Question

The distance between the first and fifth minima of a single-slit diffraction pattern is 0.400 mm with the screen 37.0 cm away from the slit, when light of wavelength 570 nm is used. (a) Find the slit width. (b) Calculate the angle θ of the first diffraction minimum.

Answer 1

Answer:

a)$d=2.1\times10^-3$

b)$\theta=0.015^\circ$

Explanation:

Given:

• Distance between the first and fifth minima $= 0.4\ nm$
• Distance of screen from the slit $D = 0.37\ m$
• Wavelength of the light $\lambda = 570\ \rm nm$

Let d be the width of the slit

a)

$\dfrac{5\lambda D}{d}-\dfrac{\lambda D}{d}=0.4\times10^{-3}\\\dfrac{4\times570\times10^{-9}\times0.37}{d}=0.4\times10^{-3}\\d=2.1\times 10^{-3}\ \rm m$

b)we know that

$\tan\theta=\dfrac{y}{D}\\\tan\theta=\dfrac{\lambda}{d}\\\tan\theta=\dfrac{570\times10^{-9}}{2.1\times10^{-3}}\\\theta=0.015^\circ$