question_answer 17)

Answer the following questions : (a) In a single-slit diffraction experiment,the width of the slit is made double the originalwidth. How does this affect the size andintensity of the central diffraction band ? (b) In what way is diffraction from each slitrelated to the interference pattern in adouble-slit experiment ? (c) When a tiny circular obstacle is placedin the path of light from a distant source, abright spot is seen at the centre of the shadowof the obstacle. Explain why ? (d) Two students are separated by a 7 mpartition wall in a room 10 m high. If bothlight and sound waves can bend aroundobstacles, how is it that the students areunable to see each other even though theycan converse easily ? (e) Ray optics is based on the assumptionthat light travels in a straight line. Diffraction effects (observed when lightpropagates through small apertures/slitsor around small obstacles) disprove thisassumption. Yet the ray optics assumptionis so commonly used in understandinglocation and several other properties ofimages in optical instruments. What is the justification?  

Answer:

(a) The width of central maxima When the width (d) of slit is doubled, thenthe width of central diffraction maximareduces to half and the intensity of thecentral band increases four times. (b) Intensity of fringes produced in doubleslit experiment is changed somewhat dueto diffraction pattern superposing due toeach slit. (c) Light waves diffract at the edges ofthe circular obstacle. These diffractedwaves interfere constructively and giverise to the bright spot at the centre of thegeometrical shadow. (d) Diffraction is observed when the wavelength of the wave is of the order of thesize of the obstacle. The wavelength of sound wave is larger than the light wave  and is also comparable to wall, so diffraction of soundwaves takes place and hence the studentscan converse easily. On the other hand,the wave length of light is very small ascompared to the obstacle i.e. 7 m highwall so the diffraction of light waves doesnot take place. (e) For diffraction to occur, size of theobstacle should nearly be equal towavelength of the waves. The wavelengthof sound waves is nearly equal to size ofthe objects such as table, certain etc. Butthe wavelength of light is very small (4000 Å to 8000 Å) to be diffracted by the object. This makes the conversation possible butnot seeing one another.