(i) Let us consider an object OO' of height h placed on the principal axis with its height perpendicular to the principal axis, The inverted real image II' is formed which has a height h' as shown in Figure.
(ii) The lateral (or) transverse magnification m is defined as the ratio of the height of the image to height of the object.
\(m=\cfrac { II' }{ OO' } \)  ...(i) 
(iii) From the two similar triangles ∆POO' and ΔPII', we can write,
\(\cfrac { II' }{ OO' } =\cfrac { PI }{ PO } \)  ...(ii) 
On applying sign convention,

\(\cfrac { -{ h' } }{ { h } } =\cfrac { v }{ -u } \) 
(iv) Substituting this in the equation (ii) for magnification,
\(m=\cfrac { -{ h }' }{ { h }} =\cfrac { v }{ -u } \) 
After rearranging,
\(m=\cfrac { { h }'}{ h } =\cfrac { v }{ u } \)   ...(iii) 
(v) The magnification is negative for real image and positive for virtual image. In the case of a concave lens, the magnification is always positive and less than one.
We can also have the other forms of equations for magnification by combining the lens equation as,
\(m=\cfrac { { h }' }{ { h }} =\cfrac { f }{ f+u } (or)m=\cfrac { { h }' }{ { h }} =\cfrac { f-v }{ f } \)  ...(iv)  

(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 (i) and (ii) 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 (i) forms image at I'.
(iii) Before it does so, it is again refracted by the surface (ii). Therefore the final image is formed at I.
(iv) The general equation for the refraction at a single spherical surface is given by the equation is,
\(\cfrac { { n }_{ 2 } }{ v } =\cfrac { { n }_{ 1 } }{ u } =\cfrac { \left( { \quad n }_{ 2 }-{ { n }_{ 1 } } \right) }{ R } \)   ....(i)
(v) For the refracting surface (1), the light goes from n₁ to n₂
 \(\cfrac { { { n }_{ 2 } } }{ v' } -\cfrac { { n }_{ 1 } }{ u } =\cfrac { \left( n_{ 2 }-{ n }_{ 1 } \right) }{ { R }_{ 1} } \)   ......(ii)

(vi) For the refracting surface (ii), the light goes from medium n₂ to n₁
 \(\cfrac { { { n }_{ 1 } } }{ v' } -\cfrac { { n }_{ 2 } }{ v' } =\cfrac { \left( n_{ 1 }-{ n }_{ 2 } \right) }{ { R }_{ 2 } } \)   ....(iii)
(vii) Adding the above two equations (ii) and (iii)
\(\cfrac { { { n }_{ 1 } } }{ v } =\cfrac { { n }_{ 1 } }{ u } =\left( { n }_{ 2 }-{ n }_{ 1 } \right) \left( \cfrac { 1 }{ { R }_{ 1 } } -\cfrac { 1 }{ { R }_{ 2 } } \right) \)
On further simplifying and rearranging
\(\cfrac { 1 }{ v } -\cfrac { 1 }{ u } =\left( \cfrac { { n }_{ 2 }-{ n }_{ 1 } }{ { n }_{ 1 } } \right) \left( \cfrac { 1 }{ { R }_{ 1 } } -\cfrac { 1 }{ { R }_{ 2 } } \right) \) 
\(\cfrac { 1 }{ v } -\cfrac { 1 }{ u } =\left( \cfrac { { n }_{ 2 } }{ { n }_{ 1 } } -1 \right) \left( \cfrac { 1 }{ { R }_{ 1 } } -\cfrac { 1 }{ { R }_{ 2 } } \right) \)    ....(iv) 
If the refractive of the lens is n₂ and is placed in air, then n₂= n and n₁= 1. So the equation (iv) becomes,
\(\cfrac { 1 }{ v } -\cfrac { 1 }{u } =(n-1) \left( \cfrac { 1 }{ { R }_{ 1 } } -\cfrac { 1 }{ { R }_{ 2 } } \right) \)  ...(v)
According to lens makers formula
\(\cfrac { 1 }{ f } =(n-1) \left( \cfrac { 1 }{ { R }_{ 1 } } -\cfrac { 1 }{ { R }_{ 2 } } \right) \)   ...(vi) 
Comparing the two equations (v) and (vi), we find
 \(\cfrac { 1 }{ v } -\cfrac { 1 }{ u }=\cfrac { 1 }{ f }\)          .....(vii)
This is called lens equation

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.

