\frac{-f_{o}}{f_{e}}\left(1+\frac{f_{e}}{D}\right)=\frac{-20}{5}\left(1+\frac{5}{25}\right)\)
⇒ m = -4.8

For a plano-convex lens, R₁ = ∞
R₂ = -R, f = 0.3 m = 30 cm
μ = 1.5
Radius of curvature of plano-convex lens, R = ?
Applying lens Maker's formula, \(\frac{1}{f}=(\mu-1)\left(\frac{1}{R_{1}}-\frac{1}{R_{2}}\right)\)
\(\Rightarrow \frac{1}{30}=(\mu-1)\left ( \frac{1}{\infty }-\frac{1}{-R} \right )=\frac{(1.5-1)}{R}\Rightarrow R=15 cm\)

When lens is immersed in a liquid, then
\(\frac{1}{f_{L}}=\left({ }^{L} \mu_{g}-1\right)\left(\frac{1}{R_{1}}-\frac{1}{R_{2}}\right)\)
where, ^lμ_g = refractive index of lens material (glass) w.r.t. liquid.
\(\therefore \quad \frac{\mu_{g}}{\mu_{L}}=\frac{1.5}{1.5}=1\)
Hence, \(\frac{1}{f_{l}}=(1-1)\left(\frac{l}{R_{1}}-\frac{1}{R_{2}}\right)=0 \Rightarrow f_{L}=\infty\)

Given, h₁ = 12m, u = -15m, v = ?,
h₂ = 12cm = 0.12 m (symbols have their usual meanings)
As, \( \frac{h_{2}}{h_{1}}=-\frac{v}{u} v=-\frac{h_{2}^{}}{h_{1}} \times u \)
⇒ \(v=-\frac{h_{2}^{}}{h_{1}} \times u \)
\(=-\frac{0.12}{12} \times-15=0.15 \mathrm{~m}=15 \mathrm{~cm}\)
Thus, the screen should be placed 15 cm from the pin hole to obtain a 12 cm tall image of the tree.

Given, magnification, m = + 3, R = - 36 cm
Object distance, u = ?
Let u = -x
\( m=\frac{h_{2}}{h_{1}}=\frac{+v}{-u}=3 v=-3 u \Rightarrow v=3 x \)⇒
⇒ \(v=-3 u \Rightarrow v=3 x\)
Applying mirror formula, we have
\( \frac{1}{u}+\frac{1}{v}=\frac{1}{f}=\frac{2}{R} \Rightarrow \frac{1}{-x}+\frac{1}{3 x}=\frac{2}{-36} \)
\(\Rightarrow \frac{-3+1}{3 x}=\frac{-1}{18} \Rightarrow 3 x=36\)
⇒ x = 12 cm or u = -12 cm

R₁ = 10 cm, R₂ = -15 cm, f = 12 cm
\(\frac { 1 }{ f } =(\mu -1)(\frac { 1 }{ R_{ 1 } } -\frac { 1 }{ R_{ 2 } } )\)
\(\frac { 1 }{ 12 } =(\mu -1)(\frac { 1 }{ 10 } -\frac { 1 }{ 15 } )\)
\( \mu = \frac { 3 }{ 2 } \)

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.

A charge moving with a circular orbit can produce electromagnetic waves because a circular motion is an accelerated motion and accelerated charges produce e.m.waves

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
\(I={ \epsilon }_{ 0 }\frac { { d\phi }_{ E } }{ dt } ={ \epsilon }_{ 0 }\frac { { d }_{ E } }{ dt } \left( \frac { q }{ 3{ \epsilon }_{ 0 } } \right) =\frac { { \epsilon }_{ 0 } }{ 3{ \epsilon }_{ 0 } } \frac { dq }{ dt } =\frac { 1 }{ 3 } \) 

When size of scatterer is much smaller than wavelength of light, then intensity of scattered light varies inversely as fourth power of wavelength of light, i.e.,\({ I }_{ s }\propto \frac { 1 }{ { \lambda }^{ 4 } } \)

Intensity of light is reduced to one fourth because the light beam spreads as it approaches into a spherical region of area \(4\pi { r }^{ 2 },i.e.,I\infty 1/{ r }^{ 2 }\) But laser beam does not spread, hence its intensity remains constant. Laser beam is unidirectional, monochromatic and coherent light, whereas the light from a bulb does not posses the above properties.

