How wide is the central diffraction peak on a screen 5m behind a 0.01mm slit illuminated by 500\,nm light source ?

Solution

The width of the central diffraction peak for a single slit is determined by the angular positions of the first minima on either side of the central maximum. For a slit of width $a$ illuminated by light of wavelength $\lambda$, the angular position $\theta$ of the first minimum is given by the condition for destructive interference:

$$a\sin \theta =\lambda$$

For small angles, $\sin \theta \approx \theta$ (in radians), so the equation simplifies to:

$$\theta \approx \frac{\lambda }{a}$$

Given:

• Wavelength of light, $\lambda =500\mathrm{nm}=500\times 10^{-9}\mathrm{m}$
• Slit width, $a=0.01\mathrm{mm}=0.01\times 10^{-3}\mathrm{m}=10^{-5}\mathrm{m}$
• Distance to the screen, $D=5\mathrm{m}$

Step 1: Calculate the angular width of the central peak

The angular width of the central peak is twice the angle to the first minimum:

$$2\theta =2\left(\frac{\lambda }{a}\right)=2\left(\frac{500\times 10^{-9}}{10^{-5}}\right)=2\times 0.05=0.1\mathrm{rad}$$

Step 2: Calculate the linear width of the central peak on the screen

The linear width $W$ of the central peak on the screen is given by:

$$W=2D\tan \theta$$

For small angles, $\tan \theta \approx \sin \theta \approx \theta$, so:

$$W\approx 2D\theta =2D\left(\frac{\lambda }{a}\right)$$

Substituting the values:

$$W=2\times 5\times \left(\frac{500\times 10^{-9}}{10^{-5}}\right)=10\times 0.05=0.5\mathrm{m}$$

Thus, the width of the central diffraction peak on the screen is 0.5 meters.