The luminous flux of a 1 W sodium vapour lamp is more than that of a 10 kW source of ultraviolet radiation. Comment.

About 5% of the power of a 100 W light bulb is converted to visible radiation. What is the average intensity of visible radiation (a) at a distance of 1m from the bulb? (b) at a distance of 10 m? Assume that the radiation is emitted isotropically and neglect reflection.

A country has a food deficit of 10% . Its population grows continuously at a rate of 3% per year. Its annual food production every year is 4% more than that of the last year. Assuming that the average food requirement per person remains constant, prove that the country will become self-sufficient in food after n years, where n is the smallest integer bigger than or equal to (ln10-ln9)/(ln(1.04)-0.03)

In a building there are 15 bulbs of 45 W, 15 bulbs of 100 W, 15 bulbs of 10 W and 2 heaters of 1 kW. The voltage of electric main is 220 V. The minimum fuse capacity (rated value ) of the building will be approximately .

A resistor of 100 Omega and a capacitor of 250 muF are connected in series to a 220 V , 50 Hz ac source. Calculate the voltage (rms) across the resistor and the capacitor . Is the algebraic sum of these voltages more than the source voltage ? If yes, resolve the paradox.

A resistor of 200 Omega and a capacitor of 15.0 muF are connected in series to a 220 V, 50 Hz ac source. (a) Calculate the current in the circuit, (b) Calculate the voltage (rms) across the resistor and the capacitor. Is the algebraic sum of these voltages more than the source voltage? If yes, resolve the paradox.

A resistor of 200 Omega and a capacitor of 15.0 muF are connected in series to a 220V, 50Hz as source . (a) Calculate the current in the circuit. (b) Calculate the voltage (rms) across the resistor and the capacitor. Is the alebraic sum of these voltages more than the source voltage? If yes resolve the paradox.

An electromagnetic wave can be represented by E = A sin (kx- omega t + phi) , where E is electric field associated with wave, According this equation, for any value of x, E remains sinusoidal for -oolt t lt oo . Obviously this corresponds to an idealised situation because radiation from ordinary sources consists of finite size wavetrains. In general, electric field remains sinusoidal only for times of order tau_(c) ' which is called coherence time. In simpler language it means that for times of order tau_(c)' a wave will have a definite phase. The finite value of coherence time could be due to many factors, for example if radiating atom undergoes collision with another atom then wave train undergoes an abrupt phase change or due to the fact that an atom responsible for emitting radiation has a finite life time in the energy level from which it drops to lower energy level, while radiating. Concept of coherence time can be easily understood using young's double slit experiment. Let interference patten is observed around point P at time t , due to superposition of waves emanting from S_(1) and S_(2) at times t =(r_(1))/(c) and (r_(2))/(c) respectively, where r_(1) and r_(2) are the distances S_(1) P & S_(2)P . Obviously if (r_(2)-r_(1))/(c) lt lt tau_(e),{"where" " "c = 3xx10^(8)m//s} then, wavetrain arriving at point P from S_(1) & S_(2) will have a definite phase relationship and an interference pattern of good contranst will be obtained. If coherence time is of order 10^(-10) second and screen is placed at a very large distance from slits in the given figure, then:-

An electromagnetic wave can be represented by E = A sin (kx- omega t + phi) , where E is electric field associated with wave, According this equation, for any value of x, E remains sinusoidal for -oolt t lt oo . Obviously this corresponds to an idealised situation because radiation from ordinary sources consists of finite size wavetrains. In general, electric field remains sinusoidal only for times of order tau_(c) ' which is called coherence time. In simpler language it means that for times of order tau_(c)' a wave will have a definite phase. The finite value of coherence time could be due to many factors, for example if radiating atom undergoes collision with another atom then wave train undergoes an abrupt phase change or due to the fact that an atom responsible for emitting radiation has a finite life time in the energy level from which it drops to lower energy level, while radiating. Concept of coherence time can be easily understood using young's double slit experiment. Let interference patten is observed around point P at time t , due to superposition of waves emanting from S_(1) and S_(2) at times t =(r_(1))/(c) and (r_(2))/(c) respectively, where r_(1) and r_(2) are the distances S_(1) P & S_(2)P . Obviously if (r_(2)-r_(1))/(c) lt lt tau_(e),{"where" " "c = 3xx10^(8)m//s} then, wavetrain arriving at point P from S_(1) & S_(2) will have a definite phase relationship and an interference pattern of good contranst will be obtained. If coherence time is of order 10^(-10) second and screen is placed at a very large distance from slits in the given figure, then:-

From source of power 4 kW,radiation emitted are such that photon are emitted at rate of 10^(20) photon/sec.This radiation will be in …..region of electromagnetic waves… []h=6.6xx10^(-34)Js

A meacury arec lamp provides 0.1 watt of ultra-violet radiation at a wavelength of lambda=2537 Å only. The photo tube (cathode of photo electric device) consists of potessium and has an effective area of 4 cm^(2) . The cathode is located at a distance of 1 m from the radiation source. The work funcation for potassium is phi_(0) = 2.22 eV . According to classical theory, the radiation from arc lamp spreads out uniformaly in space as spherical wave. What time of expource to the radiation should be required for a potassium atom (radius 2Å ) ub tge cathode to accumulate sufficient enerfy to eject a photo-electron?