we have \(\frac{1}{f}=(n-1)[\frac{1}{R_1}-\frac{1}{R_2}]\)
For equi-convex lens, R₁ = R and R₂ =-R
\(\frac{1}{f}=(n-1)[\frac{1}{R}-\frac{1}{-R}]=(n-1)[\frac{1}{R}+\frac{1}{R}]\) 
\(\frac{1}{f}=(n-1) \frac{2}{R} \Rightarrow 2f=\frac{R}{(n-1)}\)
(i) when the lens is cut into two parts vertically,
For 1^st part, R₁ = R and R₂ = ∞
\(Therefore\ \frac{1}{\mathrm{f}_{\mathrm{1}}}=(n-1)(\frac{1}{R}-\frac{1}{(-∞)})=\frac{(n-1)}{R}(or)\frac{R}{n-1}(or)f_1=2f\) 
Intensity does not change as the amount of light passing through the lens is same.
(ii) When the lens is cut into two parts horizontally, the radius of curvature remains same, therefore focal length remains same but intensity is reduced.

(i) Ampere- Maxwell law relates the magnetic field around any closed path to the conduction current and displacement current through that path.
\(\oint_{i} \vec{B} \cdot \overrightarrow{d l}=\mu_{0}i =\mu_0(i_c+i_d)\)
(ii) \(\oint_{i} \vec{B} \cdot \overrightarrow{d l}=\mu_{0} i_{C}+\mu_{0} \varepsilon_{0} \frac{dΦ_E}{d t}\)  where \(\vec{B}\) is the magnetic field.
(iii) This equation shows that both conduction current and displacement current produce magnetic field.

(i) Only in the special cases when the pass axis of (III) is parallel to (I) or (II) there shall be no light emerging.
(ii) In all other cases there shall be light emerging because the pass axis of (II) is no longer perpendicular to the pass axis of (III).

Maximum kinetic energy Ek = hv - W
Clearly, smaller the work function W, greater is the E_k This means that when work function of a metal increases, maximum kinetic energy of photoelectrons decreases.

i) The threshold wavelength is given by
\({ \lambda }=\frac { hc }{ { \phi }_{ 0 } } =\frac { 6.626\times { 10 }^{ -34 }\times 3\times { 10 }^{ 8 } }{ 4.65\times 1.6\times 10^{ -19 } } \)
= 2672 \(\mathring { A }\)
ii) Energy of the photon of wavelength 2200 \(\mathring { A }\) is
E = \(\frac { hc }{ \lambda } =\frac { 6.626\times { 10 }^{ -34 }\times 3\times { 10 }^{ 8 } }{ 2200\times 10^{ -10 } } \)
= 9.035 x 10^-19 J = 5.65 eV
We know that kinetic energy of fastest photo electron is
K_max = hv - ϕ₀ = 5.65 - 4.65
= 1 eV
From equation (7.3), K_max = eV₀
V₀ = \(\frac { { K }_{ max } }{ e } =\frac { 1\times 1.6\times { 10 }^{ -19 } }{ 1.6\times { 10 }^{ -19 } } \)
Therefore, stopping potential = 1 V

The de Broglie wavelength associated with the kinetic energy k is given as \(\lambda=\frac{h}{\sqrt{2 m k}}\) , where m is the mass of the particle.
Therefore \(\lambda_{\mathrm{e}}=\frac{\mathrm{h}}{\sqrt{2 \mathrm{~m}_{\mathrm{e}} \mathrm{k}_{\mathrm{e}}}} \text { and } \lambda_{\alpha}=\frac{\mathrm{h}}{\sqrt{2 \mathrm{~m}_{\alpha} \mathrm{k}_{\alpha}}} \text {. But } \mathrm{k}_{\mathrm{e}}=\mathrm{k}_{\alpha} . \)
Therefore \(\frac{\lambda_{\mathrm{e}}}{\lambda_{\alpha}}=\sqrt{\frac{\mathrm{m}_{\alpha}}{\mathrm{m}_{\mathrm{e}}}} \mathrm{m}_{\alpha}>\mathrm{m}_{\mathrm{e}^{*}} \text {. Therefore, } \lambda_{\mathrm{e}}>\lambda_{\alpha^{\circ}}\)

Black holes are end stage of stars whose mass ranges from 20 times to 1 million times the mass of the sun. Because of their dense mass, they have very strong gravitational force such that no particle or even light can escape from it. Therefore they are not visible to human eye. Every galaxy has black hole at its center. 

