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Cork Institute of Technology
Bachelor of Engineering (Honours) in Electronic Engineering – Stage 1
(Bachelor of Engineering in Electronic Engineering – Stage 1)
(NFQ – Level 8)
Summer 2005
Physics
(Time: 3 Hours)
Instructions Answer FIVE questions All questions carry equal marks.
Values for constants: |electron charge | = 1.602 x 10 -19^ C Electron rest mass = 9.110 x 10 –31^ kg Atomic mass unit, u = 1.661 x 10-27^ kg Planck constant = 6.626 x 10 -34^ J s Boltzmann constant = 1.381 x 10 -23^ J K - Speed of light in a vacuum = 2.998 x 10^8 m s- Universal Gravitation constant = 6.673 x 10 -11^ N m^2 kg -
Examiners: Prof. C. Burkley Mr. J. Ryan Mr. J. A. O’Doherty
Q1. (a) Explain why interference fringes are formed when thin films are illuminated with monochromatic light. Explain the difference you would expect to see in the visual appearance of the fringes if the source of illumination was white light. (8 marks) (b) An interference mirror consists of many layers of transparent material, which have alternating high and low refractive index. The refractive index and thickness of the layers, which have high refractive index, are 1.46 and 108.4 nm respectively. Calculate the wavelength for which the mirror is optimised. Also, explain why quarter-wavelength thick layers may be designed to act as either anti- reflection coatings or as reflectors (i.e. why the same coat thickness may be designed to give either destructive or constructive interference). (12 marks)
Q2. (a) Light, which has a wavelength of 546 nm in air, is absorbed in a silicon photodetector. Calculate the wavelength of the light and also the speed of light in silicon, if the refractive index of silicon is 3.5. (6 marks) (b) Calculate the focal length of a glass plano-concave lens, if the radius of curvature of the concave surface is 400 mm. The refractive index of the glass is 1.46. (8 marks) (c) Calculate the maximum and minimum magnifying powers for a single-lens magnifier that has a focal length of 12.5 mm. Explain where the final image is located for each case. (6 marks)
Q3. (a) A plane light-wave illuminates a very small rectangular aperture. Prove that the angular
width (2θ) of the central diffraction fringe which exits the slit satisfies the equation; sinθ =^ λd^ , where λ and d represent the wavelength of the light, and the slit width respectively. (12 marks) (b) Calculate the maximum viewing distance for a pair of binoculars, in order to distinguish clearly the lettering on a distant sign, if the smallest detail in the letters is 2 mm in size. Assume that the wavelength of day-light (average) is 500 nm and the diameter of the objective lens in the binoculars is 50 mm. (8 marks)
Q4. (a) Prove that the period T of a spring system of mass m , which oscillates in simple harmonic
motion, is given by the equation; T = 2 π mk , where k represents the elastic constant of
the spring. (12 marks) (b) The reference power level used in communications is 1 mW. Calculate the power in mW of a signal in a communications channel that has a level of -25 dB ( minus 25 dB ) with respect this reference. (8 marks)
Q5 (a) Prove that the kinetic energy of a solid that is rotating at an angular velocity (ω) about a
fixed axis is given by the equation; kinetic energy = 12 I ω^2 , where I represents the moment of inertia of the solid. (12 marks) (b) A spring system, which is oscillating in simple harmonic motion, has a mass of 0.5 kg, energy of 174.4 mJ and amplitude of 80 mm. Calculate the period of the system. (8 marks)
Useful Equations
Simple harmonic motion; v = ±ω(r 2 – s 2 ) 1/2^ a = -ω^2 s Period, (T); oscillating spring system; T = 2π(m/k) 1/2^ ; orbit;T = 2π[R^3 /(G M)]1/
Rotation; kinetic energy = ½ I ω^2 torque (τ) = (I) x (α)
power = (τ) x (ω) I (^) disc = M R^2 /2, I (^) sphere = 0.4 M R^2 Thermal expansion; α = (∆L/L)(∆θ) - Young’s modulus of elasticity; E = (tensile stress)/(tensile strain)
Wave equation; (^) y = a cos[ (^2) λ^ π(x - v t)]
Doppler: detector moving towards a transmitter; ∆f/f = v (^) R/v Speed of a wave on a stretched string; v = (1/2L)(T/m)1/ Refractive index, n =c/v Lens makers formula; 1/f = (n – 1)(1/R 1 –1/R 2 )
Optical fibre; numerical aperture, N A = no sinαmax = √(n 12 – n 22 ) Resolution of a microscope; d = 1.22 λ/(2 n sinα) = 1.22 λ/(2 N A)
Thickness of an interference wedge; T = 2 xL λ
Thickness of an antireflection coating, T = 4 n^ λo
Diffraction in a single slit:
- the position of the minima is given by; sinθ = m λ/w
- the position of the missing fringes mm = d/w Brewster angle, (θ (^) B); for plane polarized reflected light; tanθB = n (^) 2/ n 1
Radiated energy; Stephan-Boltzmann law; W = ε σ T^4
Wien displacement law; λmax T = constant
Photoelectric equation; (1/2)mv (^2) max = hf - φ
deBroglie equation λ = h/p
‘Uncertainty Principle’ equation; ∆p ∆x ≥ h/4π Number of surviving radioactive nuclei; (Nt ) = No exp(-λ t)
Half life; T1/2 = 0.693/λ Einstein equation; E = m c 2 ; Conversion factor; 1 u = 931.5 MeV Gain of a PM tube; G = ηn
