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Question 1 ) A) In a Young's double slit experiment the separation of the slits is 0.23 mm. The wavelength of the light is 509 nm. Determine the angle at which the third dark fringe out from the centre occurs. Give your answer as a positive value. B) In a Young's double slit experiment the separation of the slits is 0.28 mm. The distance to the screen is 1.16 m. The third bright fringe is a distance of 5.2 mm from the central fringe. Determine the wavelenght of the light.

          Question 1 )
A) In a Young's double slit experiment the separation of the
slits is 0.23 mm. The wavelength of the light is 509 nm. Determine
the angle at which the third dark fringe out from the centre
occurs. Give your answer as a positive value.
B) In a Young's double slit experiment the separation of
the slits is 0.28 mm. The distance to the screen is
1.16 m. The third bright fringe is a distance of 5.2 mm
from the central fringe. Determine the wavelenght of the light.
        
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Added by Nancy J.

University Physics with Modern Physics
University Physics with Modern Physics
Hugh D. Young 14th Edition
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Question 1 ) A) In a Young's double slit experiment the separation of the slits is 0.23 mm. The wavelength of the light is 509 nm. Determine the angle at which the third dark fringe out from the centre occurs. Give your answer as a positive value. B) In a Young's double slit experiment the separation of the slits is 0.28 mm. The distance to the screen is 1.16 m. The third bright fringe is a distance of 5.2 mm from the central fringe. Determine the wavelenght of the light.
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Transcript

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00:01 However, question says we have to determine the angle at which the third dark fringe out from the center occurs.
00:08 In the young double slit experiment, the condition for the dark fringe is given as the formula.
00:15 D sine theta is equal to 2n plus 1, lambda divided by 2.
00:26 For hard dark french n is equal to 3 minus 1 that is 2 so n is equal to 2 lambda is equal to 509 nanometer that is 509 multiplied by 10 to the power minus 9 meter d is equal to 0 .23 millimetre which is equal to 0 .23 multiplied by 10 to the bar minus 3 meter.
01:07 Now d sine theta is equal to 5 divided by 2 multiplied by lambda.
01:19 From here sine theta is equal to 5 lambda divided by 2 multiplied by lambda...
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