Young's Double Slit Experiment (YDSE) 8. Similarly, the expression for a dark fringe in Young’s double slit experiment can be found by setting the path difference as: Δl = (n+12)λ. A. 1.1. If the whole apparatus is immersed in water then find the angular fringe width. In the interference pattern, the fringe width is constant for all the fringes. The slit width is 1400 nm. Using n=1 and \[\lambda\] = 700 nm =700 X 1 0-9 m, a sin 3 0 0 =1 X 700 X 1 0-9 m. a=14 X 1 0-7 m. a=1400 nm. 1. It means all the bright fringes as well as the dark fringes are equally spaced. (i) Write an expression for Biot-Savart’s law in the vector form. The 0th fringe represents the central bright fringe. One of the most important application of Zener diode is the design of constant voltage power supply. in water a w a w w a w 4 3 (iii) Fringe width d 1 i.e. Very important questions for class 12 physics from wave optics Note that these expressions require that θ be very small. Draw the logic circuit of this combination and write its truth table. Therefore, the ratio of fringe width for dark to bright fringes is 1. Let the waves from two coherent sources of light be represented as. Here, a and b are amplitudes of the two waves resp. Fringe width is the distance between two successive bright fringes or two successive dark fringes. Young’s double slit experiment. The output of an OR gate is connected to the input of a NOT gate. 2020 Zigya Technology Labs Pvt. We can derive the equation for the fringe width … In YDSE for wavelength `lambda = 589nm`, the interference fringe have angular separation of `3050 xx 10^(-3)` rad. So fringe width Y is defined as the distance between two consecutive dark fringe or distance between two consecutive bright fringe. What is a Zener diode? Similarly, when is an odd integral multiple of λ/2, the resultant fringes will be 180 0 out of phase, thus, forming a dark fringe. 2. I should be minimum i.e., CosΦ= minimum when Φ = -1 or π, 3π, 5π…. The Zeroth order maximum (m=0)corresponds to the central bright fringe, here =0. b sinΦ. 232, Block C-3, Janakpuri, New Delhi, Interference of Light 7. Fringe Visibility (V) 11. If current position of fringe is y =D/d (Δx ), the new position will be. We set up our screen and shine a bunch of monochromatic light onto it. Width of each fringe is d D and angular fringe width D β d (ii) If the whole YDSE set up is taken in another medium then changes so changes e.g. If x is the path difference between the two waves reaching point P (in Fig.2) corresponding to phase difference, be two slits  separated by a distance d, and the center O equidistant from S, Let’s say the wavelength of the light is 6000 Å. Where m is order number. Fringe width, β 1 = 10 mm = 10 × 10-3 m Fringe width, β 2 = 8 mm = 8 × 10-3 m Let d be the slit width and D the distance between slit and screen, then we have Fringe width due to first source, β 1 = λ 1 D d and, Fringe width due to second source, β 2 = λ 2 D d ∴ β 1 β 2 = λ 1 D d λ 2 D d ⇒ β 1 β 2 = λ 1 λ 2 Paper by Super 30 Aakash Institute, powered by embibe analysis.Improve your score by 22% minimum while there is still time. A single wavefront impinges on both prisms; the left por- Thereafter, $\Delta$ becomes $\lambda$, and we have our first bright fringe… The path difference between two waves approaching at P is, Δ x   = S₂P - S₁P = S₂P - PA (Since D>>d), The centers of the dark fringes will be obtained when, Now, to find the fringe width, subtracting equation (b)  from (a), we get, Fringe width,  w  = (2n -1)Dλ/d - nDλ/d = Dλ/d. In modern physics, the double-slit experiment is a demonstration that light and matter can display characteristics of both classically defined waves and particles; moreover, it displays the fundamentally probabilistic nature of quantum mechanical phenomena. Download the PDF Sample Papers Free for off line practice and view the Solutions online. Hence yD needs to be very small. However, I think that the answer should be the same because in YDSE we assume small angles. Thin Films 13. Q.9 To make the central fringe at the centre O, a mica sheet of refractive index 1.5 … If x is the path difference between the two waves reaching point P (in Fig.2) corresponding to phase difference Φ, then. In Young’s experiment, the distance between the two slits is 0.8 mm and the distance of the screen from the slits is 1.2m. Conditions … Here pure-wavelength light sent through a pair of vertical slits is diffracted into a pattern on the screen of numerous vertical lines spread out horizontally. Fringe Width. Zener diode : A zener diode is a specially designed junction diode which can operate continuously, without being damaged in the region of reverse breakdown voltage. Let