(i) Since, Q has greater negative intercept, it will have greater \(\phi\) (work function) and hence higher threshold frequency.
(ii) To know work function of Q, we put
V = 0 in the following equation.
\(\begin{array}{rlrl} V =\frac{h v}{e}-\frac{\phi}{e} \end{array}\)
\(\begin{array}{rlrl} \Rightarrow & 0 & =\frac{h v}{e}-\frac{\phi}{e} \Rightarrow \phi=h v \\ \end{array}\)
\(\begin{array}{rlrl} \therefore & \phi & =6.6 \times 10^{-34} \times 6 \times 10^{14} \mathrm{~J} \end{array}\)
\(=\frac{6.6 \times 6 \times 10^{-20}}{1.6 \times 10^{-19}} \mathrm{eV}=2.5 \mathrm{eV}\)
(iii) From the equation, \(v \lambda=c\)
\(\begin{aligned} \Rightarrow \quad \lambda & =\frac{c}{v}=\frac{3 \times 10^8}{8 \times 10^{14}}=\frac{30}{8} \times 10^{-7} \mathrm{~m} \end{aligned}\)
\(\begin{aligned} =\frac{30}{8} \times 10^3 \times 10^{-10} \mathrm{~m} \end{aligned}\)
\(\begin{aligned} =\frac{30}{8} \times 10^3 \end{aligned}\) \(\overset{\circ}{A}\)= 3750 \(\overset{\circ}{A}\)
Energy \(=\frac{12375}{\lambda(\overset{\circ}{A})} \mathrm{eV}=\frac{12375}{3750} \mathrm{eV}=33 \mathrm{eV}\)
\(\therefore\) Maximum KE of emitted electron = 33 - 2.5 eV
= 0.8 eV

Here, \(v=6\times { 10 }^{ 14 }Hz.\mu =1.5\)
(i) In glass,
speed of light, 
\({ v }_{ g }=\frac { { v }_{ a } }{ \mu } =\frac { 3\times { 10 }^{ 8 } }{ 1.5 } =2\times { 10 }^{ 8 }m/s\)
wavelength of light,
 \({ \lambda }_{ g }=\frac { { v }_{ g } }{ v } =\frac { 2\times { 10 }^{ 8 } }{ 6\times { 10 }^{ 14 } } =3.3\times { 10 }^{ -7 }m\)
(ii) In air
speed of light,
\({ v }_{ a }=c=3\times { 10 }^{ 8 }m/s\)
Wavelength of light, 
 \({ \lambda }_{ a }=\frac { { v }_{ a } }{ v } =\frac { 3\times { 10 }^{ 8 } }{ 6\times { 10 }^{ 14 } } =5\times { 10 }^{ -7 }m\)

Here, u = -25cm, v = -50cm,
As \({1\over f}={1\over v}-{1\over u}={1\over-50}+{1\over25}={1\over 50}\)
f = +50cm

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

0.9 eV

According to the mirror equation, we have
\(\frac { 1 }{ V } +\frac { 1 }{ u } =\frac { 1 }{ f } \)
where, u = distance of the object from the mirror
v = distance of the image from the mirror
f = focal length of the mirror
From the mirror equation, we have
\(V=\frac { uf }{ u-1 } \)  ...(1)
Applying new Cartesian sign convention, we get
f = -ve and u = -ve 
Given, f < u « 2f
\(\Rightarrow v=-ve\)  [from Eq(1)]
Magnification is given by
\(m=-\left( \frac { -v }{ -u } \right) =-ve\)
Hence, the image formed is real
.From the mirror formula, when u = - 2f.
\(\Rightarrow \frac { 1 }{ -2f } +\frac { 1 }{ v } =\frac { 1 }{ -f } \)
Therefore, when the object is at f, then image is formed at infinity. 

(i) From given figure, A = r₁ + r2
As ray QR is parallel to the base BC, then
Therefore, 2r₁(or 2r₂) = A(ii) D = x + y,  r₁ = r₂ = A/2
 D = (i - r₁) + (e - r₂)
 D = (i + e) - (r₁ + r₂)
 D = 2i - A

(i) The frequency of reflected and refracted light remains same as that of incident light because frequency only depends on the source of light.
(ii) Since the frequency remains same, hence there is no reduction in energy.

