Question

Monochromatic light of wavelength \( 500 \, \text{nm} \) is used in Young's double slit experiment. An interference pattern is obtained on a screen. When one of the slits is covered with a very thin glass plate (refractive index \( = 1.5 \)), the central maximum is shifted to a position previously occupied by the 4\textsuperscript{th} bright fringe. The thickness of the glass plate is ___________ \( \mu \text{m} \).

Answer 1

Correct Answer: 4

 The optical path difference introduced by the glass plate is: 
\[ \Delta x = t(\mu - 1), \] 
where $t$ is the thickness of the plate and $\mu$ is its refractive index.
 The fringe shift is given by: 
\[ \Delta x = n\lambda, \] 
where $n = 4$ (the shift corresponds to the 4th fringe) and $\lambda = 500 \, \mathrm{nm}$. 
Equating: \[ t(\mu - 1) = n\lambda. \] 
Substituting $\mu = 1.5$, $n = 4$, and $\lambda = 500 \, \mathrm{nm}$:
 \[ t(1.5 - 1) = 4 \cdot 500. \] 

Simplify: \[ t = \frac{2000}{0.5} = 4000 \, \mathrm{nm} = 4 \, \mu \mathrm{m}. \]