(i) (c): In Young's double slit experiment, the fringe width is \(\beta=\frac{D \lambda}{d}\) where Dis the distance of the slits from the screen, d is the separation of the slits and \(\lambda\), the wavelength. Therefore the fringe width \(\beta\) can be changed either by changing the separation between the sources or the distance of the screen from the sources.
(ii) (c): As the width of one of the slits is increased to 2w, the amplitude due to slit become 2a.
(iii) (d): \(\Delta x=\left(\mu_{A}-1\right) t_{A}-\left(\mu_{B}-1\right) t_{B}\)
\(=\mu_{A} t_{A}-\mu_{B} t_{B}-t_{A}+t_{B}=t_{B}-t_{A}\)
If \(\Delta x\)> 0, then fringe pattern will shift upward.
If \(\Delta x\)< 0, then fringe pattern will shift downwards.
(iv) (b): Contrast between the bright and dark fringes will be reduced.
(v) (a): Since, one of the slit is covered, interference will not occur and fringe pattern will disappear.
 

(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.

(i) (c) : Here \(d=0.1 \mathrm{~mm}, \lambda=6000 \dot A, D=0.5 \mathrm{~m}\)
For third dark band \(d \sin \theta=3 \lambda ; \sin \theta=\frac{3 \lambda}{d}=\frac{y}{D}\)
\(y=\frac{3 D \lambda}{d}=\frac{3 \times 0.5 \times 6 \times 10^{-7}}{0.1 \times 10^{-3}}=9 \times 10^{-3} \mathrm{~m}=9 \mathrm{mn}\)
(ii) (b): Given \(d=0.2 \mathrm{~mm}=0.2 \times 10^{-3} \mathrm{~m}, D=2 \mathrm{~m}\)
\(\lambda=5000 \dot A=5 \times 10^{-7} \mathrm{~m}\)
The distance between the first minimum on other side of the central maximum
\(x=\frac{2 \lambda D}{d}=\frac{2 \times 5 \times 10^{-7} \times 2}{0.2 \times 10^{-3}} \Rightarrow x=10^{-2} \mathrm{~m}\)
(iii) (a): Here \(\lambda=600 \mathrm{nm}=6 \times 10^{-7} \mathrm{~m}\)
\(a=0.2 \mathrm{~mm}=2 \times 10^{-4} \mathrm{~m}, \theta=?\)
Angular width of central maxima
\(\theta=\frac{2 \lambda}{a}=\frac{2 \times 6 \times 10^{-7}}{2 \times 10^{-4}}=6 \times 10^{-3} \mathrm{rad}\)
(iv) (d) : When red light is replaced by blue light (\(\lambda\)B < \(\lambda\)R) the diffraction pattern bands becomes narrow and crowded together.
(v) (b) : To observe diffraction, the size of the obstacle should be of the order of wavelength.

(i) (c):The optical path difference at P is
\(\Delta x=S_{1} P-S_{2} P=d \cos \theta\)
\(\because \quad \cos \theta=1-\frac{\theta^{2}}{2} \text { for small } \theta\)
\(\therefore \quad \Delta x=d\left(1-\frac{\theta^{2}}{2}\right)=d\left[1-\frac{y^{2}}{2 D^{2}}\right], \text { where } D+d=D\)
(ii) (b) : \(\text { For } n^{\text {th }} \text { maxima, }\)
\(\Rightarrow \Delta x=n \lambda\)
\(d\left[1-\frac{y^{2}}{2 D^{2}}\right]=n \lambda\)
Y = radius of the n^th bright ring
\(=D \sqrt{2\left(1-\frac{n \lambda}{d}\right)}\)
(iii) (d): At the central maxima, \(\theta\)= 0.
\(\Delta x=d=n \lambda \)
\(\Rightarrow n=\frac{d}{\lambda}=\frac{0.5}{0.5 \times 10^{-3}}=1000\)
Hence, for the closet second bright fringe, n = 998.
(iv) (c): Light waves from two coherent sources must have a constant phase difference.
(v) (d): Interference is shown by transverse as well as mechanical waves.