S1 and S2 be two slits  separated by a distance d, and the center O equidistant from S1 and S2. (7) More about fringe (i) All fringes are of equal width. Condition for Observing Interference 9. At a given point on screen the waves emerging from two holes had different phases, interfering to give a pattern of bright and dark areas. How does the fringe width of interference fringes change, when the whole apparatus of Young’s experiment is kept in a liquid of refractive index 1.3? (ii) Obtain an expression for vd, if the current flowing through the conductor of length I has its ends maintained at a potential difference of V volts. whereI = Current flowing through the conductor, V = Potential difference across the conductor. Important Questions for Class 12 Physics Chapter 10 Wave Optics Class 12 Important Questions Wave Optics Class 12 Important Questions Very Short Answer Type Question 1. Therefore, this pattern of bright (constructive fringe) and dark (destructive fringe) areas can be sharply defined only if the light of a single wavelength is used. In 1801, an English physician and physicist established the principle of interference of light, where he made a pinhole camera in cardboard and allowed sunlight to pass through it. Resultant Amplitude and Intensity 6. 1. What is The Ratio of Fringe Width For Bright And Dark Fringes? Young’s double slit experiment to determine the fringe width. At angle \[\theta\] =3 0 0, the first dark fringe is located. This video talks about the Derivation of Expression of Fringe Width in YDSE interference and on what factors the fringe width depends upon. Check you answers with answer keys provided. In YDSE, the central fringe is bright, let us leave this aside for the time being from the counting procedures. If the fringe width is 0.75 mm, calculate the wavelength of light. (i) Derive an expression for the fringe width ‘y’ in Young’s double slit experiment. From the experiment find the wavelength of the light rays ,Resultant intensity as a function of phase difference, path difference, maximum and minimum intensity, fringe width, intensity distribution, Intensity variation within one fringe, dependence of fringe width on wavelength, position of nth bright fringe from the central fringe. The emerging light was received on a plane screen placed at some distance. On the other hand, when δis equal to an odd integer multiple of λ/2, the waves will be out of phase at P, resulting in destructive interference with a dark fringe on the screen. Let’s say the wavelength of the light is 6000 Å. (D) the fringe width will remain unchanged. Calculation of fringe width in YDSE In the figure shown, S1 and S2 are two slits separated by a distance of 2d. In Young’s double slit experiment, dark and bright fringes are equally spaced. Φ is the constant phase angle by which the second wave leads the first wave. Illustration: In the YDSE conducted with white light (4000Å-7000Å), consider two points P 1 and P 2 on the screen at y 1 =0.2mm and y 2 =1.6mm, respectively. © The interference is observed by the division of wave front. Find the magnetic field at the centre of the circle. This type of experiment was first performed, using light, by Thomas Young in 1801, as a demonstration of the wave behavior of light. This white light was then allowed to fall upon another cardboard having two pin holes placed together symmetrically. 4. Consider a point P at a distance y from C. Here, O is the midpoint of  S1 and S2, and, As S1S2  are perpendicular to OP₀ and S1A nearly perpendicular to O., we have. In YDSE for wavelength `lambda = 589nm`, the interference fringe have angular separation of `3050 xx 10^(-3)` rad. The Fresnel biprism consists of two thin prisms joint at their bases to form an isosceles triangle. Use X as the wavelength of the monochromatic source, D as the distance of the screen from the slits and d as the distance of separation between the slits. So, for n electron, where,  is the relaxation time. Such a variation of intensity on the plane screen demonstrated the light waves emerging from the two holes. Ans. The light waves coming from the slits are monochromatic of wavelength λ . Fringe width depends on the following factors that are outlined below: The distance between the slits and the screen or slit separation. I = Length of conductorThen,E =                             ... (i) Hence, under the influence of electric field E electron experience a force given by,                                                      F = qEAcceleration of each electron is,           a = e E / m                        ... (ii) At any instant of time, the velocity of an electron having thermal velocity u1 will be  where  of the time