Optics Question 857
Question: A plastic sheet (refractive index =1.6) covers one slit of a double slit arrangement meant for the Young’s experiment. When the double slit is illuminated by monochromatic light (wavelength in air $ =6600\overset{o}{\mathop{A}} $ ), the centre of the screen appears dark rather than bright. The minimum thickness of the plastic sheet to be used for this to happen is:
Options:
A) $ 3300\overset{o}{\mathop{A}} $
B) $ 6600\overset{o}{\mathop{A}} $
C) $ 2062\overset{o}{\mathop{A}} $
D) $ 5500\overset{o}{\mathop{A}} $
Show Answer
Answer:
Correct Answer: D
Solution:
[d] The path difference produced by a sheet $ \Delta =(\mu -1)t $ .
According to the given condition (for minimum thickness) $ (\mu -1)t=\frac{\lambda }{2} $
$ \therefore t=\frac{\lambda }{2(\mu -1)}=\frac{6600\times {{10}^{-10}}}{2(1.6-1)}=5500\overset{o}{\mathop{A}} $
