Light of wavelength 597 nm falls on a double slit, and the first bright fringe of the interference pattern is seen at an angle of 15.8° with the horizontal. find the separation between the slits. µm

2.2 µm

Explanation:

For constructive interference, the expression is:

Where, m = 1, 2, .....

d is the distance between the slits.

Given wavelength = 597 nm

Angle,  = 15.8°

First bright fringe means , m = 1

So,

Also,

1 nm = 10⁻⁹ m

1 µm = 10⁻⁶ m

So,

1 nm = 10⁻³ nm

Thus,

Distance between slits ≅ 2.2 µm

The separation between the slit is

Solution:

As per the question:

Wavelength of light,

Now, by Young's double slit experiment:

here,

n = 1

x = slit width

Therefore,

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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