Let us consider an object OO, of height h, placed h₁ principal on the principal axis with its height perpendicular to the principal has shown in- Figure' The ray OP passing through the pole of the lens go rated. The inverted real image II' formed has a height h,' The lateral or transverse magnification m is defined as the ratio of the height of the image to that of the object.
\(\mathrm{m}=\frac{\mathrm{II}^{\prime}}{\mathrm{OO}^{\prime}}\) ........................(1)
From the two similar uiangles POO' and PII, we can write,
\(\frac{\mathrm{II}^{\prime}}{\mathrm{OO}^{\prime}}=\frac{\mathrm{PI}}{\mathrm{PC}}\) ....................(2)
 Applying sign convention,
\(\frac{-\mathrm{h}_{2}}{\mathrm{~h}_{1}}=\frac{\mathrm{v}}{-\mathrm{u}}\) 
Substituting this in the equation for magnification,
\(\mathrm{m}=\frac{-\mathrm{h}_{2}}{\mathrm{~h}_{1}}=\frac{\mathrm{v}}{-\mathrm{u}}\) 
After rearranging,
\(\mathrm{m}=\frac{\mathrm{h}_{2}}{\mathrm{~h}_{1}}=\frac{\mathrm{v}}{\mathrm{u}}\) ........................(3)
The magnification is negative for real images and positive for virtual images- In the case of a concave lens, the magnification is always positive and less than one.
We can also have the equations for magnification by combining the lens equation with the formula for magnification as
\(\mathrm{m}=\frac{\mathrm{h}_{2}}{\mathrm{~h}_{1}}=\frac{\mathrm{f}}{\mathrm{f}+\mathrm{u}}=\frac{\mathrm{f}-\mathrm{v}}{\mathrm{f}}\) ......................(4)

Dispersion: It is splitting of white light into its constituent colours.
(i) Consider a beam of white light passes through a prism; it gets dispersed into its constituent colours as shown in Figure.
(ii) Let \(\delta_{v}, \delta_{R} \) are the angles of deviation for violet and red light. Let n_V and n_R are the refractive indices for the violet and red light respectively
(iii) The refractive index of the material of a prism is given by the equation
\(\mathrm{n}=\frac{\sin \left(\frac{\mathrm{A}+\mathrm{D}}{2}\right)}{\sin (\mathrm{A} / 2)}\)
(iv) Here A is the angle of the prism and D is the angle of minimum deviation. If the angle of prism is small of the order of 10^o, the prism is said to be a small angle prism.
(v) When rays of light pass through such prisms, the angle of deviation also becomes small. If A be the angle of a smitt angle prism and the angle of deviation then the prism formula becomes.

\(n=\frac{\sin \left(\frac{A+\delta}{2}\right)}{\sin (A / 2)}\)
For small angles of  \(A\ and \ \delta \)
\(\sin \frac{A+\delta}{A} \approx \frac{A+\delta}{A} \)
\(\sin \frac{A}{2} \approx \frac{A}{2} \)
\(n=\frac{(A+\delta / 2)}{(A / 2)}=\frac{A+\delta}{A}=1+\frac{\delta}{A} \)
Further simplifying, 
\(\frac{\delta}{A} =n-1 \)
\(\delta =(n-1) A \)    .....(1)
(vi) When white light enters the prism, the deviation is different for different colours. Thus, the refractive index is also different for different colours
For Violet colour, \( \delta_{\mathrm{v}}=\left(\mathrm{n}_{\mathrm{v}}-1\right) \mathrm{A} \)   ...(2)
For Red colour, \(\delta_{\mathrm{R}}=\left(\mathrm{n}_{\mathrm{R}}-1\right) \mathrm{A} \)   ....(3)
(vii) As, angle of deviation for violet colour \(\delta_{v}\) is greater the angle of deviation for red colour \(\delta_{\mathrm{R}}\) the refractive index for violet colour n_v is greater than the refractive index for red colour n_R Subtracting \(\delta_{v}\)  from \(\delta_{\mathrm{R}}\) we get
\(\delta_{\mathrm{v}}-\delta_{\mathrm{R}}=\left(\mathrm{n}_{\mathrm{v}}-\mathrm{n}_{\mathrm{R}}\right) \mathrm{A}\) ....(4)
(viii) The term \(\left(\delta_{v}-\delta_{R}\right)\)  is the angular separation between the two extreme colours (violet and red) in the spectrum is called the angular dispersion. If we take \(\delta\) is the angle of deviation for any middly ray (green or yellow) and the corresponding refractive index. Then,
\(\delta=(n-1) A\)    .....(5)
Dispersive power (ω):
It is the ability of the material of the prism to cause dispersion. It is defined as the ratio of the angular dispersion for the extreme colours to the deviation for any mean colour.
Dispersive power
\(\omega=\frac{\text { Angular dispersion }}{\text { Mean deviation }}=\frac{\delta_{v}-\delta_{R}}{\delta}\)  ....(6)
Substituting \(\left(\delta_{\mathrm{v}}-\delta_{\mathrm{R}}\right) \text { and }(\delta) \)
\(\omega=\frac{\mathrm{n}_{\mathrm{v}}-\mathrm{n}_{\mathrm{R}}}{(\mathrm{n}-1)} \)  .......(7)
(ix) Dispersive power is a dimensionless quality It has no unit. Dispersive power is always positive. The dispersive power of a prism depends only on the nature of material of the prism and it is independent of the angle of the prism.

