Physics: Two narrow slits spaced 100 microns apart are exposed to light of 600 nm. At what angle does the first minimum (dark space) occur in the interference pattern

For constant force in a time period, the impulse will be :I = Ft = 9.5 * 0.5 = 4.75 N`s...

Two narrow slits spaced 100 microns apart are

maleikrocks3497

31.01.2022

The angle is $\theta = 0.1719^o$

Explanation:

From the question we are told that

The distance of separation is $d = 100 * 10^{-6} \ m$

The wavelength of light is $\lambda = 600 nm = 600 *10^{-9} \ m$

Generally the condition for destructive interference is mathematically represented as

$dsin(\theta ) =[m + \frac{1}{2} ]\lambda$

Here m is the order of maxima, first minimum (dark space) m = 0

$100 *10^{-6 } * sin(\theta ) =[0 + \frac{1}{2} ]600 *10^{-9}$

=> $\theta = sin^{-1} [0.003]$

=> $\theta = 0.1719^o$

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