Question

In placing a thin sheet of mica of thickness
12 x 10-5 cm in the path of one of the interfering
beams in YDSE the central fringe shifts equal to
fringe width. Refractive index of mica (a = 600 nm)​

Answer 1

Given:

• Thickness  of mica sheet $(t) = 12 \times 10^{-5} cm = 12 \times 10^{-7} m$  

• Wavelength of light used $(\lambda) = 6 \times 10^{-9}m$

To Find:

• Refractive index of mica $( \mu)$  

Solution:

In a Young's experiment,

The shifts,    

    ⇒           $( \mu - 1 )t =n \lambda$

                                                               $n = 1$

    ⇒       $\mu = \frac{ 1 \times 6 \times 10^{-9} }{12 \times 10^{-7}} + 1$

   

    ⇒      $\mu = \frac{ 6 \times 10^{-9} + 12 \times 10^{-7} }{12 \times 10^{-7}}$

 

   ⇒          $\mu = 6 \times 10^{-7} (\frac{ 1 + 0.02 }{12 \times 10^{-7}})$

   ⇒          $\mu =\frac{1.02}{2}$

   ⇒         $\mu = 0.51$

Thus,  Refractive index of mica is $0.51$