Question

) In the figure, a slit 0.30 mm wide is illuminated by light of wavelength 426 nm. A diffraction attern is seen on a screen 2.8 m from the slit. What is the linear distance on the screen between e first diffraction minima on either side of the central diffraction maximum? Answer: 8.0 mm 30) A thin beam of laser light of wavelength 514 nm passes through a diffraction grating having 3952 lines/cm. The resulting pattern is viewed on a distant curved screen that can show all bright fringes up to and including +90.0° from the central spot. If the experiment were performed with all of the apparatus under water (which has an index of refraction of 1.33), what would be the TOTAL number of bright spots that would show up on the screen? Answer: 13 44) Treat each of your eyes as a circular aperture of diameter 3.5 mm. Light of wavelength 500 nm is used to view two point sources that are 894 m distant from you. How far apart must these two point sources be if they are to be just resolved by your eye? Assume that the resolution is diffraction limited and use Rayleigh's criterion. Answer: 16 cm 30) A thin beam of laser light of wavelength 514 nm passes through a diffraction grating havin 3952 lines/cm. The resulting pattern is viewed on a distant curved screen that can fringes up to and including +90.0° from the central spot. If the experiment were perform all of the apparatus under water (which has an index of refraction of 1.33), what would be the TOTAL number of bright spot Answer: 13 s that would show up on the screen 44) Treat each of your eyes as a circular aperture of diameter 3.5 mm. Light of wavelength 500 nm is used to view two point sources that are 894 m distant from you. How far apart must these two point sources be if they are to be just resolved by your eye? Assume that the resolution is diffraction limited and use Rayleigh's criterion. Answer: 16 cm

Answer 1

answer 1) we know the formula slit diffraction

sin$\Theta$=$\lambda$/D

$\Theta$=sin^-1($\lambda$/D)=sin^-1(426/0.3)=0.08136^o

we have ,

tan$\Theta$=y/L

y=L*tan$\Theta$=2.8*tan0.08136=0.004 m

the distance between first minima=2y=2*0.004=0.008 m=8.00 mm

so the answer is 8.0 mm