In a series LCR circuit connected to an AC source, the total opposition to current is called impedance Z, given by Z = sqrt(R^2 + (XL - XC)^2), where XL is the inductive reactance and XC is the capacitive reactance. The phase angle phi between the applied voltage and the current is given by tan(phi) = (XL - XC)/R. The current is I = V/Z.

Z = sqrt(R^2 + (XL - XC)^2)

Resonance in a series LCR circuit occurs when the inductive reactance equals the capacitive reactance (XL = XC). At resonance the impedance is minimum (Z = R), the current is maximum, and the circuit behaves as purely resistive. The condition XL = XC gives omega L = 1/(omega C), so omega^2 = 1/(LC), and the resonant frequency f = 1/(2 pi sqrt(LC)). Such a circuit is used for tuning radios and televisions.

f = 1/(2 pi sqrt(LC))

A transformer works on the principle of mutual induction: an alternating current in the primary coil produces a changing flux that induces an emf in the secondary coil. For an ideal transformer, Vs/Vp = Ns/Np = Ip/Is, where V is voltage, N is number of turns and I is current. A step-up transformer has more turns in the secondary (Ns > Np) and increases the voltage; a step-down transformer has fewer turns in the secondary (Ns < Np) and decreases the voltage.

Vs/Vp = Ns/Np = Ip/Is

For an ideal transformer, Ns/Np = Vs/Vp. Given Vp = 2200 V, Vs = 220 V, Np = 5000. So Ns = Np x (Vs/Vp) = 5000 x (220/2200) = 5000 x 0.1 = 500 turns. Since the secondary has fewer turns than the primary, it is a step-down transformer, as expected.

Ns/Np = Vs/Vp

In order of increasing frequency (decreasing wavelength): (1) Radio waves - used in radio and television communication. (2) Microwaves - used in radar and microwave ovens. (3) Infrared - used in remote controls and thermal imaging. (4) Visible light - used for vision and photography. (5) Ultraviolet - used to sterilise water and in detecting forgery. (6) X-rays - used in medical imaging and detecting fractures. (7) Gamma rays - used in cancer treatment (radiotherapy).

Using the sign convention, u = -20 cm and f = -15 cm. From the mirror formula 1/v = 1/f - 1/u = 1/(-15) - 1/(-20) = -1/15 + 1/20 = (-4 + 3)/60 = -1/60, so v = -60 cm. The magnification m = -v/u = -(-60)/(-20) = -3. The image is real (v negative), inverted (m negative) and magnified three times, formed 60 cm in front of the mirror.

1/v + 1/u = 1/f

Total internal reflection is the phenomenon in which a ray of light travelling from a denser to a rarer medium is completely reflected back into the denser medium when the angle of incidence exceeds a certain angle called the critical angle. Conditions: (1) light must travel from a denser to a rarer medium; (2) the angle of incidence must be greater than the critical angle C, where sin C = 1/n. Applications: optical fibres, sparkling of diamonds, and mirages.

sin C = 1/n

The critical angle C is given by sin C = 1/n, where n is the refractive index of the denser medium (glass). So sin C = 1/1.5 = 0.667, giving C = sin inverse (0.667) = 41.8 degrees (approximately 42 degrees). For angles of incidence greater than this, total internal reflection occurs.

sin C = 1/n

The lens maker's formula is 1/f = (n - 1)(1/R1 - 1/R2), where f is the focal length of the lens, n is the refractive index of the lens material with respect to the surrounding medium, and R1 and R2 are the radii of curvature of the two surfaces of the lens (taken with proper sign convention). It shows that the focal length depends on the material and the shape (curvature) of the lens.

1/f = (n - 1)(1/R1 - 1/R2)

Using the sign convention, u = -30 cm and f = +20 cm. From the lens formula 1/v - 1/u = 1/f, we get 1/v = 1/f + 1/u = 1/20 + 1/(-30) = 1/20 - 1/30 = (3 - 2)/60 = 1/60, so v = +60 cm. The magnification m = v/u = 60/(-30) = -2. The image is real (v positive for a lens), inverted (m negative) and magnified twice, formed 60 cm on the other side of the lens.

