Question

Light of wavelength 633 nm falls on a single slit 0.5 mm wide and forms a diffraction pattern on a screen 1.0 m away from the slit. Part A Find the position of the first dark spot on the positive side of the central maximum.

Answer 1

Answer:

The position of the first dark spot on the positive side of the central maximum is 1.26 mm.

Explanation:

Given that,

Wavelength of light is 633 nm.

Slit width, d = 0.5 mm

The diffraction pattern forms on a screen 1 m away from the slit. We need to find the position of the first dark spot on the positive side of the central maximum.

For destructive interference :

$\dfrac{dY}{D}=n\lambda$

Y is the distance of the minima from central maximum

Here, n = 1

$Y=\dfrac{n\lambda D}{d}\\\\Y=\dfrac{1\times 633\times 10^{-9}\times 1}{0.5\times 10^{-3}}\\\\Y=0.00126\ m\\\\Y=1.26\times 10^{-3}\ m\\\\Y=1.26\ mm$

So, the position of the first dark spot on the positive side of the central maximum is 1.26 mm.