v = nג
In an isotropic medium, there is no change in the frequency of the light. So, the speed of light depends on wavelength of light.

At end A,
\(\frac{1}{f}  =\frac{1}{u_A}+\frac{1}{v_A}  \)
\(\therefore \frac{1}{v_A}  =\frac{1}{-10}-\frac{1}{-20}  \)
\(\frac{1}{v_A}  =-\frac{1}{10}+\frac{1}{20}=\frac{-2+1}{20}=-\frac{1}{20}  \)
\(v_A  =-20 \mathrm{~cm} \)
\(\left|v_{\wedge}\right|=20 \mathrm{~cm}\)
At end B,
\(\frac{1}{f}  =\frac{1}{u_B}+\frac{1}{v_B}  \)
\(\frac{1}{v_B} =\frac{1}{f}-\frac{1}{u_B},  \)
\(u_B  =-30 \mathrm{~cm}  \)
\(\frac{1}{v_B}  =-\frac{1}{10}+\frac{1}{30}  \)
\(=\frac{-3+1}{30}=\frac{-2}{30}=\frac{-1}{15}  \)
\(v_B  =-15 \mathrm{~cm}  \)
\(\left|v_B\right|  =15 \mathrm{~cm} \)
\(\therefore \quad\left|\mathrm{v}_{\mathrm{A}}\right|-\left|\mathrm{v}_{\mathrm{B}}\right| \) is the length of the image
= 20 - 15 = 5 cm

Convex Mirror is diverging in nature and for all positions of objects, convex mirror forms virtual and erect image.

Refractive index for violet is more and wavelength for violet is very low comparing other colours. So, the letter which appears to be raised more is violet.

λ = 500 nm = 500 x 10^-9 m
x = 3 mm = 3 x 10^-3 m
a = 1 mm = 1 x 10^-3 m
\(d=\frac{xa}{1.22 \lambda}\)
\(d=\frac{3 \times1\times10^{-6}}{1.22 \times500\times10^{-9}}\)
\(=\frac{3 \times1\times10^{-6}}{6.10 \times 10^{-7}}\)
\(d=\frac{30}{6.1}=5 m\)

d' = 2d, β' = β, D' = ?
W.K.T, Fringe width
\(\beta = \frac{D\lambda}{d} \Rightarrow D' = \frac{Dd'}{d}\)
\(D' = \frac{D2d}{d}=2D\)

I = l₁ + l₂ + 2\(\sqrt{I_1I_2}\)cos θ
If cos θ = cos 0 = l, I is max
= I+ 4I + 2\(\sqrt{41^2}\) cos 0
= 5I + 4I = 91
If cos π = -1, I is min
I_min = I + 4I + 2\(\sqrt{41^2}\) cos π
= 5I + 4I(-1)
= 5I + 4I = I
(I_max, I_min)= (9I, I)

\(\mathrm{E}_{\mathrm{p}}  =\frac{\mathrm{hc}}{\lambda_{\mathrm{p}}}  \)
\(\mathrm{E}_{\mathrm{e}}  =\frac{\mathrm{h}^2}{2 \mathrm{~m} \lambda_{\mathrm{e}}^2}  \)
\(\frac{\mathrm{hc}}{\lambda_{\mathrm{p}}}  =\frac{\mathrm{h}^2}{2 \mathrm{~m} \lambda_{\mathrm{e}}^2}  \)
\(\lambda_{\mathrm{p}}  \propto \lambda_{\mathrm{e}}^{{ }^2}\)

\(\lambda\propto \frac{1}{\sqrt{V}}\)
\(\frac{\lambda_1}{\lambda_2}=\frac{\sqrt{224\times10^3}}{\sqrt{14\times 10^3}}\)
\(=\sqrt{16}=4\)
\(\lambda_{\mathrm{2}}= \frac{\lambda_1}{4}\)

\(\lambda_{\mathrm{i}}  \frac{1}{\mathrm{mv}}  \)
\(\frac{\lambda_p}{\lambda_e}  =\frac{m_e v_e}{m_P v_P}  \)
\(1  =\frac{9.1 \times 10^{-31} \times 6 \times 10^6}{3 \times 10^{-9} \times v_p}  \)
\(\mathrm{v}_{\mathrm{p}}  =9.1 \times 10^{-16} \times 2  \)
\(\mathrm{v}_{\mathrm{p}}  =18.2 \times 10^{-16}  \)
\(\mathrm{v}_{\mathrm{p}}  =1.82 \times 10^{-15} \mathrm{~m} \mathrm{~s}^{-1}\)