(i) (d): Path difference produced is
\(\Delta x=\frac{3}{2} \pi R-\frac{\pi}{2} R=\pi R\)
For maxima \(\Delta x=n \lambda\)
\(\begin{array}{l} \therefore \quad n \lambda=\pi R \\ \Rightarrow \lambda=\frac{\pi R}{n}, n=1,2,3, \ldots . \end{array}\)
Thus, the possible values of \(\lambda \text { are } \pi R, \frac{\pi R}{2}, \frac{\pi R}{3}, \ldots\)
(ii) (d)
(iii) (b): Maximum intensity \(I_{\max }=\left(\sqrt{I_{1}}+\sqrt{I_{2}}\right)^{2}\)
\(\text { Here, } I_{1}=I_{2}=\frac{I_{0}}{2} \text { (given) }\)
\(\therefore \quad I_{\max }=\left(\sqrt{\frac{I_{0}}{2}}+\sqrt{\frac{I_{0}}{2}}\right)^{2}=2 I_{0}\)
(iv) (d): Phase difference \(\phi=\frac{2 \pi}{\lambda}\) x path diffrerence
\(\phi=\frac{2 \pi}{\lambda} \times \frac{\lambda}{6}=\frac{\pi}{3}=60^{\circ} \text { As } I=I_{\max } \cos ^{2} \frac{\phi}{2}\)
\(\therefore \quad I=I_{0} \cos ^{2} \frac{60^{\circ}}{2}=I_{0} \times\left(\frac{\sqrt{3}}{2}\right)^{2}=\frac{3}{4} I_{0} \Rightarrow \frac{I}{I_{0}}=\frac{3}{4}\)
(v) (c): HereA² = \(a_{1}^{2}+a_{2}^{2}+2 a_{1} a_{2} \cos \delta \because a_{1}=a_{2}=a\)
\(\therefore A^{2}=2 a^{2}(1+\cos \delta)=2 a^{2}\left(1+2 \cos ^{2} \frac{\delta}{2}-1\right)\)
\(\text { or } A^{2} \propto \cos ^{2} \frac{\delta}{2}\)
\(\text { Now } I \propto A^{2} \quad \therefore I \propto A^{2} \propto \cos ^{2} \frac{\delta}{2} \Rightarrow I \propto \cos ^{2} \frac{\delta}{2}\)

(i) (b): \(\text { As } z=\frac{\lambda D}{2 d}\)
\(\text { At } S_{4}: \frac{\Delta x}{d}=\frac{z}{D}\)
\(\Rightarrow \Delta x=\frac{\lambda D}{2 d} \frac{d}{d}=\frac{\lambda}{2}\)
(ii) (c): \(z=\frac{\lambda D}{d}\)
\(\Delta x \text { at } S_{4}: \Delta x=\frac{\lambda D}{d} \frac{d}{d}=\lambda\)
Hence, maxima at S₄ as well as S₃'
Resultant intensity at S₄ I = 4I₀
\(\therefore \quad \frac{I_{\max }}{I_{\min }}=\frac{\left[\left(4 I_{0}\right)^{1 / 2}+4\left(4 I_{0}\right)^{1 / 2}\right]^{2}}{\left[\left(4 I_{0}\right)^{1 / 2}-\left(4 I_{0}\right)^{1 / 2}\right]^{2}}=\infty\)
(iii) (a): When the screen is placed perpendicular to the line joining the 'sources, the fringes will be concentric circles.
(iv) (b): Constructive interference occurs when the path difference (S₁P - S₂P) is an integral multiple of \(\lambda\).or S₁P - S₂P = n\(\lambda\), where n = 0,1,2,3, .....
(v) (d): Here, A₁ = 3A, A₂ = 2A and \(\varphi\)= 60° The resultant amplitude at a point is
\(R =\sqrt{A_{1}^{2}+A_{2}^{2}+2 A_{1} A_{2} \cos \phi} \)
\(=\sqrt{(3 A)^{2}+(2 A)^{2}+2 \times 3 A \times 2 A \times \cos 60^{\circ}} \)
\(=\sqrt{9 A^{2}+4 A^{2}+6 A^{2}}=A \sqrt{19}\)
As, Intensity \(\infty\) (Amplituder)² Therefore, intensity at the same point is
\(I \propto 19 A^{2}\)

(i) (a): Since the path difference between two waveform is equal, light traves as parallel beam in each medium.
(ii) (c): Since all points on the wavefront are in the same phase,
\(\phi_{d}=\phi_{c} \text { and } \phi_{f}=\phi_{e} \)
\(\therefore \phi_{d}-\phi_{f}=\phi_{c}-\phi_{e^{-}}\)
(iii) (a): Wavefront is the locus of all points, where the particles of the medium vibrate with the same phase
(iv) (a)
(v) (d)

(i) (a): As the beam is initially parallel, the shape of wavefront is planar.
(ii) (c): According to Huygens Principle, the surface of constant phase is called a wavefront.
(iii) (c)
(iv) (c): After refraction, the emerging wavefronts respectively become spherical. wavefront and plane wavefront as shown in figures (a) and (b).

(v) (c)

(i) (b): The wavelength of visible light is very small, that is hardly shows diffraction, so it seems to propagate in rectilinear path,
(ii) (c): Angular width of central maxima, 2\(\theta\) = 2\(\lambda\)/e. Thus, \(\theta\) does not depend on screen i.e., distance between the slit and the screen.
(iii) (c) : The intensity distribution of single slit diffraction pattern is shown in the figure. From the graph it is dear that the intensity of the central point is finite but much larger than the surrounding maxima.