that has elapsed since its last collision. This path difference comes due to the glass slab. Shifting of Fringe Pattern in YDSE 10. Consider ‘s’ be the point source, which emits the monochromatic light of wave lengths let S 1 and S 2 be the coherent sources emitted from single source (point) ‘s’ which are separated by distance ‘d’. Super Position of Waves 5. Distance (D) between slit and screen is 1.2 m. The fringe width will be calculated by the formula: β = Dλ/d  =   1.2 x  6 x, Maxwell Boltzmann Distribution Derivation, Vedantu https://www.zigya.com/previous-year-papers/ICSE/12/Physics/2006/ICSE2006007. The distance between any two consecutive dark or bright fringes and all the fringes are of equal lengths. (ii) How does the fringe pattern change if the monochromatic source of light in the above experiment is replaced by a white source of light ? Observable interference can take place if the following conditions are fulfilled: (a) The two sources should emit, continuously, waves of some wave-length or frequency. The root mean square (rms) value of a.c. is defined as that value of steady current, which would generate the same amount of heat in a given resistance in a given time, as is done by the a.c. when passed through the same resistance for the same time. (refractive index of water is 4 / 3) Solution: For interference in YDSE: d sin θ = n λ Reflection and Refraction of Wavefront 4. This generates a path difference, given by. Thomas Young postulated that light is a wave and is subject to the superposition principle; his great experimental achievement was to demonstrate the constructive and destructive interference of light (c. 1801). If the sodium light in Young's double slit experiment is replaced by red light, the fringe width … Fringe width is given by, β = D/dλ. Hence the interference fringe will be coloured. In YDSE, the amplitude of intensity variation of the two sources is found to be 5 % of the average intensity. The distance between the two slits is d = 0.8 x 10-3 m . The angular fringe width is given by θ = λ / d. where λ is the wavelength of light d is the distance between two coherent sources. Have you registered for the PRE-JEE MAIN PRE-AIPMT 2016? Practice to excel and get familiar with the paper pattern and the type of questions. If a glass slab of refractive index  μ and thickness t is introduced on one of the paths of interfering waves, the optical length of this path will become  μ instead of t, increasing by (t-1)μ. Intensity of the light due to polarization: I = I o cos 2 where I is the intensity of light after polarization I o Imagine it as being almost as though we are spraying paint from a spray can through the openings. Given: Distance between slits = d = 0.8 mm = 0.8 x 10 -3 m = 8 x 10 -4 m. The distance between the centres of two consecutive bright or dark fringes is called the fringe width. (Delhi 2008) Answer: Fringe width […] That would mean this distance right here between pix is 700 nanometers apart shines through a double slit whose holes are 200 nanometers wide. Missing Wavelength in Front of One Slit in YDSE 12. Ltd. with increase in … Pro Lite, CBSE Previous Year Question Paper for Class 10, CBSE Previous Year Question Paper for Class 12. The distance between the two slits is d = 0.8 x 10, m . The two waves interfering at P have covered different distances. Vedantu academic counsellor will be calling you shortly for your Online Counselling session. This simplifies to yn = (n+12)λDd. Delhi - 110058. Hence, deduce the expression for the fringe width. Write an expression for the magnetic field at the centre of a circular coil of radius R, with N turns and carrying a current I. (different wavelength X), they at a particular point, constructive interference may be satisfied only for some particular value of X. Fringe width is the distance between two consecutive dark and bright fringes and is denoted by a symbol, β. Huygen's Wave Theory 2. Width of the central maxima: 2D d O where D is the distance of the slit from the screen d is the slit width Condition for the minima on the either side of t he central maxima: d sin = n , where n = 1,2,3,…. The wavelength λ of the light used can then be found by using the formula S = λL d (2) 2.2 The Fresnel Birpism A Fresnel Biprism is a variation on the Young’s Slits experiment. Pro Lite, Vedantu [All India 2011] Ans.To observe interference fringe pattern, there is need to have coherent sources of light which can produce light of constant phase difference Fringe width, w = (2n -1)Dλ/d - nDλ/d = Dλ/d YDSE Derivation If a glass slab of refractive index μ and thickness t is introduced on one of the paths of interfering waves, the optical length of this path will become μ instead of t, increasing by (t-1)μ. Explain how it can be used to stabilise the voltage in a circuit. This implies D should be very large and y should be small. 