Question

The central fringe of the interference pattern produced by light of wavelength 6000 Å is found to shift to the position of 4th bright fringe after a glass plate of refractive index 1.5 is introduced in path of one of beams. The thickness of the glass plate would be -

• A. 4.8$\mu m$
• B. 8.23$\mu m$
• C. 14.98$\mu m$
• D. 3.78$\mu m$

Answer 1

The correct option is A 4.8$\mu m$

Shift in fringe pattern=$\frac{4\lambda D}{d}$
Also shift due to glass plate = $\frac{(\mu -1)tD}{d}$
$\frac{4\lambda D}{d}$= $\frac{(\mu -1)tD}{d}$
$t=\frac{4\lambda }{\mu -1}=\frac{4\times 6000\times {10}^{-10}}{1.5-1}m$
$=8\times 6\times {10}^{-7}m=4.8\mu m$