Optics Question 159
Question: In a double slit arrangement fringes are produced using light of wavelength $ 4800{AA} $ . One slit is covered by a thin plate of glass of refractive index 1.4 and the other with another glass plate of same thickness but of refractive index 1.7. By doing so the central bright shifts to original fifth bright fringe from centre. Thickness of glass plate is
Options:
A) 8 mm
B) 6 mm
C) 4 mm
D) 10 mm
Show Answer
Answer:
Correct Answer: A
Solution:
Shift $ \Delta x=\frac{\beta }{\lambda }(\mu -1)t $
Shift due to one plate $ \Delta x _{1}=\frac{\lambda }{\beta }({\mu _{1}}-1) $
Shift due to another path $ \Delta x _{2}=\frac{\beta }{\lambda }({\mu _{2}}-1)t $
Net shift $ \Delta x=\Delta x _{2}-\Delta x _{1}=\frac{\beta }{\lambda }({\mu _{2}}-{\mu _{1}})t $ …(i)
Also it is given that $ \Delta x=5\beta $ …(ii)
Hence $ 5\beta =\frac{\beta }{\lambda }({\mu _{2}}-{\mu _{1}})t $
$ \Rightarrow t=\frac{5\lambda }{({\mu _{2}}-{\mu _{1}})}=\frac{5\times 4800\times {{10}^{-10}}}{(1.7-1.4)} $
$ =8\times {{10}^{-6}}m=8\mu m. $
BETA