The output at the 1^st AND gate : A\(\overline { B } \)
The output at the 2^nd AND gate : ĀB
The output at the OR gate: Y = A. \(\overline { B } \) + Ā .B

When a PN junction diode is forward biased, the depletion region decreases and the diode conduct once after the barrier potential is crossed, when it is reverse biased the depletion region increases and the diode does not conduct sci it is called as unidirectional device.

Kinetic energies,
KE_Ai = 0
KE_Af = KE_Bt
It should obey law of conservation of energy
Total Initial KE = Total final KE
\(\mathrm{KE}_{\mathrm{A}_{i}}+\mathrm{KE}_{\mathrm{B}_{i}}=\mathrm{KE}_{\mathrm{A}_{\mathrm{t}}}+\mathrm{KE}_{\mathrm{B}_{\mathrm{t}}} \)
\(0+\mathrm{KE}_{\mathrm{B}_{i}}=\mathrm{KE}_{\mathrm{A}_{\mathrm{f}}}+\mathrm{KE}_{\mathrm{A}_{t}} \)
\(\mathrm{KE}_{\mathrm{B}_{\mathrm{i}}}=2 \mathrm{KE}_{\mathrm{A}_{\mathrm{f}}} \)
Minimum energy required for excite the hydrogen atom is 10.2eV
Hence minimum Kinetic energy of H_B is
\(\mathrm{KE}_{\mathrm{B}_{1}}=2 \times 10.2 \mathrm{eV}=20.4 \mathrm{eV}\)

The total distance d between the transmitting and receiving antennas will be the sum of the individual distances of coverage.
d = d₁ + d₂
\(=\sqrt { 2R{ h }_{ 1 } } +\sqrt { 2{ Rh }_{ 2 } } \)
\(=\sqrt { 2R } \left( \sqrt { { h }_{ 1 } } +\sqrt { { h }_{ 2 } } \right) \)
\(=\sqrt { 2\times 6.4\times { 10 }^{ 6 } } \times (\sqrt { 40 } +\sqrt { 30 } )\)
\(=16\times { 10 }^{ 2 }\sqrt { 5 } \times (6.32+5.48)\)
= 42217 m = 42.217 km

Isobars are the atoms of different elements having the same mass number A, but different atomic number Z.
Example: \({ }_{16}^{40} \mathrm{~S},{ }_{17}^{40} \mathrm{Cl},{ }_{18}^{40} \mathrm{Ar}\)

(i) In Grating, the combined width of a ruling and a slit is called 'grating element' (1.e.) e = a + b.

Quantum theory of light:
Quantum theory states that light waves consist of small packets of energy called photons. The energy associated with each photon is E = hv, Where 'h' is Planck's constant (h = 6.625 x 10^-34 J s) and v is frequency of electromagnetic radiation.

\(\mathrm{I} ∝ \frac{1}{\lambda^{4}}\)
According to Rayleigh's scattering equation, violet colour which has the shortest wavelength gets much scattered during day time. The next scattered colour is blue. As our eyes are more sensitive to blue colour than violet colour the sky appears blue during day time.

Stars appear twinkling because of the movement of the atmospheric layer with varying refractive indices due to refraction.

Infrared radiation waves are produced by hot bodies and molecules so it is referred as heat waves. (eg. Sun)
(i) Electromagnetic waves having shorter λ than Infrared radiation are visible, U - v, X-rays, and ૪ - rays.
(ii) Electromagnetic waves having longer λ than Infrared radiation are microwaves, radiowaves.

It is defined as the ability of the materials to retain the magnetism in them even magnetising field vanishes.