Optical path:
Optical path of a medium is defined as the distance d' light travels in vacuum in the same time it travels a distance d in the medium.
Equation for Optical Path:
Let us consider a medium of refractive index n and thickness d. Light travels with a speed v through the medium in a time t. The speed of light through the medium is written as,

V = d/t ⇒ t = d/v
In the same time t, light can cover a longer distance d' in vacuum as it travels with greater speed c in vacuum. It is shown in Figure. Now, we can write,
c = d'/t ⇒ t = d'/v
As the time taken in both the cases is the same, we can equate the time t,
\(\frac{d'}{c}=\frac{d}{v}\)
Rewritten for the optical path d' as,
\(d'=\frac{c}{v}d\)
\(As,\frac{c}{v}=n\)
The optical path d' =nd
The value of n is always greater than 1, for a medium. Thus, the optical path d' of a medium is always greater than d.

\(\mathrm{d} \sin \theta=n \lambda, As \ \theta\ is\ small, \ \theta=\frac{Y}{D} \)
\(d \theta =n \lambda \)
\(d \frac{Y}{D} =n \lambda \)
\(Y =\frac{D n \lambda}{d} \)
\(\mathrm{D}=2 \mathrm{~m}, \mathrm{n}=1, \lambda=600 \mathrm{~nm} \)
\(=6 \times 10^{-7} \mathrm{~m}, \mathrm{~d}=1 \times 10^{-3} \mathrm{~m} \)
\(\mathrm{Y}=\frac{\mathrm{D} \lambda}{\mathrm{d}} \text { (First minimum) } \)
\(\mathrm{Y}=\frac{2 \times 6 \times 10^{-7}}{1 \times 10^{-3}} \)
\(=12 \times 10^{-4} \)
\(Y =12 \times 10^{-4} \mathrm{~m} \)
Distance between first minima on either side of central maxima
\(\Delta \mathrm{y}=2 \mathrm{y}=2 \times 12 \times 10^{-4} \)
\(\Delta \mathrm{y}=24 \times 10^{-4} \mathrm{~m} \)

Yellow light is preferred to during fog because of penetrates deeply due to its longer wave length it is much smaller than fog particles (The scattering effect of fog is independent of wave length).

Yes, It is possible for two lens to produce zero power.
Explanation:
when the two lens (one is concave & another one is convex lens) are combined together, the focal length of the combination of the two lenses is F.
Then
\(\frac{1}{F}=\frac{1}{f_{1}}+\frac{1}{f_{2}} \)
\(\text { if } f_{1}=f_{2}=f \)
\(f_{1}=f(\text { convex }) \)
\(f_{2}=-f(\text { concave) } \)
\(P=\frac{1}{F}=\frac{1}{f}-\frac{1}{f}=0 . \ P \rightarrow \text { power } \)

The given situation is shown in the figure
Given, A = 60^o ; i₁ = 0^o; i₂ = 90^o
(a) Equation for angle of deviation,
d = i₁ + i₂ - A
Substituting the values
d = 0^o + 90^o- 60^o = 30^o
The angle of deviation produced is, d = 30°
(b) The light inside the prism must be falling on the second face at critical angle as it graces the boundary. i_c = 90° – 30° = 60°
Equation for critical angle is, \(\sin i_{c}=\frac{1}{n}\) 
\(n=\frac{1}{\sin i_{c}} ; \quad n=\frac{1}{\sin 60^{\circ}}=\frac{1}{\sqrt{3} / 2}=\frac{2}{\sqrt{3}}=1.15\) 
The refractive index of the material of the prism is, n = 1.15

