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(a) Describe briefly how a diffraction pattern is obtained on a screen due to a single narrow slit illuminated by a monochromatic source of light. Hence, obtain the conditions for the angular width of secondary maxima and secondary minima. (b) Two wavelengths of sodium light of 590 nm and 596 nm are used in turn to study the diffraction taking place at a single-slit of aperture 2xx10^(-6)m. the distance between the slit and the screen is 1.5m. calculate the separation between the positions of first maxima of the diffraction pattern obtained in the two cases. |
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Answer» Solution :(b) Here `lamda_(1)=590nm=590xx10^(-9)m,lamda_(2)=596nm=596xx10^(-9)m`, slit width `a=2xx10^(-6)`m and DISTANCE between the slit and screen D=1.5m. First maxima is obtained at a distannce x from centre point on screen such that `asintheta=(AX)/(D)=(3lamda)/(2)impliesx=(3lamdaD)/(2a)` `therefore`Separation between the positions of first maximas `DELTAX=x_(2)-x_(1)=(3D)/(2a)(lamda_(2)-lamda_(1))` `implies Deltax=(3xx1.5)/(2xx2xx10^(-6))xx[596xx10^(-9)-590xx10^(-9)]=6.75xx10^(-3)m` or 6.75mm. |
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