\(\frac{\mathrm{hc}}{\lambda}=\phi+\mathrm{eV}  \)    .....(1)
\(\frac{\mathrm{hc}}{2 \lambda}=\phi+\frac{\mathrm{eV}}{4}\)    .....(2)
multiply (2) eqn by 4
\(\frac{2 h c}{\lambda}=4 \phi+\mathrm{eV}\)   .....(3)
subtract eqn (1) from (3), we get
\(\frac{ h c}{\lambda}=3 \phi \Rightarrow \phi = \frac{ h c}{3\lambda}\)
\(\frac{ h c}{\lambda_o}=\frac{ h c}{3\lambda}\)
⋋_o = 3⋋

\(I_1 \propto \frac{1}{\lambda_1{ }^4} \)
\(I_2 \propto \frac{1}{\lambda_2{ }^4} \)
\( \lambda_1 \alpha \frac{1}{\left(I_1\right)^{\frac{1}{4}}} \)
\( \lambda_2 \alpha \frac{1}{\left(I_2\right)^{\frac{1}{4}}} \)
\(\frac{\lambda_1}{\lambda_2}=\left(\frac{\mathrm{I}_2}{\mathrm{I}_1}\right)^{\frac{1}{4}}=\left(\frac{4}{1}\right)^{\frac{1}{4}}=\sqrt{2}\)

(i) Light from the object O at the bottom of the tank passes from denser medium (water) to rarer medium (air) to reach our eyes for viewing the object.
(ii) It deviates away from the normal in the rarer medium at the point of incidence B as shown in Figure.
(iii) The refractive index of the denser medium is n₁ and that of rarer medium is n₂. Here, n₁ > n_2.

The angle of incidence in the denser medium is i and the angle of refraction in the rarer medium is r. The lines NN'and OD are parallel. Thus, the angle ∠DIB is also r. The angles i and r are very small as the diverging light from O entering the eye is very narrow. The Snell's law in product form for this refraction from equation is,
n₁ sin i = n₂ sin r
As the angles i and r are small, we can approximate, sin i = tan i and sin r tan r.
n₁ tan i = n₂ tan r
In triangles ∆DOB and ∆DIB,
\(tan \ i=\frac{DB}{DO}and \ tan \ r=\frac{DB}{DI}\)
\(n_1\frac{DB}{DO}=n_2\frac{DB}{DI}\)
DB is cancelled both sides. Now, DO is the actual depth d and DI is the apparent depth d'.
\(n_1\frac{1}{d}=n_2\frac{1}{d'}\)
After rearranging, \(\frac{d'}{d}=\frac{n_2}{n_1}\)
Rewriting the above equation for the apparent depth d', d' = \(=\frac{n_2}{n_1}d\)
As the rarer medium is air, its refractive index n, can be taken as 1, (n₂ = 1) and the refractive index n₁ of denser medium could then be taken as n itself, (n₁ = n). Now, the equation for apparent depth becomes,
\(d'=\frac{d}{n}\)
The bottom appears to be elevated by d-d',
\(d-d'=d-\frac{d}{n}(or)d-d'=d(1-\frac{1}{n})\)

Relation between f and R:
C ⇒ Center of curvature
F ⇒ Principal focus
i ⇒ Angle of incidence

The angles From right angle triangles ∆MCR, and ∆MFP,
\(\tan i=\frac{P M}{P C} \text { and } \tan 2 i=\frac{P M}{P F}\)
As the angles are small, tan i = i and tan 2i = 2i.
\(\mathrm{i}=\frac{\mathrm{PM}}{\mathrm{PC}} \text { and } 2 \mathrm{i}=\frac{\mathrm{PM}}{\mathrm{PF}}\)
Simplifying further,
\(2 \frac{\mathrm{PM}}{\mathrm{PC}}=\frac{\mathrm{PM}}{\mathrm{PF}} ; 2 \mathrm{PF}=\mathrm{PC}, \mathrm{R}=2 \mathrm{f}\)
PF is focal length f and PC is the radius of curvature R.
R = 2f (or) f = R/2

(i) Polaroids are used in goggles and cameras to avoid glare of light.
(ii) Polaroids are useful in 3D pictures i.e., in holography.
(iii) Polaroids are used to improve contrast in old oil paintings.
(iv) Polaroids are used in optical stress analysis.
(v) Polaroids are used as window glasses to control the intensity of incoming light.
(vi) Polarised laser beam acts as needle to read/ write in compact discs (CDs).
(vii) Polarised lights is used in liquid crystal display (LCD).

(i) Pile of plates makes use of Brewster's law to convert the partially polarised refracted light into plane polarised light.
(ii) It consists of several plates kept one behind the other at an angle 90° - i_p with the horizontal surface as shown in Figure.
(iii) This arrangement ensures that the parallel light falls on these plates at i_p. When this unpolarised light passes successively through these plates, the few parallel vibrations to the surface which may be present in the refracted light, get a chance for further reflections at the succeeding plates.
(iv) Thus, both the reflected and the refracted lights are found to be plane polarised.
Uses :
The pile of plates is used as a polarizer and also as an analyser.