Given, h = 5 mm = 0.5 cm, u₁ = –15 cm,
f₁ = 10 cm, f₂ = 5 cm, d = 40 cm
For the first lens, the lens equation is,
\(\cfrac { 1 }{ { v }_{ 1 } } -\cfrac { 1 }{ { u }_{ 1 } } =\cfrac { 1 }{ { f }_{ 1 } } \) 
Substituting the values,
\(\cfrac { 1 }{ { v }_{ 1 } } -\cfrac { 1 }{ -15 } =\cfrac { 1 }{ 10 } \);\(\cfrac { 1 }{ { v }_{ 1 } } +\cfrac { 1 }{ 15 } =\cfrac { 1 }{ 10 } \) 
\(\cfrac { 1 }{ { v }_{ 1 } } =\cfrac { 1 }{ 10 } -\cfrac { 1 }{ 15 } =\cfrac { 15-10 }{ 150 } =\cfrac { 5 }{ 150 } =\cfrac { 1 }{ 30 } \) 
v₁ = 30 cm
First lens forms image 30 cm to the right of first lens.
Let us find the height of this image.
Equation for magnification is, \(m=\cfrac { { h }' }{ { h } } =\cfrac { v_1 }{ u_1 } \) 
Substituting the values,\(\cfrac { { h }' }{ 0.5 } =\cfrac { 30 }{ -15 } \) 
\({ h }_{ 2 }=0.5\times \cfrac { 30 }{ -15 } =-1cm\) 
 As the height of the lens is negative, the image is inverted and real. 
This image acts as object for second lens. The object distance for second lenses, (40 – 30 = 10 cm). Hence, u₂ = –10 cm
For the second lens, the lens equation is,
\(\cfrac { 1 }{ { v }_{ 1 } } -\cfrac { 1 }{ { u }_{ 2 } } =\cfrac { 1 }{ { f }_{ 2 } } \) 
Substituting the values,
\(\cfrac { 1 }{ { v }_{ 2 } } -\cfrac { 1 }{ -10 } =\cfrac { 1 }{ 5 } ;\cfrac { 1 }{ { v }_{ 2 } } +\cfrac { 1 }{ 10 } =\cfrac { 1 }{ 5 } \) 
 \(\cfrac { 1 }{ { v }_{ 2 } } =\cfrac { 1 }{ 5 } -\cfrac { 1 }{ 10 } =\cfrac { 10-5 }{ 50 } =\cfrac { 5 }{ 50 } =\cfrac { 1 }{ 10 } \)
 v₂ = 10 cm
The image is formed 10 cm to the right of the second lens.
Let us find the height of the final image. Assume, the final height of the image formed by the second lens is h'′ and the height of the object for the second lens is h'.
Equation for magnification is m' for the
second lens is, \(m'=\cfrac { { h }^{' ' } }{ { h }^{ ' } } =\cfrac { { v }_{ 2 } }{ { u }_{ 2 } } \) 
Substituting the values,\(\cfrac { { h }^{ '' } }{ -1 } =\cfrac { 10 }{ -10 } \) 
\({ h }^{ '' }=\left( -1 \right) \times \left( \cfrac { 10 }{ -10 } \right) =1cm=10mm\) 
As the height of the image is positive, the image is erect and real.

Given, focal length of first lens, f₁ = –70 cm,
focal length of second lens, f₂ = 150 cm.
Equation for focal length of lenses in contact, \(\cfrac { 1 }{ f } =\cfrac { 1 }{ { f }_{ 1 } } +\cfrac { 1 }{ { f }_{ 2 } } \) 
Substituting the values,
\(\cfrac { 1 }{ f } =\cfrac { 1 }{ -70 } +\cfrac { 1 }{ 150 } =\cfrac { 1 }{ 70 } +\cfrac { 1 }{ 150 } \) 
\(\cfrac { 1 }{ f } =\cfrac { -150+70 }{ 70\times 150 } =\cfrac { -80 }{ 70\times 150 } =\cfrac { 80 }{ 10500 } \) 
\(f=\cfrac { -1050 }{ 8 } =131.25cm\) 
As the focal length is negative, the combination of two lenses is a diverging system of lenses
The power of the combination is,
\(P=\cfrac { 1 }{ f } =\cfrac { 1 }{ -1.3125m } =0.76D\)

This lens is called convexo-concave lens
Given, n = 1.52, R₁ = 10 cm and R₂ = 20 cm
Both R₁ and R₂ are positive
Lens makers formula,
\(\cfrac { 1 }{ f } =\left( n-1 \right) \left( \cfrac { 1 }{ { R }_{ 1 } } -\cfrac { 1 }{ { R }_{ 2 } } \right) \) 
Substituting the values,
\(\cfrac { 1 }{ f } =\left( 1.52-1 \right) \left( \cfrac { 1 }{ 10 } -\cfrac { 1 }{ 20 } \right) \) 
\(\cfrac { 1 }{ f } =\left( 0.52 \right) \left( \cfrac { 2-1 }{ 20 } \right) =\left( 0.52 \right) (\frac{1}{20})=\cfrac { 0.52 }{ 20 } \) 
\(f=\cfrac { 20 }{ 0.52 } =38.46cm\)
As the focal length is positive, the lens is a converging lens.