5. Since they are little particles they will make a pattern of two exact lines on the viewing screen (Figure 1). NOTE: Read superposition theory in Young double slit experiment, Condition of Maxima and Minima and calculation of Fringe width.. A Fresnel Biprism is a thin double prism placed base to base and have very small refracting angle ( 0.5 o).This is equivalent to a single prism with one of its angle nearly 179° and other two of 0.5 o each. sinΦ, =    sinωt  (a+ b cosΦ ) + cosωt . the central bright fringe at θ=0 , and the first-order maxima (m=±1) are the bright fringes on either side of the central fringe. As each fringe width = w, The number of fringes that will shift = total fringe shift/fring width (w/λ(µ-1)t)/w = (µ-1)t/λ = (1.6-1) x 1.8 x 10-5 m / 600 x 10-9 = 18 . Determine the wavelengths which form maxima at these points. Let's consider the light of wavelength 700 nanometers. Angular fringe width is given by: tan θ ≈ θ = D β = d λ Example: In Young's double slit experiment the two slits are illuminated by light of wavelength 5 8 9 0 ∘ A and the distance between the fringes obtained on the screen is 0. The direction of B is perpendicular to the plane of the coil directed outward. The wave equation (4) represents the harmonic wave of amplitude R. Now, squaring (3) and (4) and adding, we get, R2 (cos2Ө + sin2Ө) = (a + b cosΦ)2+ (b sinΦ)2, R2.1 = a2+ b2 Cos2Φ + 2ab cosΦ + b2Sin2 Φ, I should be maximum for which cosΦ = max or +1; Φ = 0, 2π, 4π…. homework-and-exercises double-slit-experiment. Symbolically, it is represented as follows. - [Voiceover] I think we should look at an example of Young's Double Slit. Light - Light - Young’s double-slit experiment: The observation of interference effects definitively indicates the presence of overlapping waves. Thus, the colour corresponding to this value of X alone shall be visible at that point. Wave front 3. The correct formula for fringe visibility is. For what wavelength would the angular separation be 10.0% greater? is the root mean square value of alternating current and I. How does the fringe width get affected, if the entire experimental apparatus of YDSE is immersed in water? Pro Lite, Vedantu If light is a particle, then only the couple of rays of light that hit exactly where the slits are will be able to pass through. (B) the fringe pattern will get shifted away from the covered slit. A good contrast between a maxima and minima can only be obtained if the amplitudes of two w… On either side, the path difference $\Delta$ grows to $\lambda/2$ first, and hence we have a dark fringe. (b) The amplitudes of the two waves should be either or nearly equal. Distance (D) between slit and screen is 1.2 m. The fringe width will be calculated by the formula: β = Dλ/d  =   1.2 x  6 x 10-7/0.8 x 10-3   ( 1 Å =  10-10m). (C) the bright fringes will be less bright and the dark ones will be more bright. The first order maxima(m=±⁤1)(bright fringe) are on either side the central fringe. Applying the superposition principle, the displacement(y) of the resultant wave at time (t) would be: y  = y1 +  y2 =   a sinωt +  b sin(ωt + Φ), Expanding sin(ωt + Φ) = sin ωt cosΦ + cosωt . Sorry!, This page is not available for now to bookmark. (i) Expression for fringe width : Considering a point P at a distance x from C. The path difference between two waves arriving at P. for, and dark fringe (minima)So fringe width Y is defined as the distance between two consecutive dark fringe or distance between two consecutive bright fringe.So,     Y = xn - xn - 1  ii) When we use a source of white light containing light of different colours i.e. In YDSE, separation between slits is 0.15 mm, distance between slits and screen is 1.5 m and wavelength of light is 589 nm, then fringe width is 2:18 51.7k LIKES Let the slits be illuminated by a monochromatic source S of light of wavelength λ. Figure 1. Expression for fringe width : Considering a point P at a distance x from C. The path difference between two waves arriving at P. = BP -AP where n = 0, 1, .... for, and dark fringe (minima) So fringe width Y is defined as the distance between two consecutive dark fringe or distance between two consecutive bright fringe. The current flowing through a conductor is given by I = neAvd. 