Question

A lens that is "optically perfect" is still limited by diffraction effects. Suppose a lens has a diameter of 150 mm and a focal length of 620 mm. A. Find the angular width (that is, the angle from the bottom to the top) of the central maximum in the diffraction pattern formed by this lens when illuminated with 520 nm light.B. What is the linear width (diameter) of the central maximum at the focal distance of the lens?

Answer 1

Answer:

a

  w = 8.46 *10^{-6} \  rad

b

  D  =5.24 *10^{-6}

Explanation:

From the question we are told that

   The  diameter of the lens is  $d =  150 \ mm  = 150*10^{-3} \  m$

    The  focal length is  $f =  620 mm =  620 *10^{-3} \  m$

    The wavelength is  $\lambda  =  520 \ nm  =  520 *10^{-9} \  m$

   

Generally the angular width is mathematically represented as

       $w =  2 \theta$

Here  $\theta$ is the angular radius of the central maxima which is mathematically represented as

        $\theta =  \frac{1.22* \lambda }{ d}$

=>    $\theta  =  \frac{1.22 *  520 *10^{-9}}{ 150 *10^{-3}}$

=>    $4.23 *10^{-6} \  rad$

Hence

      $w =  2 *  4.23 *10^{-6}$

      =>  $w = 8.46 *10^{-6} \  rad$

Generally linear diameter is mathematically represented as

      $D  =  2 *  R$

Where  $R$ is the linear radius  which is  mathematically represented as

        $R =  \frac{1.22 *  f  *  \lambda  }{ d}$

=>      $R= \frac{1.22 *  620 *10^{-3} *  520 *10^{-9}}{150 *10^{-3}}$

=>     $R  =  2.62*10^{-6} \  m$

Thus  

    $D =  2 *  2.62 *10^{-6}$

     $D  =5.24 *10^{-6}$