For a biconvex lens, radius of curvature of the first surface is positive and that of the second surface is negative as shown in the figure.

Given, n = 1.5, R₁ = 20 cm and R₂ = –15 cm
(a) Lensmaker’s formula \(\frac{1}{f}=(n-1)\left(\frac{1}{R_{1}}-\frac{1}{R_{2}}\right)\) 
Substituting the values,
\(\frac{1}{f}=(1.5-1)\left(\frac{1}{20}-\frac{1}{-15}\right)=(1.5-1)\left(\frac{1}{20}+\frac{1}{15}\right)\) 
\(\frac{1}{f}=(0.5)\left(\frac{1}{20}+\frac{1}{15}\right)=(0.5)\left(\frac{3+4}{60}\right)=\left(\frac{1}{2} \times \frac{7}{60}\right)=\frac{7}{120}\) 
\(f=\frac{120}{7}\) = 17.14 cm
As the focal length is positive the lens is a converging lens. 
(b) When the lens is flipped by the side,

Now, R₁= 15 cm and R₂ = –20 cm, n = 1.5
Substituting the values in the lens maker's formula,
\(\cfrac { 1 }{ f } =\left( 1.5-1 \right) \left( \cfrac { 1 }{ 15 } -\cfrac { 1 }{-20 } \right) \) 
\(\cfrac { 1 }{ f } =\left( 1.5-1 \right) \left( \cfrac { 1 }{ 15 } +\cfrac { 1 }{ 20 } \right) \) 
This will also result in, f = 17.14 cm
Thus, it is concluded that the focal length of the lens will not change if it is flipped by the side. This is true for any lens. Students can verify this for any kind of lens.

Given, u = –40 cm, R = –20 cm, n₁ = 1 and n₂ = 1.33
Equation for single spherical surface is
\(\cfrac { { n }_{ 2 } }{ v } -\cfrac { { n }_{ 1 } }{ u } =\cfrac { \left( { n }_{ 2 }-{ n }_{ 1 } \right) }{ R } \)
Substituting the values
\(\cfrac { 1.33 }{ v } -\cfrac { 1 }{ -40 } =\cfrac { \left( 1.33-1 \right) }{ -20 } ;\cfrac { 1.33 }{ v } +\cfrac { 1 }{ 40 } =\cfrac { \left( 0.33 \right) }{ -20 } \) 
\(\cfrac { 1.33 }{ v } =\cfrac { \left( 0.33 \right) }{ 20 } -\cfrac { 1 }{ 40 } ;\) 
\(\cfrac { 1.33 }{ v } =\cfrac { 0.66-1 }{ 40 } =\cfrac { 1.66 }{ 40 } \) 
\(v=-40\times \cfrac { 1.33 }{ 1.66 } =-32.0cm\)
The equation for magnification is,\(m=\cfrac { { h }_{ 2 } }{ { h }_{ 1 } } =\cfrac { { { n }_{ 1 }v } }{ { n }_{ 2 }u } \) 
\(\cfrac { { h }_{ 2 } }{ 1.0 } =\cfrac { \left( 1.0 \right) \times \left( -32 \right) }{ \left( 1.33 \right) \times \left( -40 \right) } =0.6cm\)  (or) h₂ = 0.6cm
The erect virtual image of height 0.6 cm is formed at 32.0 cm to the left of the single spherical surface.