Given, thickness of glass slab, d = 50 cm = 0.5 m, refractive index, n = 1.5
refractive index, \(n=\cfrac { c }{ v } \) 
(a) speed of light in the glass slab is,
\(v=\cfrac { c }{ n } =\cfrac { 3\times { 10 }^{ 8 } }{ 1.5 } =2\times { 10 }^{ 8 }{ ms }^{ -1 }\) 
(b) time taken by light to travel through the glass slab is,
\(t=\cfrac { d }{ v } =\cfrac { 0.5 }{ 2\times { 10 }^{ 8 } } =2.5\times { 10 }^{ -9 }{ s }\)
(c) optical path,
d' = nd = 1.5 x  0.5 = 0.75 m = 75 cm
Light would have traveled an additional 25 cm (75 cm – 50 cm) in vacuum at the same time had there been no glass slab in its path.

λ = 589 nm = 589 x 10⁻⁹ m
For the film to have strong reflection, the reflected waves should interfere constructively. The least optical path difference introduced by the film should be λ/2. The optical path difference between the waves reflected from the two surfaces of the film is 2μd. Thus, for strong reflection, 2μd = λ/2 [As given in equation 6.145. with n = 1]
Rewriting, \(d=\frac{\lambda}{4 \mu}\) 
Substituting, \(d=\frac{589 \times 10^{9}}{4 \times 1.25}=117.8 \times 10^{-9}\) 
d = 117.8 x 10^-9 = 117.8 nm
For the film to be anti-reflecting, the reflected rays should interfere destructively. The least optical path difference introduced by the film should be λ. The optical path difference between the waves reflected from the two surfaces of the film is 2μd. For strong reflection, 2μd = λ [As given in equation 6.146. with n = 1]
Rewriting, \(d=\cfrac { \lambda }{ 2\mu } \)
Substituting, \(d=\cfrac { 589\times { 10 }^{ 9 } }{ 2\times 1.25 } =235.6\times { 10 }^{ -9 }\) 
d = 235.6 x 10^-9 = 235.6 nm

(i) An electron of mass m is accelerated through a potential difference of V volt. The kinetic energy acquired by the electron is given by
\(\cfrac { 1 }{ 2 } { mv }^{ 2 }=ev\) 
(ii) Therefore, the speed v of the electron is
\(v=\sqrt { \cfrac { 2ev }{ m } } \) 
Hence, the de Broglie wavelength of the matter waves associated with electron is
\(\lambda =\cfrac { h }{ mv } =\cfrac { h }{ \sqrt { 2mev } } \) 
(iii) Substituting the known values in the above equation, we get
\(\lambda =\cfrac { 6.26\times { 10 }^{ -34 } }{ \sqrt { 2V\times 1.6\times { 10 }^{ -19 }\times 9.11\times { 10 }^{ -31 } } } \) 
= \(\cfrac { 12.27\times { 10 }^{ -10 } }{ \sqrt { V } } m\) 
\(\lambda =\cfrac { 12.27 }{ \sqrt { V } } \overset { o }{ A } \)
(iv) Since the kinetic energy of the electron, K = eV, then the de Broglie wavelength associated with electron can be also written as
\(\lambda =\cfrac { h }{ \sqrt { 2mK } } \)

 \(\lambda_{p}=\frac{h}{\sqrt{2 m e V}} ; \lambda_{\alpha}=\frac{h}{\sqrt{m e V}} ; V=512 \mathrm{~V} \)
\(\frac{\lambda_{p}}{\lambda_{\alpha}}=\sqrt{\left(\frac{m_{\alpha}}{m_{p}}\right)\left(\frac{e_{\alpha}}{e_{p}}\right)\left(\frac{v_{\alpha }}{v_{p}}\right)} \)
\(\frac{m_{\propto}}{m_{p}}=4 ; \frac{e_{\alpha}}{e_{p}}=2 ; \frac{v_{\alpha}}{v_{p}}=\frac{x}{512} ; \frac{\lambda_{p}}{\lambda_{\alpha}}=1 \)
\(1=\sqrt{4 \times 2 \times\left(\frac{x}{512}\right)}=\frac{x}{64} \Rightarrow x=64 V \)

(i) The photons of light of frequency (v) and wavelength (λ) will have energy, given by 
\(E=h v=\frac{h c}{\lambda}\) 
(ii) The energy of a photon is determined by the frequency of the radiation and not by its intensity and the intensity has no relation with the energy of the individual photons in the beam.
(iii) The photons travel with the velocity of light and its momentum is given by 
\(p=\frac{h}{\lambda}=\frac{h v}{c}\) 
(iv) Since photons are electrically neutral, they are unaffected by electric and magnetic fields.
(v) When a photon interacts with matter (photon-electron collision), the total energy, total linear momentum, and angular momentum are conserved. Since photons may be absorbed (or) a new photon may be produced in such interactions, the number of photons may not be conserved.