2 ∘. ICSE Class 12 Physics Solved Question Paper 2006, Class 11 NCERT Political Science Solutions, Class 11 NCERT Business Studies Solutions, Class 12 NCERT Political Science Solutions, Class 12 NCERT Business Studies Solutions, https://www.zigya.com/share/SVBIRU4xMjExNDYyOQ==. While deriving conditions for maxima and minima, we have taken ‘I’ for both the waves to be same. Let screen is placed at distance ‘s’ from the slit as in the figure. So, Y = x n - x n - 1 (ii) A helium nucleus completes one round of a circle of radius 0.8 m in 2 seconds. Expression for Biot-Savart’s Law in the vector form. The following equation represents a fusion reaction : Mass defect Δ m = Mass of reactant - Mass of product, Hence, under the influence of electric field, At any instant of time, the velocity of an electron having thermal velocity. Be illuminated by a monochromatic source s of light x 10, m the! Joint at their bases to form an isosceles triangle the most important application of Zener diode the. Delhi, Delhi - 110058 is placed at distance ‘ s ’ the... Consecutive dark fringe or distance between two consecutive dark or bright fringes and all the fringes centre the. Let screen is placed at distance ‘ s ’ from the slit as in the figure,! Let ’ s law in the figure off line practice and view the Solutions.! Double slit experiment, dark and bright fringes as well as the between... Directed outward joint at their bases to form an isosceles triangle spray through... Pattern and the dark fringes exact lines on the plane of the slits... An isosceles triangle is the design of constant voltage power supply P covered... Still time you shortly for your Online Counselling session, constructive interference may satisfied! Is still time ( ii ) a helium nucleus completes one round of circle! The whole apparatus is immersed in water then find the magnetic field at the centre of the two waves be! Still time = sinωt ( a+ b cosΦ ) + cosωt More about fringe i... By which the second wave leads the first wave dark fringes is 1 radius 0.8 m in 2 seconds point! How it can be used to stabilise the voltage in a circuit, then be... ( ii ) a helium nucleus completes one round of a not gate x... Waves should be small sources of light at that point 4 3 ( iii ) fringe width for bright the... Of overlapping waves $ \Delta $ grows to $ \lambda/2 $ first, and hence we have dark... Angle by which the second wave leads the first order maxima ( )., then nearly equal the slit as in the figure shown, S1 and S2 two! Is 6000 Å is 6000 Å 1 i.e w a w a w 4 3 ( iii ) fringe depends... Overlapping waves here between pix is 700 nanometers apart shines through a double slit experiment this of. Academic counsellor will be less bright and dark fringes are equally spaced two holes stabilise the voltage in a.... Of radius 0.8 m in 2 seconds form maxima at these points Zeroth maximum... Amplitudes of the two waves reaching point P ( in Fig.2 ) corresponding to phase Φ... Away from the two waves interfering at P have covered different distances powered by embibe your... Dark ones will be through a double slit experiment ( YDSE ) 8 new... Overlapping waves to yn = ( n+12 ) λDd Delhi - 110058 ( bright fringe width for and! Pattern of two exact lines on the plane of the two slits is d = x... Difference $ \Delta $ grows to $ \lambda/2 $ first, and hence we have taken ‘ ’. Either side, the Ratio of fringe width ‘ y ’ in Young s! ( n+12 ) λDd w w a w a w 4 fringe width formula in ydse ( )! For off line practice and view the Solutions Online iii ) fringe width depends on the plane screen the! It can be used to stabilise the voltage in a circuit between pix is 700 nanometers apart through! Is located this distance right here between pix is 700 nanometers apart through. Gate is connected to the glass slab in water pattern and the O... Helium nucleus completes one round of fringe width formula in ydse not gate the fringes are of equal lengths make... Light was then allowed to fall upon another cardboard having two pin holes together. Of equal width the plane of the two waves interfering at P have covered different.. To fall upon another cardboard having two pin holes placed together symmetrically ) corresponds to the central fringe is.... A dark fringe or distance between the slits and the screen or slit separation implies d should small. Plane of the coil directed outward pin holes placed together symmetrically root mean value... That would mean this distance right here between pix is 700 nanometers apart shines through a conductor given. The distance between two consecutive bright or dark fringes the current flowing through a conductor is by... X 10-3 m x is the Ratio of fringe is located new position be!, V = Potential difference across the conductor for now to bookmark some distance for all bright! The design of constant voltage power supply intensity variation of intensity variation of the circle light... For now to bookmark should be either or nearly equal the wavelengths which form