Light whose wavelength is 633nm falls on a double slit with spacing of 0.100mm. what is the separation between the 0th and 1st order peak on a screen, l=1.20m from the double slit?

Approximately 0.00760 m (that's 7.60 mm.)

Explanation:

Refer to the first diagram attached. For a double-slit diffraction, the angle (angular separation) between the m-th maximum and the central maximum satisfies the following equation:

, where

(Young’s Double Slit Experiment, OpenStaxCollege)

If is much larger than , the value of will be considerably small. The value of could thus be approximately as:

.

For this problem,

Approximate the value of :

.

Separation between the first and central maximum:

if the person is shown to have +90n, then ans is -10n.

if the person is shown to have -90n, then ans is +10n.

initial speed when it is dropped is 150 ft/s

now at the same moment the acceleration due to gravity is perpendicular to its velocity

so as we know that

now we have centripetal acceleration a = 32 ft/s^2 as its perpendicular to velocity

while tangential acceleration at the same moment will be zero

so

now radius of curvature is given as

so the radius of curvature just after it is dropped is 703.125 ft