Case i) \(\cfrac { { n }_{ 2 } }{ v } -\cfrac { { n }_{ 1 } }{ u } =\cfrac { \left( { n }_{ 2 }-{ n }_{ 1 } \right) }{ R } \)
applying sign convention, u = –10 cm, R = 15 cm
\(\cfrac { 2 }{ v } -\cfrac { 1 }{ -10 } =\cfrac { \left( 2-1 \right) }{ 15 } ;\cfrac { 2 }{ v } +\cfrac { 1 }{ 15 } =\cfrac { \left( 1 \right) }{ 10 } \) 
∴ =− 60 cm
[a virtual image is formed 60 cm, to the left of the surface]

Case ii) \(\cfrac { { n }_{ 2 } }{ v } -\cfrac { { n }_{ 1 } }{ u } =\cfrac { \left( { n }_{ 2 }-{ n }_{ 1 } \right) }{ R } \)
applying sign convention, u = –30 cm, R = 15 cm
\(\cfrac { 2 }{ v } -\cfrac { 1 }{ -30 } =\cfrac { \left( 2-1 \right) }{ 15 } ;\cfrac { 2 }{ v } +\cfrac { 1 }{ 30 } =\cfrac { \left( 1 \right) }{ 15 } \)
∴ = 60 cm [a real image is formed 60 cm, to the right of the surface]

Given, thickness of the slab, t = 0.25 m, refractive index, n = 1.5, angle of incidence, i = 60^o.
Using Snell’s law, 1 sin i = n sin r
\(sinr=\cfrac { sini }{ n } =\cfrac { sin60^o }{ 1.5 } =0.58\) 
\(r={ sin }^{ -1 }(0.58)=35.25^{ 0 }=35^o15'0''\) 
Lateral displacement is, \(L=t\left( \cfrac { sin\left( i-r \right) }{ cos\left( r \right) } \right) \) 
\(L=\left( 0.25 \right) \times \left( \cfrac { sin\left( 60-35.25 \right) }{ cos\left( 35.25 \right) } \right) =0.1281m\)
The  lateral displacement is, L = 12.81 cm

Given, n = 4/3, d = 10 m.
Radius of illumination, \(R=\cfrac { d }{ \sqrt { { n }^{ 2 }-1 } } \) 
\(R=\cfrac { 10 }{ \sqrt { \left( 4/3 \right) ^{ 2 }- } 1 } =\cfrac { 10\times 3 }{ \sqrt { 16-9 } } \) 
\(R=\cfrac { 30 }{ \sqrt { 7 } } =11.32cm\) 
To find the critical angle,
\({ i }_{ c }={ sin }^{ -1 }\left( \cfrac { 1 }{ n } \right) \) 
\({ i }_{ c }={ sin }^{ -1 }\left( \cfrac { 1 }{ 4/3 } \right) ={ sin }^{ -1 }\left( \cfrac { 3 }{ 4 } \right) =48.6^{ o }\)
The total angle of view of the cone is, \({ 2i }_{ c }=2\times { 48.6 }^{ 0 }={ 97.2 }^{ 0 }\) 

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.

\(\text{ longitudinal magnifcation}(m_l)=\frac { length\ of\ image\left( l' \right) }{ length\ of\ object\left( l \right) } \)
Given: length of object, \(l=\cfrac { f }{ 3 } \) 
For the given condition, the image formation is shown in the figure.

Let, l' be the length of the image, then
\(m=\cfrac { l' }{ l } =\cfrac { l' }{ f/3 } \) (or) \(l=\cfrac { m_lf }{ 3 } \) 
Image of one end coincides with the object. Thus, the coinciding end must be at center of curvature.
 \(u_B=u_A-\cfrac { f }{ 3 } =2f-\cfrac { f }{ 3 } =\cfrac { 5f }{ 3 } \)
\(v_B=u_B+l+l'\) 
\(v_b =\cfrac { 5f }{ 3 } +\cfrac { f }{ 3 } +\cfrac { mf }{ 3 } =\cfrac { f(6+m) }{ 3 } \)
 Mirror equation,\(\cfrac { 1 }{ v } +\cfrac { 1 }{ u } =\cfrac { 1 }{ f } \) 
\(\cfrac { 1 }{ -\left( \cfrac { f(6+m_l) }{ 3 } \right) } +\cfrac { 1 }{ -\left( \cfrac { 5f }{ 3 } \right) } =\cfrac { 1 }{ -f } \)
After simplifying,
 \(\cfrac { 3 }{ f(6+m_l) } +\cfrac { 3 }{ 5f } =\cfrac { 1 }{ f } ;\cfrac { 3 }{ (6+m_l) } =\cfrac { 2 }{ 5 } \) 
\(6+m_l=\cfrac { 15 }{ 2 } ;m_l=\cfrac { 15 }{ 2 } -6\) 
\(m_l=\cfrac { 3 }{ 2 } =1.5\) 

