Given: a = 2 unit, b = 2 unit
(a) Resultant amplitude,
\( r=\sqrt{a^{2}+b^{2}+2 a b \cos \theta} \)
\(r=\sqrt{2^{2}+2^{2}+2.2 .2 \cdot \cos 60^{\circ}} \quad\left(\therefore \theta=60^{\circ}\right) \)
\(r=\sqrt{12}=2 \sqrt{3} \text { unit } \)
(b) The path difference of λ/2 corresponds to a phase difference of n, thus, the resultant amplitude,
\(r=\sqrt{2^{2}+2^{2}-8}=0 \text { unit }(\because \cos \pi=-1)\)

Wavefront A wavefront is the locus of points (wavelets) having the same phase (a surface of constant phase) of oscillations. A wavelet is the point of disturbance due to propagation of light. A line perpendicular to wavefront is called a ray.

When a point source is placed at the focus of a convex lens, the rays emerging from the lens are parallel. Therefore, the wavefront must be plane wavefront.

\(Using \ x=\frac { \lambda D }{ d } \)
\( { x }_{ 2 }-{ x }_{ 1 }=\left( 3\lambda -\lambda \right) \frac { D }{ d } =\frac { 2\lambda D }{ d } \)
\(d=\frac { 2\lambda D }{ { x }_{ 2 }-{ x }_{ 1 } } =\frac { 2\times 6000\times { 10 }^{ -10 }\times 0.5 }{ 3\times { 10 }^{ -3 } }\)
\(d=2\times { 10 }^{ -4 }m=0.2mm\)

Conditions for constructive interference is that the path difference between two rays must be \(=n\lambda \)  Where n is a whole number
\(HereD=1.5;d=0.3mm=0.3\times{ 10 }^{ -3 }m\) 
\( =3\times{ 10 }^{ -4 }m\)
\( \lambda =6,400\mathring { A } =6.4\times{ 10 }^{ -7 }m;\beta =?\)
\( \beta =\frac { \lambda D }{ d } =\frac { 6.4\times{ 10 }^{ -7 }\times1.5 }{ 3\times{ 10 }^{ -4 } } \)
\( \beta =3.2\times{ 10 }^{ -3 }m \ or \ 3.2 \ mm.\)

As the order increases only \(\frac{1}{n^{\text {th }}}\) (where n is an odd number) of the slit will contribute in producing brightness at a point in the diffraction. So, the higher order maxima are not so bright as the central.

The frequency and wavelength of reflected wave will not change. The refracted wave will have same frequency. The velocity of light in water is given by v = f \(\lambda\)
where, v = velocity of light
f = frequency of light
\(\lambda\)= wavelength of light
If velocity will decrease, then wavelength (\(\lambda\))will also decrease.

Resultant amplitude = b - a = 5 - 3 = 2 nm.

(a) Two sources of light having same frequency and a constant or zero phase difference are said to be coherent.
A light wave emitted from an ordinary source (like a sodium lamp) undergoes the abrupt phase changes in times of the order of 10^-10 seconds. Thus, two independent sources of light will not have a fixed phase relationship and would be incoherent.
(b) When waves from the slits meet at a point on the screen with same phase, the maxima are obtained and with a phase difference of n, the minima are obtained.
According to the Young's experiment, the path difference between the waves is given by
\( \Delta P =S_{2} Q-S_{1} Q=S_{2} M \)
\(\text { i.e. } \quad \sin \theta =\frac{\Delta P}{a} \)
\(\text { From } \Delta Q O O^{\prime}, \tan \theta=\frac{Y}{D}\)
 For a small angle, \(\sin \theta \approx \tan \theta \text { i.e. } \Delta P=\frac{Y a}{D}\)
(i) For bright fringes, ∆P = nλ
Thus, \(\frac{Y_{n} a}{D}=n \lambda \Rightarrow Y_{n}=\frac{n \lambda D}{a}\)
(ii) For dark fringes, \(\Delta P=(2 n-1) \frac{\lambda}{2}\)
Thus, \(\frac{Y_{n}^{\prime} a}{D}=(2 n-1) \frac{\lambda}{2} \Rightarrow Y_{n}=\frac{(2 n-1) D \lambda}{2 a}\)
The separation between two consecutive dark or bright fringes is called fringe width.
\( \beta =Y_{n}-Y_{n-1} \)
\(=\frac{n \lambda D}{a}-\frac{(n-1) \lambda D}{a}=\frac{\lambda D}{a} \)
Here β is the fringe width.
(c) For interference fringes to be seen, the condition \(\frac{s}{d}<\frac{\lambda}{a}\) should be satisfied.

Here,  \(I_{ 0 }=K, \ where \ path \ diff. \ is \ \lambda \)
\(I=?, \ where \ path \ diff.=\lambda /3\)
Phase diff., ​​​​\( \phi =\frac { 2\pi }{ 3 } =120\)
From  \(I=I_{ 0 }cos^{ 2 }\phi /2\)
\( I=K\left( cos\frac { 120° }{ 2 } \right) =K\left( \frac { 1 }{ 2 } \right) ^{ 2 }=K/4\)

(i) (b): The condition for possible interference maxima on the screen is, dsin\(\theta\) = nA
where d is slit separation and Ais the wavelength.
As d = 2\(\lambda\) (given) \(\therefore\) 2\(\lambda\)sin\(\theta\)= n\(\lambda\) or 2sin\(\theta\) = n
For number of interference maxima to be maximum,
sin\(\theta\) = 1 \(\therefore\) n = 2
The intprference maxima will be forgied when
n = 0, ± 1, ± 2
Hence the maximum number of possible maxima is 5.
(ii) (c): Fringe width, \(\beta=\frac{\lambda D}{d}\)
\(\therefore\) If we replace yellow light with blue light, i.e., longer wavelength with shorter one, therefore the fringe width decreases.
(iii) (d): \(d^{\prime}=\frac{d}{2} \text { and } D^{\prime}=2 D\)
Fringe width, \(\beta=\frac{\lambda D}{d}\)
New fringe width \(\beta^{\prime}=\lambda\left(\frac{2 D}{d / 2}\right)=4 \beta\)
(iv) (c): When Young's double slit experiment is repeated in water, instead of air \(\lambda^{\prime}=\frac{\lambda}{\mu}\) i.e., wavelength decreases.\(\beta=\frac{\lambda^{\prime} D}{d}\) i.e..,fringe width decreases.
\(\therefore\) The fringe become narrower.
(v) (a): Using white light, we get white fringe at the centre i.e., white fringe is the central maximum. When the screen is moved, its position is not changed.

(b) If both Assertion and Reason are correct but Reason is not a correct explanation of the Assertion.

(d): When a light wave travel from a rarer to a denser medium it loses speed, but energy carried by the wave does not depend on its speed. Instead, it depends on the amplitude of wave. The frequency also remain constant.