(iv) (a): Resolving power of telescope \(=\frac{a}{1.22 \lambda}\)
\(\therefore\) It increases when wavelength of light decreases and/or objective lens of greater diameter is used.
(v) (a): Width of central maxima = 2\(\lambda\) DIe
width of other secondary maxima = \(\lambda\)DI e
\(\therefore\)​​​​​​Width of central maxima: width of other secondary maxima
= 2 : 1

(i) (c) :  Given   \(I_{1}=10 \mathrm{~W} / \mathrm{m}^{2} \text { and } I_{2}=25 \mathrm{~W} / \mathrm{m}^{2}\)
\(\frac{I_{1}}{I_{2}}=\frac{a_{1}^{2}}{a_{2}^{2}}=\frac{10}{25} \Rightarrow \frac{a_{1}}{a_{2}}=\frac{3.16}{5} \text { or } a_{1}=\frac{3.16}{5} a_{2}=0.6324 a_{2}\)
\(\frac{I_{\max }}{I_{\min }}=\frac{\left(a_{1}+a_{2}\right)^{2}}{\left(a_{1}-a_{2}\right)^{2}}=\frac{\left[0.6324 a_{2}+a_{2}\right]^{2}}{\left[0.6324 a_{2}-a_{2}\right]^{2}}=19.724\)
(ii) (b): In an excessively thin film, the thickness of the film is negligible. Thus the path difference between the reflected rays becomes \(\lambda\)/2 which produces a minima.
(iii) (a): Since, \(\beta=\frac{\lambda D}{d} \text { for } d=2 d\)
\(\beta^{\prime}=\frac{\lambda D^{\prime}}{2 d}=\beta(\text { Gives })\)
\(\therefore \quad D_{1}=2 D\)
(iv) (b): The condition for possible interference maxima on the screen is, dsin \(\theta\) = n\(\lambda\)
where d is slit separation and Ais the wavelength.
\(\text { As } d=2 \lambda \text { (given) } \quad \therefore 2 \lambda \sin \theta=n \lambda \text { or } 2 \sin \theta=n\)
For number of interference maxima to be maximum,sin\(\theta\) = 1 :. n = 2
The interference maxima will be formed when n = 0, ± 1, ± 2
Hence the maximum number of possible maxima is 5.
(v) (d): \(y_{1}=a \sin \left(\omega t+\frac{\pi}{3}\right) \text { and } y_{2}=a \sin \omega t\)
\(A=\sqrt{a_{1}^{2}+a_{2}^{2}+2 a_{1} a_{2} \cos \phi}, \text { where } \phi=\frac{\pi}{3}\)
\(=\sqrt{a^{2}+a^{2}+2 a a \cos \frac{\pi}{3}}=\sqrt{3} a\)

(i) Light travels as a parallel beam in each medium, as wavefronts are perpendicular to a beam of light, therefore wavefronts are plane parallel wavefronts in each media.
(ii) All points on a wavefront are at the same phase.
\(\phi_{d}=\phi_{c} \text { and } \phi_{f}=\phi_{e}\)
\(\therefore \phi_{d}-\phi_{f}=\phi_{c}-\phi_{e}\)
(iii) In medium 2, wavefront bends away from the normal after refraction. Therefore, ray of light which is perpendicular to wavefront bends towards the normal in medium 2 during refraction. So, medium 2 is denser or its speed in medium I is more or v₁ > v ₂ . So, the speed of light in two media is not same.

If part of wavefront reach from d to f with velocity speed v₁, during this time another part will travel from c to e with speed v ₂ .
If df = \(\lambda\)₁ then ce = \(\lambda\)₂ in medium 2
df = v₁ x t             ........(i)
ce = v ₂ x t          ...........(ii)
v₁ and v₂ are speeds in media 1 and 2 respectively.​​​​​​
then Dividing eq. (i) by eq. (ii) is \(\frac{v_{1} t}{v_{2} t}=\frac{\lambda_{1}}{\lambda_{2}}\) or \(\frac{v_{1}}{\lambda_{1}}=\frac{v_{2}}{\lambda_{2}}\) or v₁ = v₂ here v₁, v₂ are frequenecies in medium 1 and 2 respectively.