Angle of deviation Produced by Prism:
(i) Let light ray PQ is incident on one of the refracting faces of the prism.
(ii) The angles of incidence and refraction at the first face AB are i₁ and r_l. The path of the light inside the prism is QR.
(iii) The angle of incidence and refraction at the second face AC is r₂ and i₂ respectively.
(iv) RS is the ray emerging from the second face. Angle i₂ is also caned angle of emergence.
(v) The angle between the direction of the incident ray PQ and the emergent ray RS is called the angle of deviation d.
(vi) The two normals drawn at the point of incidence Q and emergence R meet at point N. They meet at point N.
(vii) The extended incident ray and the emergent ray meet at a point M.

The angle of deviation d₁ at the surface AB is,
ㄥRQM = d = i₁ - r₁ ...(1)       
The angle of deviation d₂ at the surface AC is
ㄥQRM = d₂ = i₂ - r₂          .......(2)
Total angle of deviation d produced is, 
d = d₁ + d₂             .....(3)
Substituting for d₁ and d₂ in equation (3)
d = (i₁ - r₁) + (i₂ - r₂)
After rearranging,
d = (i₁ - r₁) + (i₂ - r₂)                    ........(4)
 In the quadrilateral AQNR, two of the angles (at the vertices Q and R) are right angles. Therefore, the sum of the other angles of the quadrilateral is 180°.
\(\angle A+\angle QNR={ 180 }^{ 0 }\)         .........(5)
From the triangle ΔQNR 
\({ r }_{ 1 }+{ r }_{ 2 }+\angle QNR={ 180 }^{ o }\)                ......(6)
Comparing these two equations (5) and (6) we get,
r₁ + r₂ = A              .......(7)
Substituting this in equation (4) for angle of deviation,
d = i₁+ i₂ - A                .............(8)
(viii) Thus, the angle of deviation depends on the angle of incidence i₁, angle of emergence i₂ and the angle for the prism A. 
(ix) For a given angle of incidence the angle of emergence is decided by the refractive index of the material of the prism. Hence the angle of deviation depends on these following factors.
(i) the angle of incidence
(ii) the angle of the prism.
(iii) the refractive index of the material of the prism (which decides the angle of emergence).
Refractive index of the material of the prism:

At minimum deviation, i₁ = i₂ = i and r₁ = r₂ = r
Now, the equation (8) becomes,
D - i₁ + i₂ - A = 2_i - A (or) \(i=\cfrac { \left( A+D \right) }{ 2 } \) 
The equation (7) becomes
r₁ + r₂ = A ⇒ 2r = A (or) \(r=\cfrac { A }{ 2 } \)
Substituting i and r in Snell's law
\(n=\cfrac { sini }{ sinr } \) 
\(n=\cfrac{\cfrac{sin(A+D)}{2}}{sin(A/2)}\)