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.

No, image will be formed at the same position. This follows from lens maker's formula:
\(\frac { 1 }{ f } =(u-1)\left( \frac { 1 }{ { R }_{ 1 } } -\frac { 1 }{ { R }_{ 2 } } \right) \)
When we interchange R₁ and R₂, the value of f does not change except for the sign. Hence the image will be formed at the same position.

1.76 \(\times\) 10^-3 m

X-rays: They are generated by bombarding a target of high atomic number Z with a beam of fast moving electrons. Uses : (i) radio theapy for curing skin diseases. (ii) Medical diagnosis locating fracture TB etc.

Distance between the object and the image, d = 3 m
Maximum focal length of the convex lens = f_max
For real images, the maximum focal length is given as:
f_max = \(\frac{d}{4}\)
= \(\frac{3}{4}\) = 0.75 m
Hence, for the required purpose, the maximum possible focal length of the convex lens is 0.75 m.

Given, focal length of convex lens, f₁ = 30 cm
Focal length of the concave lens, f₂ = -20 cm
Using the formula of combination of lenses,
\(\frac{1}{f}=\frac{1}{f_{1}}+\frac{1}{f_{2}}=\frac{1}{30}-\frac{1}{20}=\frac{2-3}{60}=-\frac{1}{60}\)
\(\Rightarrow\) f = -60 cm
Since, the focal length of combination is negative in nature. so, the combination behaves like a diverging lens, i.e. as a concave lens.

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.

(a) Equation second shows that the e.m.wave travels along the positive x-axis
(b) Wavelength of wave, \(\lambda =\frac { c }{ v } =\frac { c }{ (\omega /2\pi ) } =\frac { 2\pi c }{ \omega } =\frac { 2\pi \times (3\times { 10 }^{ 8 }) }{ 2\pi \times { 10 }^{ 8 } } =3.0m\)
(c) since the magnetic field is perpendicular to an electric field as well as the direction of propagation of e.m.wave, hence magnetic field must be varying along a z-axis. Therefore
\(\overset { \rightarrow }{ { B }_{ x } } =0; \ \overset { \rightarrow }{ { B }_{ y } } =0; \ and \ \overset { \rightarrow }{ { B }_{ z } } =\frac { 0.5 }{ 3\times { 10 }^{ 8 } } cos\left[ 2\pi \times { 10 }^{ 8 }(t-x/c) \right] \ \therefore [E/B=c]\)

\(Here\ { E }_{ 0 }=200 \ N{ C }^{ -1 }\)
\( A=20 \ { cm }^{ 2 }=20\times { 10 }^{ -4 }{ m }^{ 2 };l=0.10 \ m\)
Vol. of cylinder, \(V=Al=(20\times { 10 }^{ -4 })\times 0.40=8\times { 10 }^{ -4 }{ m }^{ 2 }\)
Energy contained in cylinder is
U = volume × energy density
\(=v\times \frac { 1 }{ 2 } { \epsilon }_{ 0 }{ E }_{ 0 }^{ 2 }\)
\( =(8\times { 10 }^{ -4 })\times \frac { 1 }{ 2 } (8.85\times { 10 }^{ -12 })\times { (200) }^{ 2 }\)
\( =1.42\times { 10 }^{ -10 }J\)

Given, amplitude of the magnetic field part of harmonic electromagnetic wave,
B₀ = 510 nT = 510 \(\times\)10^-9 T
Speed of light in a vacuum, c = 3 × 10⁸ m/s
Amplitude of electric field of the electromagnetic wave is given by the relation,
E = cB₀
= 3 × 10⁸ × 510 × 10⁻⁹ = 153 N/C
Therefore, the electric field part of the wave is 153 N/C.