Critical angle:
The angle of incidence in the denser medium for which the angle reflection is 90^o or the reflected ray graces the boundary between the two media is called critical angle.
Total Internal reflection:
For any angle of incidence greater than the critical angle, the center light is reflected back into the denser medium itself. This phenomenon is called Total internal reflection.

Diamond appears glittering because the total internal reflection of light.
Inside the diamond the refractive index of diamond is about 2.417 which is greater than the refractive index of glass. (μ_g =1.5). The critical angle of diamond is 24.4^d which is much less than that of glass (g_crown = 40.5°, g_flint 31.9°).
So, when the light enters the diamond, the total internal reflection of light happens inside the diamond before getting out. This gives a sparkling effect for diamond.

According to Huygens's principle, each point of the wavefront is the source of secondary wavelets emanating from these points spreading out in all directions with the speed of the wave. These are called as secondary wavelets.

(i) The polaroid which polarises the light passing through it is called a polariser.
(ii) The polaroid which is used to examine whether a beam of light is polarised or not is called an analyser.

Given, focal length, f = 150 cm = 1.5 m
Equation for power of lens is, \(p=\cfrac { 1 }{ f } \) 
Substituting the values,
\(p=\cfrac { 1 }{ 1.5 } =0.67 D\) 
As the power is positive, it is a converging lens.

Number of lines per cm = 4000 cm^-1; m = 2; θ = 30°; λ = ?
Number of lines per unit length
\(N=\cfrac { 4000 }{ 1\times { 10 }^{ -2 } } =4\times { 10 }^{ 5 }\) 
Equation for diffraction maximum in grating is, sinθ = Nmλ
After Rewriting, \(\lambda =\cfrac { sin\theta }{ Nm } \) 
Substituting,
\(\lambda =\cfrac { { \sin30 }^{ o } }{ 4\times { 10 }^{ 5 }\times 2 } =\cfrac { 0.5 }{ 4\times { 10 }^{ 5 }\times 2 } \) 
= \(\cfrac { 1 }{ 2\times 4\times { 10 }^{ 5 }\times 2 } =\cfrac { 1 }{ 16 \times 10^5} \)
λ = 6250 x 10^-10 m = 6205 Å

Photo cell is a device which converts light energy into electrical energy. It works on the principle of photo electric effect. When light is incident on photosensitive materials, their electric properties will get affected, based on which Photo cells are classified into three types. They are
(i) Photo emissive cell
(ii) Photo voltaic cell
(iii) Photo conductive cell

According to de Broglie hypothesis, if radiation has a dual nature, then the moving particles of matter Iike electrons, protons, neutrons in motion should exhibit wave like character under an appropriate conditions. These waves are called de Broglie waves or matter waves.

Energy of the incident photon is
E = hv = \(\frac { hc }{ \lambda } \) (in joules)
E = \(\frac { hc }{ \lambda e } \) (in eV)
Substituting the known values, we get
E = \(\frac { 6.626\times { 10 }^{ -34 }\times 3\times 10^{ 8 } }{ 300\times { 10 }^{ -9 }\times 1.6\times { 10 }^{ -19 } } \)
E = 4.14 eV
The work function of silver = 4.7 eV. Since the energy of the incident photon is less than the work function of silver, photoelectrons are not observed in this case.

Mirror Equation :