(i) Let us consider two transparent media with refractive indices n, and n, which are separated by a spherical surface. Let C be the centre of curvature of the spherical surface. Let a point object O be in the medium n.
(ii) The line OC cuts the spherical surface at the pole P of the surface. As the rays considered are paraxial rays, the perpendicular dropped for the point of incidence to the principal axis is very close to the pole (or) passes through the pole itself.
(iii) Light from O falls on the refracting surface at N. The normal drawn at the point of incidence passes through the centre of curvature C.
(iv)  As n₂ > n₁ light in the denser medium deviates towards the normal and meets the principal axis at I where the image is formed.
(v) Snell's law in product form for the refraction at the point N can be written from the cquation,
n₁ sin i = n₂ sin r   ...(1)
(vi) As the angles are small, sine of the angle could be approximated to the angle itself,
n₁ i = n₂r                .........(2)
Let the angles be,
\(\angle NOP=\alpha ,\angle NCP=\beta ,\angle NIP=\gamma \)
From the right angle triangles, ∆NOP, ∆NCP and ∆NIP 
\(tan\alpha =\cfrac { PN }{ PO } ;tan\beta =\cfrac { PN }{ PC } ;tan\gamma =\cfrac { PN }{ PI } \) 
As these angles are small, tan of the angle could be approximated to the angle itself.
\(\alpha =\cfrac { PN }{ PO } ;\beta =\cfrac { PN }{ PC } ;\gamma =\cfrac { PN }{ PI } \)            ................(3)
For the triangle, ΔONC,
\(i=\alpha +\beta \)                 ......(4)
For the triangle, ΔINC,
\(\beta =r+\gamma (or)r=\beta -\gamma \)               ...............(5)
Substituting for i and r from equations (4) and (5) in equation (2),
\({ n }_{ 1 }(\alpha +\beta )={ n }_{ 2 }\left( { \beta -\gamma } \right) \) 
After rearranging,
\({ n }_{ 1 }a+{ n }_{ 2 }\gamma =\left( { n }_{ 2 }-{ n }_{ 1 } \right) \beta \) 
Substituting for α, β and y from equation
\({ n }_{ 1 }\left( \cfrac { PN }{ PO } \right) +{ n }_{ 2 }\left( \cfrac { PN }{ PI } \right) ={ (n }_{ 2 }-{ n }_{ 1 })\left( \cfrac { PN }{ PC } \right) \) 
Further simplifying by cancelling PN,
\(\cfrac { { n }_{ 1 } }{ PO } +\cfrac { { n }_{ 2 } }{ PI } =\cfrac { { n }_{ 2 }-{ n }_{ 1 } }{ PC } \)          .............(6)
Following sign conventions, PO = -u, PI = +v and PC = +R in equation (6)
\(\cfrac { { n }_{ 1 } }{ -u } +\cfrac { { n }_{ 2 } }{ v } =\cfrac { \left( { n }_{ 2 }-{ { n }_{ 1 } } \right) }{ R } \) 
After rearranging, finally we get,
\(\cfrac { { n }_{ 1 } }{ -u } +\cfrac { { n }_{ 2 } }{ v } =\cfrac { \left( { n }_{ 2 }-{ { n }_{ 1 } } \right) }{ R } \)               ..................(7)
(vii) If the first medium is air then, n₁ = 1 and the second medium is taken just as n₂ = n, then the equation (7) is reduced to,
\(\cfrac { n }{ v } -\cfrac { 1 }{ u } =\cfrac { \left( n-1 \right) }{ R } \)             ....(8)

(i) Consider a glass slab of thickness t and refractive index n is kept in air medium.
(ii) If path of the light is ABCD and the refractions occur at two points B and C in the glass slab.
(iii) The angles of incidence i and refraction r are measured with respect to the normal N₁ and N₂ at the two points Band C respectively. The lateral displacement 'L' is the perpendicular distance CE drawn between the path of light and the undeviated light at point C. In the right angle triangle ΔBCE,
\(sin(i-r)=\frac{1}{BC};BC=\cfrac { L }{ sin(i-r) } \)   ..(1)
In the right angle triangle ΔBCF,
\(cos(r)=\cfrac { t }{ BC } ;BC=\cfrac { t }{ cos(r) } \) 
Equating equation (1) and (2),
\(\cfrac { L }{ sin(i-r) } =\cfrac { t }{ cos(r) } \) 
After rearranging,
\(L=t\left( \cfrac { sin(i-r) }{ cos(r) } \right) \) 
(iv) Lateral displacement depends upon
(a) the thickness of the slab
(b) the angle of incidence 
(c) the refractive index of the slab.
(v) Thicker the slab, larger will be the lateral displacement. Greater the angle of incidence, larger will be the lateral displacement.
(vi) Higher the refractive index, larger will be the lateral displacement.

