Discussion Forum

 The image of an object, formed by a plano-convex lens at a distance of 8 m behind the lens, is real and is one -third the size of the object. The wavelength of light inside the lens is 2/3 times the wavelength in free space. The radius of the curved surface of the lens is 

Answer:

Explanation:

$\mu=\frac{\lambda_{air}}{\lambda_{medium}}=\frac{1}{(2/3)}=\frac{3}{2}$

 Further , $|m|=\frac{1}{3}=|\frac{v}{u}|$

    $\therefore$         $|v|=\frac{|u|}{3}$

                 u=-24 m     (real object)

      $\therefore$    v=+8 m (real image)

  Now,   $\frac{1}{v}-\frac{1}{u}=\frac{1}{f}=(\mu-1)(\frac{1}{+R}-\frac{1}{\infty})$

   $\therefore$      $\frac{1}{8}+\frac{1}{24}=(\frac{3}{2}-1)(\frac{1}{R})$

   $\therefore$     R= 3m