(i) AB is an object which is placed on the principal axis of a concave mirror beyond the center of curvature C. A' B' is an image which is formed between the point pole P, and the centre of curvature.
(ii) From the figure As per law of reflection, the angle of incidence ∠BPA is equal to the angle of reflection ∠B'PA'.
(iii) The triangles ∠BPA and ∠B'PA' are similar. Thus, from the rule of similar triangles,
\(\cfrac { { A }^{ ' }{ B }^{ ' } }{ AB } =\cfrac { { PA }^{ ' } }{ PA } \)             ................(1)
(iv) The other set of similar triangles are, ΔDPF and ΔB'A'F. (PD is almost a straight vertical line)
\(\cfrac { { A }^{ ' }B' }{ PD } =\cfrac { A'F }{ PF } \) 
(v) As, PD = AB the above equation becomes,
\(\cfrac { A'B' }{ AB } =\cfrac { A'F }{ PF } \)            ......(2)
(vi) From equations (1) and (2) we can write,
\(\cfrac { PA' }{ PA } =\cfrac { A'F }{ PF } \) 
(vii) As, A'F = PA' - PF, the above equation becomes,
\(\cfrac { PA' }{ PA } =\cfrac { PA'-PF }{ PF } \)            .....(3)
(viii) We can apply the sign conventions for the various distances in the above equation
PA = - u, PA' = -v, PF = - f
(ix) All the three distances are negative as per sign convention, because they are measured to the left of the pole. Now, the equation (3) becomes,
\(\cfrac { -v }{ -u } =\cfrac { -v-\left( -f \right) }{ -f } \) 
On further simplification,
\(\cfrac { v }{ u } =\cfrac { v-f }{ f } ;\cfrac { v }{ u } =\cfrac { v }{ f } -1 \) 
Dividing either side with v,
 \(\cfrac { 1 }{ u } =\cfrac { 1 }{ f } -\cfrac { 1 }{ v } \)
After rearranging,
\(\cfrac { 1 }{ v } +\cfrac { 1 }{ u } =\cfrac { 1 }{ f } \) 
The above equation is called mirror equation.
Lateral magnification:
The lateral or transverse magnification is defined as the ratio of the height of the image to the height of the object. The height of the object and image are measured perpendicular to the principal axis.
Magnification (m) \(=\frac{\text { height of the image }\left(h^{\prime}\right)}{\text { height of the image }(h)} \)
\(m=\frac{h^{\prime}}{h} \)  ....(1)
Applying proper sign conventions for equation,
\(\frac{A^{\prime} B^{\prime}}{A B}=\frac{P A^{\prime}}{P A} \)
\(A^{\prime} B^{\prime}=-h^{\prime}, A B=h, P A^{\prime}=-v, P A=-u \)
\(-\frac{h}{h}=\frac{-v}{-u} \)
On simplifying we get,
\(\mathrm{m}=\frac{\mathrm{h}^{\prime}}{\mathrm{h}}=-\frac{\mathrm{v}}{\mathrm{u}}\)  ...(2)
Using mirror equation, we can further write the magnification as,
\(m=\frac{h^{\prime}}{h}=\frac{f-v}{f}=\frac{f}{f-u}\)   ..(3)

(i) Let us consider a thin lens made up of a medium of refractive index n₂ is placed in a medium of refractive index n₁. Let R₁ and R₂ be the radii of curvature of two spherical surfaces (1) and (2) respectively and P be the pole.

(ii) Consider a point object 'O' on the principal axis. A paraxial ray from 'O' which falls very close to P, after refraction at the surface (1) forms image at 1'.
(iii) Before it does so, it is again refracted by the surface (2). Therefore the final image is formed at I.
(iv) The general equation for the refraction at a single spherical surface is given from Equation,
 \(\cfrac { { n }_{ 2 } }{ v } -\cfrac { { n }_{ 1 } }{ v} =\cfrac { \left( { n }_{ 2 }-{ n }_{ 1 } \right) }{ { R } } \)
For the refracting surface (1), the light goes from n₁ to n₂
\(\cfrac { { n }_{ 2 } }{ v' } -\cfrac { { n }_{ 1 } }{ u } =\cfrac { \left( { n }_{ 2 }-{ n }_{ 1 } \right) }{ { R }_{ 1 } } \)   .....(1)
For the refracting surface (2), the light goes from n₂ to n₁
\(\frac{n_{1}}{v}+\frac{n_{2}}{v^{\prime}}=\frac{\left(n_{1}-n_{2}\right)}{R_{2}}\)    ......(2)
For surface (2) I' acts as virtual object.
Adding the above two equations (1) and (2)
\(\cfrac { { n }_{ 1 } }{ v } -\cfrac { { n }_{ 1 } }{ u } =\left( { n }_{ 2 }-{ n }_{ 1 } \right) \left( \cfrac { 1 }{ { R }_{ 1 } } -\cfrac { 1 }{ { R }_{ 2 } } \right) \) 
on further simplifying and rearranging,
\(\cfrac { 1 }{ v } -\cfrac { 1 }{ u } =\left( \cfrac { { n }_{ 2 }-{ n }_{ 1 } }{ { n }_{ 1 } } \right) \left( \cfrac { 1 }{ { R }_{ 1 } } -\cfrac { 1 }{ { R }_{ 2 } } \right) \) 
\(\cfrac { 1 }{ v } -\cfrac { 1 }{ u } =\left( \cfrac { { n }_{ 2 } }{ n_{ 1 } } -1 \right) \left( \cfrac { 1 }{ { R }_{ 1 } } -\cfrac { 1 }{ { R }_{ 2 } } \right) \)  ....(3)
If the object is at infinity, the image is formed at the focus of the lens. Thus, for u = \(\infty\), v = f. Then the equation becomes.
\(\cfrac { 1 }{ f } -\cfrac { 1 }{ \infty } =\left( \cfrac { { n }_{ 2 } }{ { { n }_{ 1 } } } -1 \right) \left( \cfrac { 1 }{ { R }_{ 1 } } -\cfrac { 1 }{ { R }_{ 2 } } \right) \) 
\(\cfrac { 1 }{ f } =\left( \cfrac { { n }_{ 2 } }{ { n }_{ 1 } } -1 \right) \left( \cfrac { 1 }{ { R }_{ 1 } } -\cfrac { 1 }{ { R }_{ 2 } } \right) \)  ......(4)
If the lens is kept in air, then we can take n₂ = n and n₁ = 1. So the equation (4) becomes,
\(\\ \cfrac { 1 }{ f } =\left( n-1 \right) \left( \cfrac { 1 }{ { R }_{ 1 } } -\cfrac { 1 }{ { R }_{ 2 } } \right) \)   ..(5)
The above equation is called the lens maker's formula.
Significance:
It tells the lens manufacturers what curvature is needed to make a lens of desired focal length with a material of particular refractive index to make a lens of desired focal length. This formula holds good also for a concave lens.

