Answer to Question #162540 in Optics for Samanta samira

Question

Helium-neon laser light of wavelength 632.8 nm is sent through a 0.300mm wide single slit. What is the width of central maximum on a screen 1.00 m from the slit?

Answer 1

Let's write the condition for the minimum (or destructive interference):

$$wsin\theta_1=m\lambda,$$

here, $w$ is the width of the slit, $\theta_1$ is the half-angular width of the central maximum, $\lambda$ is the wavelength.

For $m=1$ (the first minimum) our formula becomes:

$$wsin\theta_1=\lambda,$$ $$sin\theta_1=\dfrac{\lambda}{w}=\dfrac{632.8\cdot10^{-9}\ m}{0.3\cdot10^{-3}\ m}=2.11\cdot10^{-3}.$$

Then, we can find the position of the minimum on the screen from the geometry:

$$tan\theta_1\approx sin\theta_1\approx \theta_1\approx \dfrac{y}{L},$$ $$y=Lsin\theta_1=1\ m\cdot2.11\cdot10^{-3}=2.11\cdot10^{-3}\ m.$$

Finally, we can find the width of the central maximum:

$$w_{central}=2y=2\cdot2.11\cdot10^{-3}\ m=4.22\cdot10^{-3}\ m=4.22\ mm.$$