(i) For points of maxima, the slit is to be divided into an odd number of equal parts so that one part remains un-canceled making the point P appear bright.
The condition for the first maximum is,
\(\cfrac { a }{ 3 } sin\theta =\cfrac { \lambda }{ 2 } \) (or) \(asin\theta =\cfrac { 3\lambda }{ 2 } \) 
The condition for the second maximum is
\(\cfrac { a }{ 5 } sin\theta =\cfrac { \lambda }{ 2 } \) (or) \(asin\theta =\cfrac { 5\lambda }{ 2 } \) 
The condition for the third maximum is,
\(\cfrac { a }{ 7 } sin\theta =\cfrac { \lambda }{ 2 } \) (or) \( asin\theta =\cfrac { 7\lambda }{ 2 } \)
In the same way, the condition for n^th maximum is 
\(asin\theta =\left( 2n+1 \right) \cfrac { \lambda }{ 2 } \) (n^th maximum)
where, n = 0, 1, 2, 3, ... , is the order of diffraction maximum.
(ii) The central maximum is called the order maximum. The points of the maximum intensity lie nearly midway between the successive minima.

(i) The line joining the pole and the centre of curyature is called the principal axis of the mirror.
(ii) The light ray travelling along the principal axis towards the mirror after reflection travels back along the same principal axis. It is also called optical axis.

From Einstein's photoelectric equation, stopping potential V_o is
eVo = hv - hv_o V_o = \(\frac{h}{e}(v-v_o)\)
Given > v_o, so with increase of frequency v, stopping potential increases.

There are two ways of preparing the nanomateials, top down and bottom up approaches

i) Search engine: The search engine is basically a web-based service tool used to search for information on World Wide Web.
ii) Communication: It helps millions of people to connect with the use of social networking: emails, instant messaging services and social networking tools.
iii) E-Commerce: Buying and selling of goods and services, transfer of funds are done over an electronic network.

The radius of the nth orbit 
\({ r }_{ n }=\frac { { n }^{ 2 }{ h }^{ 2 }{ \epsilon }_{ 0 } }{ \pi m{ Ze }^{ 2 } } \)
r₁ = 0.53Å
∴ The diameter of H₂ = 2 x r₁
= 0.53 x 2 = 1.06Å

(a) X - rays are used as diagnostic tool in medicine. X-rays are used extensively in studying structures of inner atomic electron shells and crystal structures. It is used in detecting fractures, diseased organs, formation of bones and stones, observing the progress of healing bones. Further, in a finished metal product, it is used to detect faults, cracks, flaws and holes.
(b) Radio are produced by oscillators in electric circuits. It obeys reflection and diffraction. It is used in radio and communication system and cellphones.

Right-hand thumb rule is used :
This rule is used to determine the direction of the magnetic field. If we rotate aright-handed screw using a screwdriver, then the direction of current is the same as the direction in which the screw advances and the direction of rotation of the screw gives the direction of the magnetic field.

(i) The electromagnetic oscillations of LC system can be compared with the mechanical oscillations of a spring-mass system.
(ii) There are two forms of energy involved in LC oscillations. One is electrical energy of the charged capacitor; the other magnetic energy of the inductor carrying current.
(iii) Likewise, the mechanical energy of the spring-mass system exists in two forms; the potential energy of the compressed or extended spring and the kinetic energy of the mass. The Table lists these two pairs of energy.
(iv) By examining, the analogies between the various quantities can be understood and these correspondences.
(v) The angular frequency of oscillations of a spring-mass is given by equation
\(\omega =\sqrt { \frac { k }{ m } } \)
k ⟶ \(\frac { 1 }{ C } \) and m ⟶ L. Therefore, the angular frequency of LC oscillations is given by
ω = \(\frac { 1 }{ \sqrt { LC } } \)

(i) Due to this internal resistance, the power delivered to the circuit is not equal to power rating mentioned in the battery.
(ii) For a battery of emf \({ \xi }_{ 1 }\) with an internal resistance r, the power delivered to the circuit of resistance R is given by
\(P=I\xi =I(V+Ir)\)
Here V is the voltage drop across the resistance R and it is equal to IR.
Therefore, P = I (IR +Ir)
P = I² R + I² r
(iii) Here Pr is the power delivered to the internal resistance and PR is the power delivered to the electrical device (here it is the resistance R). For a good battery, the internal resistance r is very small, then for P < r < P