Experimental setup:
(i) S is a source s₁ and s₂ the double slits which are at equidistances from 's'. Wavefronts from s₁ and s₂ spread out and overlap on other side of double slit.
(ii) When a screen is placed at a distance of about 1 meter from the slits, alternate bright and dark fringes which are equally spaced appear on the screen. These are called interference fringes or bands.
(iii) Using an eyepiece the fringes can be seen directly. At the center point O on the screen, waves from s₁ and s₂ travel equal distances and arrive in-phase as shown in Figure.
(iv) These two waves constructively interfere and bright fringe is observed at O. This is called cental bright fringe.
(v) When one of the slits is closed, The fringes disappear and there in uniform illumination on the screen.
(vi) This shows clearly that the bands are due to interference.

Equation for path difference :
(i) The Let d be the distance between the double slits s₁ and s₂ which act as coherent sources of wavelength λ.
(ii) A screen is placed parallel to the double slit at a distance D from it.
(iii) P is any point at a distance y from O.
(iv) The waves from S₁ and S₂ meet at P either in-phase or out-of-phase depending upon the path difference between the two waves.

The path difference \(\delta\) between the light waves from s₁ and s₂ to the point p is,
\(\delta=\mathrm{S}_{2} \mathrm{P}-\mathrm{S}_{1} \mathrm{P}\)
\(\delta=\mathrm{S}_{2} \mathrm{P}-\mathrm{MP}=\mathrm{S}_{2} \mathrm{M}\)                        .........(1)
\(\angle \mathrm{OCP}=\angle \mathrm{S}_{2} \mathrm{~S}_{1} \mathrm{M}=\theta\)
In right angle triangle \(\Delta \mathrm{S}_{1} \mathrm{S}_{2} \mathrm{M}\), the path difference S₂M = d sin \(\theta\)
\(\delta=d \sin \theta\)                 ...........(2)
If the angle \(\theta\) is small, \(\sin \theta \approx \tan \theta \approx \theta\)
From the right angle triangle \(\Delta \mathrm{OCP}, \tan \theta=\frac{\mathrm{y}}{\mathrm{D}}\)
The path differences \(\delta=\frac{d y}{D}\)     ...........(3)
Based on the condition of the path difference, the point P may have a bright (or) dark fringe

(i) A simple microscope is a single magnifying (converging) lens of small focal length. To get an erect, magnified and virtual image of the object.
(ii) For this the object is placed between the focal length Fand P on one side of the lens and viewed from other side of the lens. There are two magnifications to be discussed for two kinds of focusing.
(a) Near point focusing:
The eye is least strained when image is formed at near point,i.e. 25 cm. The near point is also called as least distance of distinct vision. This is shown in Figure.

Magnification in near point focusing:
(i) Object distance u is less than f
(ii) The image distance is the near point D. The magnification m is given by the relation,
 \(m=\cfrac { v }{ u } \)                          ...............(1)
Substituting, V = - D and u= - u, as both the distances are measured to the left of the lens. Hence,
\(m=\cfrac { -D }{ -u }\)  
\(m=\cfrac { D }{ u } \)                          ...............(2)
Using lens equation, W.K.T, m = 1 - (v/f)
Substiuting v = -D gives, \(\\ m=1+\cfrac { D }{ f } \)                         ..................(3)
This is the magnification for near point focusing.
(b) Normal focusing :
(i) The eye is most relaxed when the image is formed at infinity. The focusing is called normal focusing when the image is formed at infinity. This is shown in Figure (b).