Ф = LI
For same current Ф_A > Ф_B
So L_A > L_B
i.e conductor a has a larger value of self-induction
For a series LCR circuit power factor is
\(\frac { R }{ z } =\frac { R }{ \sqrt { { R }^{ 2 }+\left( \omega L-\frac { 1 }{ \omega c } \right) } } \)
The maximum value of power factor = 1
The minimum value of power factor = 0

The convention is that the current entering the junction is taken as positive and current leaving the junction is taken as negative.

When an external electric field is applied, the centers of positive and negative charges are separated by a small distance which induces dipole moment in the direction of the external electric field. Then the dielectric is said to be polarized by an external electric field.

Q factor is defined as the ratio of voltage across L or C to resonance to the applied voltage
Q - factor = \(\frac{Voltage \ across \ L \ or \ C \ resonance }{Applied \ voltage}\)
\(Q-factor=\frac{X_{L}}{R}=\frac{1}{R}\sqrt\frac{{L}}{C}\)

The displacement current can be defined as the current which comes into play in the region in which the electric field or the electric flux is changing with time.

From right hand rule, the magnetic field by the straight wire is directed into the plane of the square loop perpendicularly and its magnetic flux is decreasing. The decrease in flux is opposed by the current induced in the loop by producing a magnetic field in the same direction as the magnetic field of the wire. Again from right hand rule, for this inward magnetic field, the direction of the induced current in the loop is clockwise.

Apply Kirchoff's 1^st law at junction B
I_I - I₂ - I₃ = 0
I₃ = I₁ - I₂          ....(1)
Applying Kirchoff's II^nd law at path ABEFA,
100I₃ + 100 I₁ = 15   ....(2)
Substitute equation (1) in equation (2),
100 (I₁ - I₂) + 100 I₁ = 15
100 I₁ - 100 I₂ + 100 I₁ = 15
200 I₁ - 100 I₂ = 15  ....(3)
Apply Kirchhoffs 2^nd law ar path BCDEB,
100 I₂ - 100 I₃ = -9    ....(4)
Substitute equation (1) in above equation (4),
100 I₂ - 100(I₁ - I₂) 100 = -9
100 I₂ - 100 I₁ + 100 I₂ = -9
200 I₂ - 100 I₁ = - 9 ....(5)
Solve equation (3) and (5), we get
Eqn (3)       ⇒ 200 I₁ - 100 I₂ = 15
2 x Eqn (5) ⇒ -200 I₁ + 400I₂ = -18
                        ________________
                                      300 I₂ = -3
I₂ = -3 /300 = -0.01 A
I₂ = -0.01 A
Substitute I₂ value in equation (5)
200 I₂ -100 I₁ = - 9
200 x (- 0.01) - 100 I₁ = - 9
- 2 - 100 I₁ = -9
-100 I₁ = -9 + 2
-100 I₁ = -7
I₁ = 7/100
I₁ = 0.07A
Substitute I₁, I₂ value in equation (1)
I₃ = I₁ - I₂
= 0.07 + 0.01
I₃ = 0.08 A
I₁ = 0.07 A, I₂ = - 0.01 A and I₃ = 0.08 A

The geometrical length of the bar magnet is 12 cm
Magnetic length = \(\frac { 5 }{ 6 } \times \)(geometrical length)
= \(\frac { 5 }{ 6 } \times \)12 = 10 cm
In this figure, the dot implies the pole points.