Given, f = –15 cm, u = –20 cm
(a) Mirror equation,\(\frac { 1 }{ v } +\frac { 1 }{ u } =\frac { 1 }{ f } \) 
Rewriting to find v,\(\cfrac { 1 }{ v } =\cfrac { 1 }{ f } -\cfrac { 1 }{ u } \) 
Substituting for f and u,\(\cfrac { 1 }{ v } =\cfrac { 1 }{ -15 } -\cfrac { 1 }{ -20 } \) 
\(\cfrac { 1 }{ v } =\cfrac { \left( -20 \right) -\left( -15 \right) }{ 300 } =\cfrac { -5 }{ 300 } =\cfrac { -1 }{ 60 } \) 
v = -60.0cm
The screen is to be placed at distance 60.0 cm to the left of the concave mirror.
(b) Magnification, \(m=\cfrac { h' }{ h } =-\cfrac { v }{ u } \) 
\(m=\cfrac { h' }{ h } =\cfrac { \left( -60 \right) }{ \left( -20 \right) } =-3\) 
As the sign of magnification is negative, the image is inverted.
As the magnitude of magnification is 3, the image is enlarged three times.
As the image is formed to the left of the concave mirror, the image is real.

\(\mathrm{n}_{1} \lambda_{1}=\mathrm{n}_{2} \lambda_{2} \)
\(\mathrm{n}_{1}=62 \text { fringes } \)
\(\lambda_{1}=5893 \mathrm{~A}^{0} \)
\(\lambda_{2}=4359 \mathrm{~A}^{0} \)
\(\mathrm{n}_{2}=? \)
\(\mathrm{n}_{2}=\frac{\mathrm{n}_{1} \lambda_{1}}{\lambda_{2}} \)
\(=\frac{5893 \times 62}{4359} \)
= 83.81 = 84 fringes.

Given data:
λ = 900 nm = 900 x 10^-9 m
λ ₂ = 750 nm = 750 x 10^-9 m
D = 2 m d = 2 nm = 2 x 10^-3 m
Let n^th order bright fringe of λ₁,
Coincides with (n + 1)^th order bright fringe of λ ₂
\(y_{n}=\frac{n \lambda_{1} D}{d}, Y_{n+1}=\frac{(n+1) \lambda_{2} D}{d} \)
\(\frac{n \lambda_{1} D}{d}=\frac{(n+1) \lambda_{2} D}{d} \)
\(n \lambda_{1}=(n+1) \lambda_{2} \)
\(\frac{n+1}{n}=\frac{\lambda_{1}}{\lambda_{2}}=\frac{900 \times 10^{-9}}{750 \times 10^{-9}}=\frac{18}{15}=\frac{6}{5} \)
\(1+\frac{1}{n} =\frac{6}{5} \)
\(\frac{1}{n} =\frac{6}{5}-1=\frac{6-5}{5} \)
\(\frac{1}{n} =\frac{1}{5} \)
n = 5
n + 1 = 6
5^th bright fringe of  λ ₁ coincides with 6^th bright fringe of λ ₂ in the least distance of y
\(y =\frac{n \lambda_{1} D}{d}=\frac{5 \times 900 \times 10^{-9} \times 2}{2 \times 10^{-3}} \)
\(=4500 \times 10^{-6}=4.5 \times 10^{-3} \)
y = 4.5 mm

From figure,
D = u + v
d = r - u
D + d = u + v + v - u
D + d = 2v

\(v=\frac{D+d}{2}\)
D - d = u + v - v + u = 2u
\(v=\frac{D-d}{2}\)
\(\frac{1}{f}=\frac{1}{v}-\frac{1}{u} \)
\(\frac{1}{f} =\frac{1}{\frac{D+d}{2}}-\frac{1}{\frac{D-d}{2}} \)
\(=\frac{2}{D+d}-\frac{2}{D-d} \)
\(=\frac{2[D-d+D+d]}{D^{2}-d^{2}} \)
\(=\frac{2 \times 2 D}{D^{2}-d^{2}} \)
\(\frac{1}{f} =\frac{4 D}{D^{2}-d^{2}} \quad \therefore f=\frac{D^{2}-d^{2}}{4 D} \)

