Question

When a thin transparent plate of refractive index 1.5 is introduced in one of the interfacing beams, a shift of 20 fringes is observed. If it is replaced by another thin plate of half the thickness and of the refractive index 1.7 the number of fringes that undergo shift will be
(a) 23
(b) 14
(c) 28
(d) 7

Answer 1

Hint: Firstly we will find the formula that relates the number of fringes that undergo a shift, the refractive index and the thickness of the slab. Then, we will compute the number of fringes that undergo shift by substituting the given values in the formula obtained.

Formula used:
\[N=\dfrac{\left( \mu -1 \right)t}{\lambda }\]

Complete step by step solution:
From the given information, we have the data as follows.
The refractive index of a first glass slab, \[{{\mu }_{1}}=1.5\]
The refractive index of a second glass slab, \[{{\mu }_{2}}=1.7\]
The number of fringes, \[{{N}_{1}}=20\]
The relation between the number of fringes, their fringe width and the shift in the fringes is given by the formula as follows.
\[\begin{align}
  & N=\dfrac{\text{Shif}t}{\text{Fringe width}} \\
 & \Rightarrow N=\dfrac{\left( \mu -1 \right)t}{\lambda } \\
\end{align}\]
Where \[N\]is the number of fringes which undergo shift, \[\mu \]is the refractive index, \[t\] is the thickness and \[\lambda \]is the wavelength.
Considering the wavelength of the source light remains the same, we get the relation between the number of fringes that undergo shift and the refractive index. So, we have,
\[N\propto \left( \mu -1 \right)t\]
Now, we will consider two situations with the given values of the refractive indices and the thickness transparent plate.
\[\dfrac{{{N}_{1}}}{{{N}_{2}}}=\dfrac{\left( {{\mu }_{1}}-1 \right){{t}_{1}}}{\left( {{\mu }_{2}}-1 \right){{t}_{2}}}\]
Let the thickness of the transparent glass slab having the refractive index of 1.5 be ‘t’. Then, the thickness of the transparent glass slab having the refractive index of 1.7 becomes ‘\[\dfrac{t}{2}\]’ according to the given conditions.
Substitute these values of the refractive indices, the thickness of the glass slabs and the number of fringes in the above formula.
\[\dfrac{20}{{{N}_{2}}}=\dfrac{\left( 1.5-1 \right)t}{\left( 1.7-1 \right)\left( {}^{t}/{}_{2} \right)}\]
Continue further computation.
\[\begin{align}
  & {{N}_{2}}=20\times \dfrac{0.7}{0.5}\times \dfrac{1}{2} \\
 & \therefore {{N}_{2}}=14 \\
\end{align}\]
\[\therefore \]The number of fringes that undergo shift when the glass slab of refractive index 1.5 is replaced by the glass slab of refractive index 1.7, is 14, thus, the option (b) is correct.

Note: The relation between the number of fringes that undergo a shift, the refractive index and the thickness of the slab should be known to solve this problem, as, in this case, we have considered the wavelength to be constant to obtain the formula.