\(From\frac { 1 }{ f } =(u-1)\left( \frac { 1 }{ { R }_{ 1 } } -\frac { 1 }{ { R }_{ 2 } } \right) ,\)
\(\\ taking \ { R }_{ 1 }=R \ and \ { R }_{ 2 }=-R\)
\( \frac { 1 }{ 15 } =(\mu -1)\left( \frac { 1 }{ R } +\frac { 1 }{ R } \right) =\frac { 2 }{ R } (u-1)\)
For each half(plano-convex) lens
\({ R }_{ 1 }=R,{ R }_{ 2 }=\infty\)             
\( \therefore \frac { 1 }{ { f }^{ ' } } =(u-1)\left( \frac { 1 }{ R } -\frac { 1 }{ \infty } \right) =\frac { \mu -1 }{ R } =\frac { 1 }{ 30 }\)                      
\(\therefore \ { f }^{ ' }=30 \ cm\)

A photoelectric cell is called an electric eye as the photoelectric current set up in the photoelectric cell corresponding to incident light provides the information about the objects as has been seen by our eye in the presence of light.

Momentum and Energy.

(a) Refractive index of glass, μ = 1.5
Speed of light, c = 3 x 10⁸ m/s
Speed of light in glass is given by the relation
\(v=\frac{c}{\mu}\)
\(=\frac{3 \times 10^{8}}{1.5}=2 \times 10^{8} \mathrm{~m} / \mathrm{s}\)
Hence, the speed of light in glass is 2 x 10⁸ m/s.
(b) The speed of light in glass is not independent of the colour of light
The refractive index of a violet component of white light is greater than the refractive index of a red component. Hence, the speed of violet light is less than the speed of red light in glass. Hence, violet light travels slower than red light in a glass prism.

K.E. of electron,
\(K=\frac { 1 }{ 2 } { mv }^{ 2 }=\frac { 1 }{ 2 } \frac { { m }^{ 2 }{ v }^{ 2 } }{ m } =\frac { 1 }{ 2 } \frac { { h }^{ 2 } }{ m{ \lambda }^{ 2 } } \left[ \therefore \lambda =\frac { h }{ mv } \right] \)
Energy of a photon, \(E=\frac { hc }{ \lambda } \)
\(\therefore \frac { E }{ K } =\frac { hc/\lambda }{ \frac { 1 }{ 2 } \frac { { h }^{ 2 } }{ { m\lambda }^{ 2 } } } =\frac { 2\lambda mc }{ h }\)
\( \\ or \ E=\frac { 2\lambda mc }{ h } K=\frac { 2\lambda mc }{ h } \times K.E.of \ electron\)

(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) (a)
(ii) (a): According to Einstein's photoelectric equation,
\(e V_{0}=\frac{h c}{\lambda}-\frac{h c}{\lambda_{0}}\)
As \(\lambda\)₀ is constant, so when \(\lambda\). is decreased, stopping potential (V₀) increases.
(iii) (b): It indicates that threshold frequency is greater than that of ultraviolet light. As X-rays have greater frequency than ultraviolet rays, so they can cause photoelectric effect.
(iv) (b): \(\mathrm{K.E}_{\cdot 1}=h \mathrm{v}-\phi\)
\({K.E}_{\cdot 2}=n h v-\phi=n(h v-\phi)+(n-1) \phi\)
\({K.E.} 2=n \mathrm{KE}_{1}+(n-1) \phi \)
\(\text { K.E. }_{2}>n \mathrm{KE}_{1}\)
(v) (b): Stopping potential is the measurement of maximum kinetic energy of emitted photoelectrons and kinetic energy of emitted photoelectrons is linearly related with the frequency of incident light corresponding (i.e., corresponding to shortest wavelength, KE. is maximum). Stopping potential is independent of intensity.

(a): When light waves from two coherent sources superimpose at any particular point, crest of one wave falls on trough of the other and trough falls on the crest, the amplitude of the resultant wave is zero. Hence resultant intensity is zero. This is the phenomenon of destructive interference. Thus destructive interference produces darkness.