Question

Light of wavelength 578.0 nm is incident on a narrow slit. The diffraction pattern is viewed on a screen 62.5 cm from the slit. The distance on the screen between the third order minimum and the central maximum is 1.35 cm . What is the width a of the slit in micrometers (μm)?

Answer 1

Answer:

$80.6\mu m$

Explanation:

When light passes through a narrow slit, it produces a diffraction pattern on a distant screen, consisting of several bright fringes (constructive interference) alternated with dark fringes (destructive interference).

The formula to calculate the position of the m-th maximum in the diffraction pattern produced in the screen is:

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

where

y is the distance of the m-th maximum from the central maximum (m = 0)

$\lambda$ is the wavelength of light used

D is the distance of the screen from the slit

d is the width of the slit

In this problem, we have:

$\lambda=578.0 nm = 578\cdot 10^{-9} m$ is the wavelength

D = 62.5 cm = 0.625 m is the distance of the screen

We know that the distance between the third order maximum (m=3) and the central maximum is 1.35 cm (0.0135 m), which means that

$y_3 = 0.0135 m$

For

m = 3

Therefore, rearranging the equation for d, we find the width of the slit:

$d=\frac{m\lambda D}{y_3}=\frac{(3)(578\cdot 10^{-9})(0.625)}{0.0135}=80.3\cdot 10^{-6} m=80.6\mu m$