Question

Monochromatic light of wavelength 385 nm is incident on a narrow slit. On a screen 3.00 m away, the distance between the second diffraction minimum and the central maximum is 1.40 cm. (a) Calculate the angle of diffraction θ of the second minimum.(b) Find the width of the slit.

Answer 1

To solve this problem it is necessary to apply the concepts related to the concept of overlap and constructive interference.

For this purpose we have that the constructive interference in waves can be expressed under the function

$a sin\theta = m\lambda$

Where

a = Width of the slit

d = Distance of slit to screen

m = Number of order which represent the number of repetition of the spectrum

$\theta =$ Angle between incident rays and scatter planes

At the same time the distance on the screen from the central point, would be

$sin\theta = \frac{y}{d}$

Where y = Represents the distance on the screen from the central point

PART A ) From the previous equation if we arrange to find the angle we have that

$\theta = sin^{-1}(\frac{y}{d})$

$\theta = sin^{-1}(\frac{1.4*10^{-2}}{3})$

$\theta = 0.2673\°$

PART B) Equation both equations we have

$a sin\theta = m\lambda$

$a \frac{y}{d} = m\lambda$

Re-arrange to find a,

$a = \frac{(2)(385*10^{-9})(3)}{(1.4*10^{-2})}$

$a = 1.65*10^{-4}m$