1/v - 1/u = 1/f

When a ray passes through a prism, it is deviated; the deviation is least at the angle of minimum deviation Dm, when the ray passes symmetrically through the prism. The refractive index of the prism material is given by n = sin((A + Dm)/2)/sin(A/2), where A is the refracting angle of the prism and Dm is the angle of minimum deviation. This formula is used to determine the refractive index of a transparent material.

n = sin((A + Dm)/2)/sin(A/2)

An astronomical telescope has an objective lens of large focal length fo and an eyepiece of small focal length fe. In normal adjustment (final image at infinity), the magnifying power is M = fo/fe, and the length of the telescope is fo + fe. For high magnification, the objective should have a large focal length (and large aperture) and the eyepiece a small focal length. The final image is inverted.

M = fo/fe

In Young's double slit experiment, two coherent sources produce alternate bright and dark fringes on a screen. The fringe width (the distance between two consecutive bright or dark fringes) is beta = lambda D/d, where lambda is the wavelength of light used, D is the distance between the slits and the screen, and d is the separation between the two slits. All fringes are of equal width, and the width increases with wavelength and D but decreases as d increases.

beta = lambda D/d

Given lambda = 600 nm = 600 x 10^-9 m, d = 1 mm = 1 x 10^-3 m, D = 1 m. Fringe width beta = lambda D/d = (600 x 10^-9 x 1)/(1 x 10^-3) = 600 x 10^-6 = 6 x 10^-4 m = 0.6 mm.

beta = lambda D/d

Diffraction is the bending of light around the edges of an obstacle or aperture and its spreading into the geometrical shadow region. In diffraction at a single slit of width a, a central bright maximum is flanked by alternate dark and bright fringes. The condition for the dark fringes (minima) is a sin theta = n lambda, where n = 1, 2, 3, ... and theta is the angle of diffraction. The central maximum is the brightest and widest.

a sin theta = n lambda

Einstein's photoelectric equation is h f = W0 + KE(max), where h f is the energy of the incident photon (h is Planck's constant and f the frequency of light), W0 is the work function (the minimum energy needed to eject an electron from the metal surface), and KE(max) is the maximum kinetic energy of the emitted photoelectron. It expresses conservation of energy: the photon's energy is used partly to free the electron and the rest appears as its kinetic energy. Thus KE(max) = h f - W0.

h f = W0 + KE(max)

The energy of the incident photon E = hc/lambda = 1240/400 = 3.1 eV. Using Einstein's equation, the maximum kinetic energy KE(max) = E - W0 = 3.1 - 2.0 = 1.1 eV. Since the photon energy (3.1 eV) is greater than the work function (2 eV), emission occurs and the photoelectrons have a maximum kinetic energy of 1.1 eV.

KE(max) = hc/lambda - W0

Bohr's postulates are: (1) The electron revolves around the nucleus only in certain fixed circular orbits called stationary states, in which it does not radiate energy. (2) Only those orbits are allowed for which the angular momentum of the electron is an integral multiple of h/(2 pi), that is, m v r = n h/(2 pi) (quantisation of angular momentum). (3) Energy is emitted or absorbed only when the electron jumps from one orbit to another, the energy of the emitted or absorbed photon being h f = E2 - E1.

m v r = n h/(2 pi)

Energy of the electron in n = 3: E3 = -13.6/9 = -1.51 eV. Energy in n = 2: E2 = -13.6/4 = -3.4 eV. The energy of the emitted photon = E3 - E2 = -1.51 - (-3.4) = 1.89 eV. This corresponds to the H-alpha line of the Balmer series (visible red light).

E(photon) = E(higher) - E(lower)

The binding energy is the energy equivalent of the mass defect. BE = (mass defect in u) x 931 MeV = 0.0304 x 931 = 28.3 MeV. Thus the binding energy of the helium nucleus is about 28.3 MeV, and the binding energy per nucleon = 28.3/4 = 7.1 MeV.

BE (MeV) = (delta m in u) x 931