\(P_{a}=\frac{1}{f_{a}}=\left(\frac{\mu_{g}}{\mu_{a}}-1\right)\left(\frac{1}{R_{1}}-\frac{1}{R_{2}}\right) \)  ..(1)
\(P_{w}=\frac{1}{f_{w}}=\left(\frac{\mu_{g}}{\mu_{w}}-1\right)\left(\frac{1}{R_{1}}-\frac{1}{R_{2}}\right) \)  ...(2)
\(p_{a}=5 D, f_{w}=-100 \text { (Diverging lens) } \)
\(\mu_{\mathrm{g}}=1.5, \mu_{\mathrm{a}}=1 \)
\(5=(1.5-1)\left(\frac{1}{R_{1}}-\frac{1}{R_{2}}\right) \)   ...(3)
\(\left(\frac{1}{R_{1}}-\frac{1}{R_{2}}\right)=\frac{5}{0.5}=\frac{50}{5}=10 \)
\(\frac{1}{f_{w}}=\left(\frac{1.5}{n_{w}}-1\right)\left(\frac{1}{R_{1}}-\frac{1}{R_{2}}\right) \)    ...(4)
\(\frac{-1}{100 \times 10^{-2}}=\left(\frac{1.5}{n_{w}}-1\right)\left(\frac{1}{R_{1}}-\frac{1}{R_{2}}\right) \)
\(-1=\left(\frac{1.5}{n_{w}}-1\right)(10) \Rightarrow \frac{-1}{10}=\frac{1.5}{n_{w}}-1 \)
\(\frac{1.5}{n_{w}}=\frac{-1}{10}+1 \)
\(\frac{1.5}{n_{w}}=\frac{9}{10} \)
\(n_{w}=\frac{1.5 \times 10}{9}=\frac{15}{9}=\frac{5}{3} \)
\(n_{w}=\frac{5}{3} \)

r - Radius of surface of water
h - Depth of the water
\(\sin c=\frac{1}{n} \)
\(n=\frac{4}{3} \)
h = 80 cm = 0.80 m
\(\sin c =\frac{1}{\frac{4}{3}}=\frac{3}{4} n=\frac{1}{\sin c} \)
\(\tan c =\frac{\sin c}{\cos c} \)
\(=\frac{\sin c}{\sqrt{1-\sin ^{2} c}} \)
\(=\frac{1}{n \sqrt{1-\frac{1}{n^{2}}}} \)
\(\tan c =\frac{1}{\sqrt{n^{2}-1}} \)
\(\frac{r}{h} =\frac{1}{\sqrt{n^{2}-1}} ; r=\frac{h}{\sqrt{n^{2}-1}} \)
\(=\frac{0.80}{\sqrt{\left(\frac{4}{3}\right)^{2}}-1} \)
\(=\frac{0.80}{\sqrt{\frac{16}{9}-1}}=\frac{0.80 \times 3}{\sqrt{7}} \)
\(=\pi r^{2}=\frac{22}{7} \times \frac{0.80 \times 3}{\sqrt{7}} \times \frac{0.80 \times 3}{\sqrt{7}} \)
\(=\frac{22}{49} \times 2.4 \times 2.4 \)
\(=\frac{22 \times 5.76}{49}=\frac{126.72}{49}=2.586 \)
A = 2.6 m²

Given data :
Magnification of compound microscope m = 30
Focal length of eye piece f. = 5 cm
Image distance D = 25 cm
To find:
Magnification of objective m_o = ?
Formula :
Magnification m = m_om_e
\(m={ m }_{ 0 }\left( 1+\cfrac { D }{ { f }_{ e } } \right) \) 
\(30={ m }_{ 0 }\left( 1+\cfrac { 25 }{ 5 } \right) \) 
⇒ m_o = 5

(i) Resolution: Resolution is the quality of image which is decided by diffraction effect and Rayleigh criterion.
(ii) Resolving power: The ability of an optical instrument to separate or distinguish small or closely adjacent objects through the image formation is said to be resolving power of the instrument.