(i) Merits of Huygens Wave theory of light:
(a) Wave theory correctly predicted that velocity of light in an optically denser medium is less than that in the rarer medium which is in agreement with the experimental results.
(b) On the basis of wave theory phenomenon of reflection, refraction, interference, diffraction, polarization of light could be explained.
Demiritssof Huygens wave theory of light
(a) Huygens wave theory assumes the existence of luminiferous ether. However,experimentally it couldn't be proved.
(b) This theory couldn't explain rectilinear propagation of light.
(ii) (1) Each pointon a given primary wavefront acts as a source of secondary wavelets, sending out disturbances (waves) in all directions in a similar manner as the original source of light does.
(2) The new position of the wavefront at any instant (secondary wavefront) is given by the forward envelope to the secondary wavelets at that instant.
Huygens' construction

Using this principle the laws of reflection and refraction can be verified.
(iii) Primary source of light: It is a real source of light. It generates light itself and sends primary wavefronts in all directions.
Secondary source of light: It is a fictitious source of light presents on the wavefront and sends out secondary waves only in forward direction.

(i) By the principle of interference, condition for constructive interference is the path difference = x_n d/D = n\(\lambda\) here n = 0, 1,2 ... indicate the order of bright fringes. Therefore position of n^th bright is given by
\(x_{n}=\frac{n \lambda \mathrm{D}}{d}\) and the dark fringe width
\(\beta=\frac{(n+1) \lambda D}{d}-\frac{n \lambda D}{d}=\frac{\lambda D}{d}\)        ..........(i)
Similarly, condition for destructive inetreference is the path difference
\(=\frac{x_{n}^{\prime} d}{\mathrm{D}}=(2 n-1) \frac{\lambda}{2}\) therefore the position of n^th dark fringe is given by \(x_{n}^{\prime}=(2 n-1) \frac{\lambda \mathrm{D}}{2 d}\) here n = 1, 2, 3 ... shows the order of dark fringes, and the bright fringe width
\(\beta=\frac{[2(n+1)-1] \lambda D}{2 d}-\frac{(2 n-1) \lambda D}{2 d}=\frac{\lambda D}{d} \)     ..........(ii)
From eqs (i) and (ii) we see that width of bright and dark fringes are equal.
Let, A and a be the wavelength and slit width of the diffracting system respectively. Let O be the position of the central maximum.

Condition for the first minimum is given by
\(a\sin { \theta } =m\lambda \)     ...(i)
Let \(\theta\) be the angle of diffraction.
As, diffraction angle is small
\(\therefore\) \(\sin { \theta } \approx \theta \)
For first diffraction minimum,
\(\theta ={ \theta }_{ 1 }\)
For the first minimum, take m = I
\(a{ \theta }_{ 1 }=\lambda \quad \Rightarrow { \theta }_{ 1 }=\frac { \lambda }{ a } \)
Now, angular width, \(AB={ \theta }_{ 1 }\)
Angular width, \(BC={ \theta }_{ 1 }\)
Angular width, \(AC=2{ \theta }_{ 1 }\)
(ii) The maxima become weaker and weaker with increasing n...This is because the effective part of the wavefront, contributing to the maxima, becomes smaller and smaller, with increasing n.
(iii) In case of a single slit of width a, the wavelets from two halves of slit produce a minimum as the corresponding wavelets have a path difference of \(\frac{\lambda}{2}\).
In the second case, the overlapping of the wavefronts from the two slits produce first maximum because they meet with a path difference of \(\lambda\).

(i) n the double-slit experiment, the pattern on the screen is actually a superposition of single-slit diffraction from each slit or hole and the double-slit interference pattern as shown in the given figure. It shows a broader diffraction peak in which there appear several fringes of smaller width due to double-slit interference. The number of interference fringes occuring in the broad diffraction peak depends on the ratio d/a, which is the ratio of the distance between the two slits to the width of a slit. In the limit of a becoming very small, the diffraction pattern will become flat and we will observe the two slit interference pattern.

(ii) (a) \(\lambda\)Given  = 400 nm
= 4 x 10^{- 7} m,
d = 20 \(\mu\)m = 2 x 10^{- 5} m,
a = 4\(\mu\)m = 4 x 10^-6 m
If n = no. of bright fringes within central peak of diffraction envelope
Then \(\frac{2 \lambda \mathrm{D}}{a}=n \lambda \frac{\mathrm{D}}{d}\)
\(\Rightarrow \ \frac{2}{a}=\frac{n}{d}\)
\(\Rightarrow \quad n=\frac{2 d}{a}=\frac{2 \times 2 \times 10^{-5}}{4 \times 10^{-6}}=10\)
(b) \(\frac{\lambda \mathrm{D}}{a}=n \lambda \frac{\mathrm{D}}{d}\)
\(\Rightarrow \ \frac{1}{a}=\frac{n}{d}\)
\(\Rightarrow \ n=\frac{d}{a}=\frac{2 \times 10^{-5}}{4 \times 10^{-6}}=5\)​​​​