From the Snell's law in product form, n₁ sini = n₂ sinr  ....(1)
The equation for this refraction at the point A is as shown in the Figure 


n₃sin i_a = n₁sin r_a  ........(1)
(vi) To have the total internal reflection inside optical fibre, the angle of incidence at the core-cladding interface at B should be at least critical angle i_e'  From equation (1), for equation for the refraction at point B is,
n₁ sin i_c = n₂ sin 90^o      ...(3)
n₁sin i_c = n₂          ∵ sin 90^o=1 
\(sin{ i }_{ c }=\cfrac { { n }_{ 2 } }{ { n }_{ 1 } } \)     ...(4) 
From the right angle triangle Δ ABC,
i_c = 90^o - r_a
Now, equation (4) becomes
\(sin\left( { 90 }^{ o }-{ r }_{ a } \right) =\cfrac { { n }_{ 2 } }{ { n }_{ 1 } } \) (or) \(cos{ r }_{ a }=\cfrac { { n }_{ 2 } }{ { n }_{ 1 } } \)  ...(5)   
 \(sin{ r }_{ a }=\sqrt { 1-{ cos }^{ 2 }{ r }_{ a } } \) 
Substituting for cos r_a
\(sin{ r }_{ a }=\sqrt { 1-\left( \cfrac { { n }_{ 2 } }{ { n }_{ 1 } } \right) ^{ 2 } } =\sqrt { \cfrac { { n }_{ 1 }^{ 2 }-{ n }_{ 2 }^{ 2 } }{ { n }_{ 1 }^{ 2 } } } \)     ...(6) 
Substituting this in equation
\({ n }_{ 3 }{ sini }_{ a }={ n }_{ 1 }\sqrt { \cfrac { { n }_{ 1 }^{ 2 }-{ n }_{ 2 }^{ 2 } }{ { n }_{ 1 }^{ 2 } } } =\sqrt { { n }_{ 1 }^{ 2 }-{ n }_{ 2 }^{ 2 } } \)    ...(7) 
On further simplification
\(sini_{ a }=\cfrac { \sqrt { { n }_{ 1 }^{ 2 }-{ n }_{ 2 }^{ 2 } } }{ n_{ 3 } } \) (or) \(\quad { i }_{ a }=\sqrt { \cfrac { { n }_{ 1 }^{ 2 }-{ n }_{ 2 }^{ 2 } }{ { n }_{ 3 }^{ 2 } } } \)  ...(8) 
\(\therefore{ i }_{ a }={ sin }^{ -1 }\left( \sqrt { \cfrac { { n }_{ 1 }^{ 2 }-{ n }_{ 2 }^{ 2 } }{ { n }_{ 3 }^{ 2 } } } \right) \)     ...(9)
If outer medium is air, then n₃ = 1. The acceptance angle i_a becomes
\({ i }_{ a }={ sin }^{ -1 }\left( \sqrt { { n }_{ 1 }^{ 2 }-{ n }_{ 2 }^{ 2 } } \right) \)   ..(10)
Light can have any angle of incidence from 0 to i_a with the normal at the end of the optical fibre forming a conical shape called acceptance cone. In the equation (6), the term (n₃sini_a) is called numerical aperture NA of the optical fibre.
\(NA={ n }_{ 3 }{ sini }_{ a }=\sqrt { { n }_{ 1 }^{ 2 }-{ n }_{ 2 }^{ 2 } } \)
If outer medium is ab then n₃ = 1 
The numerical aperture NA becomes,
\(NA={ sini }_{ a }=\sqrt { { n }_{ 1 }^{ 2 }-{ n }_{ 2 }^{ 2 } } \)

(i) The angle of view for water animals is restricted to twice the critical angle 2i_c. The critical angle for water is 48.6°. Thus the angle of view is 97.2°.
(ii) The radius R of the circular area depends on the depth d from which it is seen and also the refractive indices of the media.
(iii) The radius R of Snell's window can be deduced with the illustration as shown in Figure.