Consider two lenses 1 and 2 of focal length f₁ and f₂ are placed coaxially in contact with each other so that they have a common principal axis.
O be the object which is placed beyond the focus of the first lens on the principal axis. I' is the image of object O which is formed beyond the lens 2. Then, I' acts as an object for the lens 'P' is the common optical centre of the two lenses.
From the figure, PO =u, PI' = v' for lens I
PI' =  v' (object distance) PI = v (image distance) for lens 2

For lens 1,
\(\cfrac { 1 }{ v' } -\cfrac { 1 }{ u } =\cfrac { 1 }{ { f }_{ 1 } } \)   .........(1) 
For lens 2,
\(\cfrac { 1 }{ v } -\cfrac { 1 }{ u' } =\cfrac { 1 }{ { f }_{ 2 } } \)    .........(2)
Adding (1) of (2)
 \(\cfrac { 1 }{ v } -\cfrac { 1 }{ u } =\cfrac { 1 }{ f_1 }+\cfrac{1}{f_2} \)   .........(3)
(vi) If the combination acts as a single lens of focal length f so that for an object at the position O it forms the image at I,
Then,
\(\cfrac { 1 }{ v } -\cfrac { 1 }{ u } =\cfrac { 1 }{ f } \)   .........(4) 
Comparing equations (3) and (4) we can write, 
\(\cfrac { 1 }{ F } =\cfrac { 1 }{ { f }_{ 1 } } +\cfrac { 1 }{ { f }_{ 2 } } \) .........(5)
The above equation can be extended for any number of lenses in contact as, 
\(\cfrac { 1 }{ f } =\cfrac { 1 }{ { f }_{ 1 } } +\cfrac { 1 }{ { f }_{ 2 } } +\cfrac { 1 }{ { f }_{ 3 } }+\cfrac { 1 }{ { f }_{ 4 } } +..........\)

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) The image formed by a plane mirror is virtual, erect, and laterally inverted.
(ii) The size of the image is equal to the size of the object.
(iii) The image distance behind the mirror is equal to the object distance in front of the mirror.
(iv) If an object is placed between two plane mirrors inclined at an angle 0, then the number or
images n formed is given in Table.

Snell's law in the product form for critical angle incidence becomes,
\(n_{1} \sin i_{c}=n_{2} \sin 90^{\circ} \)
\(n_{1} \sin i_{c}=n_{2} \)
\(\sin i_{c}=\frac{n_{2}}{n_{1}} \)
\(Here, \mathrm{n}_{1}>\mathrm{n}_{2}\)
n₁ - refractive index of denser medium = n
n₂ - refractive index of rarer medium = 1 (for air)
\(\sin i_{c}=\frac{1}{n}(o r) i_{c}=\sin ^{-1}\left[\frac{1}{n}\right]\)
Critical angle i_c depends on the refractive index n of the mediunm.

(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',
\(d-d'=d-\frac{d}{n}(or)d-d'=d(1-\frac{1}{n})\)

The law of reflection states that the incident ray, the reflected ray, and the normal to the surface of the mirror all lie in the same plane. The angle of reflection is equal to the angle of incidence.

(i) Let us Consider an object OO' of height h placed on the principal axis with its height perpendicular to the principal axis. The inverted real image Il' is formed which has a height h' as shown in Figure.
(ii)The lateral (or) transverse magnification m is defined as the ratio of the height of the image to height of the object.
\(m=\cfrac { II' }{ OO' } \)   .......(1) 
From the two similar triangles ΔPOO' and  ΔlPII'', we can write,
\(\cfrac { II' }{ OO' } =\cfrac { PI }{ PO } \)    ......(2) 
On applying sign convention,
\(\cfrac { -{ h }{ ' } }{ { h } } =\cfrac { v }{ -u } \) 
Substituting this in the equation (2) for magnification,

\(m=\cfrac { -{ h }{ ' } }{ { h } } =\cfrac { v }{ -u } \) 
After rearranging
\(m=\cfrac { { h }' }{ { h } } =\cfrac { v }{ u } \)  ......(3) 
(iii) The magnification is negative for real image and positive for virtual image. In the case of a concave lens, the magnification is always positive and less than one.
(iv) The equations for magnification by combining the lens equation with the formula for magnification as,
\(m=\cfrac { { h }' }{ { h } } =\cfrac { f }{ f+u } \quad m=\cfrac { { h }' }{ h } =\cfrac { f-v }{ f } \)   ......(4)