Light of wavelength 610 nm is incident on a slit of width 0.20 mm. and a diffrac

Question

Light of wavelength 610 nm is incident on a slit of width 0.20 mm. and a diffraction pattern is produced on a screen that is 1.5 m from the slit. What is the distance of the second dark fringe from the center of the bright fringe (1 nm = 10'* m) 0.68 cm 0.92 cm 1.2 cm 1.4 cm A diffraction grating with 10 000 lines'em will exhibit the first order maximum for light of wave length 510 nm at what angle (1 nm = 10 * m) 0.51 degree 0.62 degree 15.3degree 31degree In a two-slit experiment, the slit separation is 3.00 * 10^-5 m. The interference pancr is created on a screen that is 2.00 m away from the slits. If the 7th bright fringe on the screen is 10.0 cm away from the central fringe, what is the wavelength of the light 100 nm 204 nm 214 nm 224 nm 234 nm Using monochromatic light of 410 nm wavelength, a single thin slit forms a diffraction pattern, with the first minimum at an angle of 40.0 degree from central maximum. The same slit, when illuminated by a new monochromatic light source, produces a diffraction pattern with the second minimum at a 60.0degree angle from the central maximum What is the wavelength of this new light 276 nm 293 nm 309 nm 326 nm 342 nm A single slit. 1400 nm wide, forms a diffraction pattern when illuminated by monochromatic light of 490-nm wavelength. What is the largest angle from the central maximum at w hich the intensity of the light is zero 44 degree 41 degree 38 degree 35 degree

Explanation / Answer

35. for dark fringe
y = (2n - 1)lambda*D/d*2 = 3*610*10^-9*1.5/2*0.2*10^-3 = 0.0068625 m
36. theta = n*lambda/d = 1*510*10^-9*10000/10^-2 = 0.51 rad = 0.51*180/3.14 = 29.235 deg
37. y = n(lambda)D/d = 7*lambda*2/3*10^-5 = 0.1
lambda = 2.14*10^-7 m
38. theta = n*lambda/d [ for bright fringe]
40*3.13/180 = 1*410*10^-9/d
d = 5.894*10^-7 m

now, theta = (2n-1)lambda/2d [ for dark fringe]
60*3.14/180 = 3*lambda/2*5.894*10^-7
lambda = 4.1127*10^-7 m

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