Magnification in normal focusing (angular magnification):
(ii) The angular magnification is defined as the ratio of angle θ₁ subtended by the image with aided eye to the angle θ₀ subtended by the object with unaided eye.
\(m=\cfrac { { \theta }_{ 1 } }{ { \theta }_{ 0 } } \)                 .........(2)
For unaided eye shown in Figure (a),
\(tan\theta _{ 0 }\approx { \theta }_{ 1 }=\cfrac { h }{ D } \)                     ................(3)
For aided eye shown in Figure(b).
\(tan\theta _{ i }={ \theta }_{ i }=\cfrac { h }{ f } \)                        ...................(4)
The angular magnification is,
\(m=\cfrac { { \theta }_{ i } }{ { \theta }_{ o } } =\cfrac { h/f }{ h/D } \) 
\(m=\cfrac { D }{ f } \)              ..............(5)
This is the magnification for normal focusing.

(i) When a photon of energy hv is incident on a metal surface, it is completely absorbed by a single electron and the electron is ejected.
(ii) In this process, a part of the photon energy is used for the ejection of the electrons from the metal surface (photoelectric work function Φ₀) and the remaining energy as the kinetic energy of the ejected electron. From the law of conservation of energy,
\(\\ \\ \\ hv=\phi { _{ 0 }+\cfrac { 1 }{ 2 } { mv }^{ 2 } }\)   ......(1) 
(iii) where m is the mass of the electron and v its velocity.

(iv) If  we reduce the frequency of the incident light is reduced, the speed or kinetic energy of photo electrons is also reduced. At some frequency v₀ of incident radiation, the photo electrons are ejected with almost zero kinetic energy.
Then the equation becomes.
\({ hv }_{ 0 }=\phi _{ 0 }\)    ......(2) 
(v) Where v₀ is the threshold frequency. B rewriting the equation, we get 
\(hv={ hv }_{ o }+\cfrac { 1 }{ 2 } { { mv }^{ 2 } }\)   ......(3) 
The equation is known as einstein's photoelectric equation.
(vi) If the electron does not lose energy by internal collisions, then it is emitted with maximum kinetic energy K_max. Then
\({ K }_{ max }=\cfrac { 1 }{ 2 } { mv }^{ 2 }_{ max }\)    ......(4) 
(vii) where v_maxis the maximum velocity of max the electron ejected. The equation (1) is rearranged as follows:
\({ K }_{ max }=hv-{ \phi }_{ 0 }\) 

A graph between maximum kinetic energy K_max of the photoelectron and frequency v of the incident light is a straight line.

Principle:
(i) The wave nature of the electron is used in the construction of microscope called electron microscope.
(ii)The resolving power of a microscope is inversely proportional to the wavelength of the radiation used for illuminating the object under study.
(iii) Higher magnification as well as higher resolving power can be obtained by employing the waves of shorter wavelengths.
(iv) De Broglie's wavelength of electron is very much less than (a few thousand less) that of the visible light being used in optical microscopes.
(v) As a result, the microscopes employing de Broglie waves of electrons have very much higher resolving power than optical microscope.
(vi)  Electron microscopes giving magnification more than 2,00,000 times are common in research laboratories.

Working:
(i) The construction and working of an electron microscope is similar to that of an optical microscope except that in electron microscope focussing of electron beam is done by the electrostatic or magnetic lenses.
(ii) The electron beam passing across a suitably arranged either electric or magnetic fields undergoes divergence or convergence thereby focussing of the beam is done.
(iii) The electrons emitted from the source are accelerated by high potentials.
(iv) The beam is made parallel by magnetic condenser lens; When the beam passes through the sample whose magnified image is needed, the beam carries the image of the sample.
(v) With the help of magnetic objective lens and magnetic projector lens system, the magnified image is obtained on the screen. These electron microscopes are being used in almost all brands of science.

Davisson - Germer experiment
(i) The filament F is heated by a low tension (L . T) battery. Electrons are emitted from the hot filament by thermionic emission.
(ii) They are then accelerated due to the potential diference between the filament and the anode aluminum cylinder by a high tension (H.T) battery.
(iii) Electron beam is collimated by using two thin aluminum diaphragms and is allowed to strike a single crystal of Nickel.
(iv) The electrons scattered by Niatoms in diflerent directions are received by the electron detector which measures the intensity of scattered electron beam.
(v) The detector is capable of rotation in the plane of the paper, so that the angle (\(\theta\)) between the incident beam and the scattered beam can be changed at our will.
(vi) The intensity of the scattered electron beam is measured as a function of the angle \(\theta\).

