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3 years ago

14

Light from a 600 nm source goes through two slits 0.080 mm apart. What is the angular separation of the two first order maxima o

ccurring on a screen 2.0 m from the slits

1 answer:

Vera_Pavlovna [14]3 years ago

6 0

Answer:

The angular separation is $k = 0.8594^o$

Explanation:

From the question we are told that

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

The distance of separation between the slit is $d = 0.080 \ mm = 0.080 *10^{-3} \ m$

The distance from the screen is

Generally the condition for constructive interference is mathematically represented as

$d \ sin(\theta) = n \lambda$

=> $\theta = sin ^{-1} [ \frac{n * \lambda }{ d } ]$

here $\theta$ is the angular separation between the central maxima and one side of the first order maxima

given that we are considering the first order of maxima n = 1

=> $\theta = sin ^{-1} [ \frac{1 * 600*10^{-9} }{ 2.0 } ]$

=> $\theta = sin ^{-1} [ 0.0075 ]$

=> $\theta = 0.4297^o$

So the angular separation of the two first order maxima is

$k = 2 * \theta$

$k = 2 * 0.4297$

$k = 0.8594^o$

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