Question

Chapter 36, Problem 007 Light of wavelength 586 nm is incident on a narrow slit. The angle between the first diffraction minimum on one side of the central maximum and the first minimum on the other side is 1.15°. What is the width of the slit?

Answer 1

Answer:

$d =4.77\times 10^{-5}\ m$

Explanation:

given

wavelength of light  λ = 479 nm

                                   = 479 x 10⁻⁹ m

the angle

θ = 1.15 / 2 = 0.575°                

using                                              

condition for diffraction minimum ,

         d sinθ = m λ                        

for first minimum m = 1

       d sinθ = λ                      

therefore ,

slit width                                  

$d =\dfrac{\lambda}{sin\theta}$          

$d =\dfrac{479\times 10^{-9}}{sin 0.575^0}$

$d =\dfrac{479\times 10^{-9}}{0.01}$              

$d =4.77\times 10^{-5}\ m$                                                    

hence, the width of the slit is equal to $d =4.77\times 10^{-5}\ m$