Applying Kirchoff’s rule to the point P in the circuit,
The arrows pointing towards P are positive and away from P are negative.
Therefore, 0.2A - 0.4A + 0.6A - 0.5A + 0.7A - I = 0
0.7 A - I = 0
1.5A - 0.9A – I = 0
0.6A - I = 0
I = 0.6 A

(i) The electric field lines start at q₂ and end at q₁. In figure (a), q₂ is positive and q₁ is negative. The number of lines starting from q₂ is 18 and number of the lines ending at q₁ is 6. So q₂ has greater magnitude. The ratio of \(\left| \frac { { q }_{ 1 } }{ { q }_{ 2 } } \right| =\frac { { N }_{ 1 } }{ { N }_{ 2 } } =\frac { 6 }{ 18 } =\frac { 1 }{ 3 } \). It implies that |q₂| = 3|q₁|.
(ii) In figure (b), the number of field lines emanating from both positive charges are equal (N = 18). So the charges are equal. At point A, the electric field lines are denser compared to the lines at point B. So the electric field at point A is greater in magnitude compared to the field at point B. Further, no electric field line passes through C, which implies that the resultant electric field at C due to these two charges is zero.
(iii) In the figure (c), the electric field lines start at q₁ and q₃ and end at q₂. This implies that q₁ and q₃ are positive charges. The ratio of the number of field lines is \(\left| \frac { { q }_{ 1 } }{ { q }_{ 2 } } \right| =\frac { 8 }{ 16 } =\left| \frac { { q }_{ 3 } }{ { q }_{ 2 } } \right| =\frac { 1 }{ 2 } \), implying that q₁ and q₃ are half of the magnitude of q₂. So q₁ = q₃ = +10 nc.

By using superposition principle, the net electric field at point A is
\(\vec { E_{ A } } =\frac { 1 }{ 4\pi { \varepsilon }_{ 0 } } \frac { { q }_{ 1 } }{ { r }_{ 1A }^{ 2 } } \hat { { r }_{ 1A } } +\frac { 1 }{ 4\pi { \varepsilon }_{ 0 } } \frac { { q }_{ 1 } }{ { r }_{ 2A }^{ 2 } } \hat { { r }_{ 2A } } \)
where r_1A and r_2A are the distances of point A from the two charges respectively
\(\vec { E_{ A } } =\frac { 9\times 10^{ 4 }\times 1\times { 10 }^{ -6 } }{ (2\times { 10 }^{ -3 })^{ 2 } } (\hat { j } )+\frac { 9\times 10^{ 9 }\times 1\times { 10 }^{ -6 } }{ (2\times 10^{ -3 })^{ 2 } } (\hat { i } )\)\(\)
= 2.25 x 10⁹\(\hat { j } \) + 2.25 x 10⁹\(\hat { i } \) = 2.25 x 10⁹ (\((\hat { i } +\hat { j } )\)
The magnitude of electric field
\(|\vec { { E }_{ A } } |=\sqrt { (2.25\times { 10 }^{ 9 })^{ 2 }+(2.25\times 10^{ 9 })^{ 2 } } \)
= 2.25 x \(\sqrt { 2 } \) x 10⁹ NC^-1
The direction of \(\vec { { E }_{ A } } \) is given by \(\frac { \vec { { E }_{ A } } }{ |\vec { { E }_{ A } } | } =\frac { 2.25\times 10^{ 9 }(\hat { i } +\hat { j } ) }{ 2.25\times \sqrt { 2 } \times { 10 }^{ 9 } } =\frac { (\hat { i } +\hat { j } ) }{ \sqrt { 2 } } \), which is the unit vector along OA as shown in the figure.

The acceleration experienced by an electron placed at point A is
\(\vec { a_{ A } } =\frac { \vec { F } }{ m } =\frac { q\vec { { E }_{ A } } }{ m } \)
=\(\frac { (-1.6\times 10^{ -19 })\times (2.25\times 10^{ 9 })(\hat { i } +\hat { j } ) }{ 9.1\times { 10 }^{ -31 } } \)
= -3.95 x 10²⁰ \((\hat { i } +\hat { j } )\)Nkg^-1
The electron is accelerated in a direction exactly opposite to \(\vec { { E }_{ A } } \).

The relation between drift velocity of electrons and current in a wire of cross- sectional area A is
\({ v }_{ d }=\frac { I }{ neA } =\frac { 0.2 }{ 8.4\times { 10 }^{ 28 }\times 1.6\times { 10 }^{ -19 }\times 0.5\times { 10 }^{ -6 } }\)
v_d = 0.03 x 10^-3 m s^-1