f₁=7.5×10⁶ Hz
f₂=12×10⁶ Hz
λ₁=c/f₁=40 m
λ₂=c/f₂=25 m
So the range is 40m to 25m

Using Eq, the magnitude of B is
\(B=\frac { E }{ c } \) 
\(=\frac { 6.3V/m }{ 3\times { 10 }^{ 8 }m/s } =2.1\times { 10 }^{ -8 }T\)
To find the direction, we note that E is along y-direction and the wave propagates along x-axis. Therefore, B should be in a direction perpendicular to both x- and y-axes. Using vector algebra, E × B should be along x-direction.
Since, \((+\overrightarrow{\mathbf{j}}) \times(+\hat{\mathbf{k}})=\overrightarrow{\mathbf{i}}, \mathbf{B}\) is along the z-direction.
Thus, \(\mathbf{B}=2.1 \times 10^{-8} \hat{\mathbf{k}} \mathrm{T}\)

It is given that a plane electromagnetic wave travels in vacuum along z-direction and the frequency of the electromagnetic wave is 30MHz.
We can say that electric field and magnetic field will be in x-plane because the electromagnetic wave travels along the z-direction and both fields are mutually perpendicular to each other.
The formula of the wavelength of a wave is,
λ=c/ν
Substitute the values in the above expression,
λ=(3×10⁸)/(30×10⁶)=10m.
Thus, the value of wavelength is 10 m and the direction of electric and magnetic fields will be in x-y plane.

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
\({ E }_{ Z }={ B }_{ Z } \ c=24\times 10^{ 14 }sin\left[ 2\times 10^{ 11 }t+300\pi x \right] Vm^{ -1 }\)

Magnetic field between the plates of a capacitor at distance \(r\) having varying electric field is given by 
\(B=\frac { \mu _{ 0 } }{ 4\pi } \frac { 2I_{ D } }{ r } =\frac { \mu _{ 0 }\varepsilon _{ 0 } }{ 2\pi r } \frac { d\phi }{ dt }\)
\(=\frac { \mu _{ 0 }\varepsilon _{ 0 } }{ 2\pi r } \times \frac { d }{ dt } (E\pi r^{ 2 })\)
\( B=\frac { \mu _{ 0 }\varepsilon _{ 0 } }{ 2\pi r } \times \pi r^{ 2 }\left[ \frac { dE }{ dt } \right] =\frac { \mu _{ 0 }\varepsilon _{ 0 }r }{ 2 } \frac { dE }{ dt } \)
\(=\frac { r }{ { 2C }^{ 2 } } \frac { dE }{ dt } =\frac { 1 }{ 2\times 9\times 10^{ 16 } } \times { 10 }^{ 10 }=5.6\times 10^{ -8 }T\)

Coherent sources of light are those sources which emit continuous light waves of the same wavelength and same frequency either having same phase or constant phase difference.
\( \beta =2.4\times{ 10 }^{ -4 }m;\lambda =6,400\mathring { A } \)
\(\lambda '=4,000\mathring { A } ;\beta '=?\)
\( \beta =\frac { D }{ d } \lambda \)
\( Since \ \beta \propto \lambda\)
\(\therefore \ \frac { \beta ' }{ 2.4\times{ 10 }^{ -3 } } =\frac { 4,000 }{ 6,400 }\)
\(or \ \beta '=\frac { 5 }{ 8 } \times2.4\times{ 10 }^{ -3 }\)
\( =1.5\times{ 10 }^{ -3 }m \ or \ \beta '=1.5mm.\)

\(\lambda =4,800\mathring { A } =4,800\times{ 10 }^{ -8 }cm\)
\( =4.8\times{ 10 }^{ -5 }cm;\)
\( D=150cm;8\beta =2\)
\(so\quad \beta =\frac { 1 }{ 4 } =0.25cm\)
\( \beta =0.25cm;d=?\)
\( \beta =\frac { D }{ d } \lambda =\frac { D\lambda }{ \beta } \)
\( =\frac { 1504.8\times{ 10 }^{ -5 } }{ 0.25 } \)
\( or =0.0288cm.\)

\({ I }_{ av }=\frac { c }{ 2 } \frac { { B }^{ 2 }_{ 0 } }{ { \mu ^{ 2 } }_{ 0 } } =\frac { 3\times 10^{ 8 }\times (12\times 10^{ 8 })^{ 2 } }{ 2\times 4\pi \times 10^{ -7 } } =1.7Wm^{ -2 }\)