Chapter 24, Problem 17P

To determine

The full angular width of central diffraction peak.

Answer to Problem 17P

  $1.51°$ .

Explanation of Solution

Given:

Width of slit, $d=0.0440mm=0.000044m$

Order, $m=1$

Wavelength, $\lambda =580nm=580\times {10}^{-9}m$

Formula used:

It is known that $\sin \theta =\frac{m\lambda }{d}$

Where,

  $m$ is an integer

  $\lambda$ is wavelength

  $d$ is a distance

Calculation:

Rearrange the given formula,

  $\sin \theta =\frac{m\lambda }{d}$

So,

  $\theta =\sin {}^{-1}\frac{m\lambda }{d}$

The path difference is an integer multiple of the wavelength in constructive interference.

Substitute the given values in formula,

  ${\theta }_{1\min }=\sin {}^{-1}\frac{m\lambda }{d}$

  ${\theta }_{1\min }={\sin }^{-1}\frac{580\times {10}^{-9}}{0.000044}=0.755°$

It is known that width $=2\theta$

Now, calculate the angular width of the central diffraction

  $\Delta {\theta }_1=2{\theta }_{1\min }$

  $=0.755°\times 2=1.51°$

Conclusion:

The required angle is $1.51°$ .