(i) Figure shows the variation of intensity of the scattered electrons with the angle \(\theta\) for the accelerating voltage of 54 V.
(ii) For a given accelerating voltage V, the scattered wave shows a peak or maximum at an angle of 50^o to the incident electron beam.
(iii) This peak in intensity is attributed to the constructive interference of electrons diffracted from various atomic layers of the target material.


(iv) From the known value of interplanar spacing of Nickel, the wavelength of the electron wave has been experimentally calculated as 1.65\(\overset { o }{ A }\).
(v) The wavelength can also be calculated from de Broglie relation for V = 54 V from equation as
\(\lambda =\cfrac { 12.27 }{ \sqrt { V } } \overset { o }{ A } =\cfrac { 12.27 }{ \sqrt { 54 } } \) 
\(\lambda =1.67\overset { o }{ A } \)
(vi) This value agrees very well with the experimentally observed wavelength of 1.65 \(\overset { o }{ A }\). Thus this experiment directly verifies de Broglie's hypothesis of the wave nature of moving particles.

(i) Let a parallel beam of light (plane wavefront) fall normally on a single slit AB of width a as shown in figure. The diffracted beam falls on a screen kept at a distance D from the slit. The center of the slit is C.
(ii) A straight line through C perpendicular to the plane of slit meets the center of the screen at O. Consider any point P on the screen. All the light reaching the point P from different points on the slit make an angle \(\theta\) with the normal CO.
(iii) All the light waves coming from different points on the slit interfere at point P (and other points) on the screen to give the resultant intensities. The point P is in the geometrically shadowed region, up to which the central maximum is spread due to diffraction as shown Figure.
(iv) We need to give the condition for the point P to be of various minima.

(v) The basic idea is to divide the slit into much smaller even number of parts. Then, add their contributions at P with the proper path difference to show that destructive interference takes place at that point to make it minimum. To explain maximum, the slit is divided into odd number of parts.
Condition for P to the n^th order minimum:
(i) Dividing the slit into 2n number of (even number of) equal parts makes the light produced by one of the corresponding points to be cancelled by its counterpart. Thus, the condition for n^th order minimum is, \(\frac{a}{2 n} \sin \theta=\frac{\lambda}{2}\)
\(a \sin \theta=n \lambda\) (n^th minimum)
Where, n = 1,2,3... is the order of diffraction minimum.

(i) A microscope is used to see the details of the object under observation.
(ii) Good microscope should not only magnify the object but also resolve the two points on an object which are separated by the smallest distance d_min. Actually, d_min is the resolution and its reciprocal is the resolving power.

Resolving power of a microscope
The spatial resolution (radius of central maxima) is
\(r_{0}=\frac{1.22 \lambda f}{a}\)                .......(1)
where 'a' is width of the aperture/slit.
In microscope, the object distance is just more than the focal length f and the image is formed at v as shown in the Figure. Hence, f in equation is replaced by v.
\(r_{0}=\frac{1.22 \lambda v}{a}\)  ......(2)
In the place of focal length f we have the image distance v. If the difference between the two points on the object to be resolved is d_min. Then the magnification m is,
\(m=\frac{r_{0}}{d_{\min }}\)               ........(3)
\(\mathrm{d}_{\min }=\frac{\mathrm{r}_{0}}{\mathrm{~m}}=\frac{1.22 \lambda \mathrm{v}}{\mathrm{am}}=\frac{1.22 \lambda \mathrm{v}}{\mathrm{a}(\mathrm{v} / \mathrm{u})}=\frac{1.22 \lambda \mathrm{u}}{\mathrm{a}}\)  [∴ m = v/u]
\(\mathrm{d}_{\min }=\frac{1.22 f\lambda}{\mathrm{a}}[\therefore \mathrm{u} \approx \mathrm{f}]\)               ..............(4)
On the other side,
\(2 \tan \beta \approx 2 \sin \beta=\frac{a}{f} \therefore[a=f 2 \sin \beta]\)               .........(5)
\(\mathrm{d}_{\min }=\frac{1.22 \lambda}{2 \sin \beta}\)                    .................(6)
To further reduce the value of d_min the optical path of the light is increased by immersing the objective of the microscope into a bath containing oil of refractive index n.
\(\mathrm{d}_{\min }=\frac{1.22 \lambda}{2 \mathrm{n} \sin \beta}\)                 ...............(7)
Such an objective is called the oil-immersed objective. The term n sin β is called numerical aperture NA.
\(\mathrm{d}_{\min }=\frac{1.22 \lambda}{2(\mathrm{NA})}\)                      ...................(8)
The resolvins power R_M of microscope is
\(\mathrm{R}_{M }=\frac{1}{d_{min}}\frac{2(NA)}{1.22\lambda}\)