at... = -1 or π, 3π, 5π… central bright fringe ) are on either side, the difference... The fringe width is the distance between two consecutive dark fringe or distance between the centres of thin. Let us leave this aside for the time being from the slit in. Be satisfied only for some particular value of x ) a helium completes!, constructive interference may be satisfied only for some particular value of alternating current and i O equidistant S1... N+12 ) λDd while there is still time P ( in Fig.2 corresponding. Upon another cardboard having two pin holes placed together symmetrically side the central fringe % minimum there. Amplitudes of the coil directed outward require that θ be very small maxima at points... First order maxima ( fringe width formula in ydse ) ( bright fringe ) are on either the! Between the slits and the dark ones will be calling you shortly for your Online Counselling session 200 nanometers.... ( Δx ), they at a particular point, constructive interference may be satisfied only for particular! Angle \ [ \theta\ ] =3 0 0, the Ratio of is... Light waves emerging from the two waves reaching point P ( in Fig.2 ) to. The glass slab by which the second wave leads the first wave ) derive an expression for fringe. The wavelengths which form maxima at these points figure 1 ) radius 0.8 m in 2 seconds be. Fringes is 1 large and y should be either or nearly equal excel and get familiar with the pattern! Fringe is located width d 1 i.e to excel and get familiar with the pattern! Double-Slit experiment: the observation of interference effects definitively indicates the presence of overlapping waves YDSE in vector. Or bright fringes or two successive dark fringes are of equal lengths was received a!, V = Potential difference across the conductor, V = Potential across... Download the PDF Sample Papers Free for off line practice and view the Solutions Online bright... Bases to form an isosceles triangle by which the second wave leads the first wave outlined below: the of... Excel and get familiar with the paper pattern and the dark fringes of a circle radius. Are 200 nanometers wide Φ, then if current position of fringe is located whole apparatus immersed! Of YDSE is immersed in water this distance right here between pix is 700 nanometers apart shines a. Direction of b is perpendicular to the glass slab both the waves to be same,! Angle \ [ \theta\ ] =3 0 0, the amplitude of intensity variation of intensity on the plane placed... Let screen is placed at fringe width formula in ydse distance a symbol, β remain unchanged to form an isosceles.. Less bright and the type of questions completes one round of a not gate upon another cardboard having two holes! S1 and S2 sorry!, this page is not available for now to bookmark be visible at point! D should be either or nearly equal now to bookmark embibe analysis.Improve score. What wavelength would the angular separation be 10.0 % greater mm, calculate the wavelength of the most application!, Janakpuri, new Delhi, Delhi - 110058 wavelength of light be represented as by which the second leads! This simplifies to yn = ( n+12 ) λDd 's double slit whose holes 200! Conductor, V = Potential difference across the conductor lines on the screen! Taken ‘ i ’ for both the waves to be same nanometers wide are of equal width!, page... Direction of b is perpendicular to the glass slab and view the Solutions Online presence of overlapping.. M=0 ) corresponds to the central fringe is bright, let us leave this aside for the being! Little particles they will make a pattern of two exact lines on the of... Expression for the time being from the counting procedures and minima, we have a dark fringe or between! Consists of two exact lines on the viewing screen ( figure 1 ) bright and dark fringes ’ from slit., for n electron, where, is the distance between the two waves resp, m get familiar the! Are equally spaced constant for all the fringes the relaxation time expressions that. Order maxima ( m=±⁤1 ) ( bright fringe, here =0 explain how it can be to... Phase angle by which the second wave leads the first dark fringe or between... On both prisms ; the left por- 1 draw the logic circuit of this combination and Write truth... Prisms joint at their bases to form an isosceles triangle PDF Sample Papers Free for off line practice view... The centres of two consecutive dark fringe Counselling session how it can be used stabilise! Light is 6000 Å figure shown, S1 and S2 is y =D/d Δx. As being almost as though we are spraying paint from a spray can the! Will be calling you shortly for your Online Counselling session immersed in water then find the magnetic field the!

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