(iv) Light is seen from a point A at a depth 'd'.
(v) From the Snell's law in product form, n₁ sini = n₂ sinr
(vi) The equation for the refraction happening at the point B on the boundary between the two media is,
n₁ sin i_c = n₂ sin90^o    ..(1)
n₁sini_c = n₂ (∵ sin90^o = 1)
\(sin{ i }_{ c }=\cfrac { { n }_{ 2 } }{ { n }_{ 1 } } \)     ...(2)
From the right angle triangle ΔABC,
\({ sini }_{ c }=\cfrac { CB }{ AB } =\cfrac { R }{ \sqrt { { d }^{ 2 }+{ R }^{ 2 } } } \)      ....(3)
Equating the above two equation 
\(\cfrac { R }{ \sqrt { { d }^{ 2 }+{ R }_{ 2 } } } =\cfrac { { n }_{ 2 } }{ { n }_{ 1 } } \) 
Squaring on both sides
\(\cfrac { { R }^{ 2 } }{ { R }^{ 2 }+d^{ 2 } } \left( \cfrac { { n }_{ 2 } }{ { n }_{ 1 } } \right) ^{ 2 }\) 
Taking reciprocal,
\(\cfrac { { R }^{ 2 }+{ d }^{ 2 } }{ { R }^{ 2 } } =\left( \cfrac { { n }_{ 1 } }{ { n }_{ 2 } } \right) ^{ 2 }\) 
On further simplifying
\(1+\cfrac { { d }^{ 2 } }{ { R }^{ 2 } } =\left( \cfrac { { n }_{ 1 } }{ { n }_{ 2 } } \right) ^{ 2 };\cfrac { { d }^{ 2 } }{ { R }^{ 2 } } =\left( \cfrac { { n }_{ 1 } }{ { n }_{ 2 } } \right) ^{ 2 }-1;\) 
\(\cfrac { { d }^{ 2 } }{ { R }^{ 2 } } =\cfrac { { n }_{ 1 }^{ 2 } }{ { n }_{ 1 }^{ 2 } } -1=\cfrac { { n }_{ 1 }^{ 2 }-{ n }_{ 2 }^{ 2 } }{ { n }_{ 2 }^{ 2 } } \) 
Again taking reciprocal and rearranging
\(\cfrac { { R }^{ 2 } }{ { d }^{ 2 } } =\cfrac { { { n }_{ 2 }^{ 2 } } }{ { n }_{ 1 }^{ 2 }-{ n }_{ 2 }^{ 2 } } { R }^{ 2 }={ d }^{ 2 }\left( \cfrac { { n }_{ 2 }^{ 2 } }{ { n }_{ 1 }^{ 2 }-{ n }_{ 2 }^{ 2 } } \right) \) 
∴ The radius of illumination is,
\(R=d\sqrt { \cfrac { { n }_{ 2 }^{ 2 } }{ \left( n_{ 1 }^{ 2 }-{ n }_{ 2 }^{ 2 } \right) } } \)   ...(4)
If the rarer medium outside is air, then, n₂ = 1, and we can take n₁ = n
\(R=d\left( \cfrac { 1 }{ \sqrt { { n }^{ 2 }-1 } } \right) \) or \(R=\cfrac { d }{ \sqrt { { n }^{ 2 }-1 } } \)   ....(5)

Apparatus:
The apparatus used by Fizeau for determining speed of light in air is shown in Figure 

(i) The light from the source S was first allowed to fall on a partially silvered glass plate G kept at an angle of 45° to the incident light.
(ii) The light then was allowed to pass through a rotating toothed-wheel with N teeth and N cuts of equal widths whose speed of rotation could be varied through an external mechanism.
(iii) The light passing through one cut in the wheel will get reflected by a mirror M kept at a long distance d, about 8 km from the toothed wheel.
(iv) The toothed wheel was not rotating, the reflected light from the mirror would again pass through the same cut and reach the eyes of the observer, through the partially silvered glass plate.
Working:
(i) The angular speed of rotation of the toothed wheel was increased from zero to a value w until light passing through one cut would completely be blocked by the adjacent tooth.
(ii) This is ensured by the disappearance of light while looking through the partially silvered glass plate.
Expression for speed of light:
The speed of light in air 'v' is equal to the ratio of the distance 2d (the distance light travelled from the toothed wheel to the mirror and back) to the time taken 't'.
\(v=\cfrac { 2d }{ t } \)   ...(i)
(v) The distance d is a known value from the arrangement. The time taken 't' for the light to travel the distance (2d) is calculated from the angular speed 'ω' of the toothed wheel.
(vi) The angular speed 'ω' of the toothed wheel when the light disappeared for the first time is,
\(\omega =\cfrac { \theta }{ t } \)  ...(2)
(vii) Here, θ is the angle between the tooth and the next slot which is turned within that time t.
\(\theta =\cfrac { total \ angle\ of \ the\ circle \ in \ radian }{ number\ of\ teeth+number\ of\ cuts } \)
\(\theta =\cfrac { 2\pi }{ 2N } =\cfrac { \pi }{ N } \) 
Substituting for θ in the equation for ω,
\(\omega =\cfrac { \pi /N }{ t } =\cfrac { \pi }{ Nt } \) 
Rewriting the above equation for t,
\(t=\cfrac { \pi }{ N\omega } \)    ....(3) 
Substituting 't' in equation (1)
\(V=\cfrac { 2d }{ \pi /N\omega } \)  
After rearranging,
\(v=\cfrac { 2dN\omega }{ \pi } \)
From this, the speed of light in air was determined as, v = 2.997 x 10⁸ m/s.

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)