question_answer

Light of wavelength 500 nm is used to form interference pattern in Youngs double slit experiment. A uniform glass plate of refractive index 1.5 and thickness 0.1 mm is introduced in the path of one of the interfering beams. The number of fringes which will shift the cross wire due to this is

A)  400                                       

B)  300

C)  200                                       

D)  100

Correct Answer: D

Solution :

                 On placing a uniform glass plate of refractive index \[(\mu )\], the additional path difference introduced is \[(\mu -1)t\], where t is thickness of glass plate.                   \[\therefore \] Number of fringes shifted                                 \[N=\frac{(\mu -1)}{\lambda }\,t\] Given,   \[\mu =1.5\],                                 \[t=0.1\,mm=0.1\times {{10}^{-3}}m\],                 \[\lambda =500\,\,nm=500\times {{10}^{-9}}m\] \[\therefore \]  \[N=\frac{(1.5-1)}{500\times {{10}^{-9}}}\times 0.1\times {{10}^{-3}}\] \[\Rightarrow \]               N = 100 fringes