Question

In YDSE a glass slab of thickness \(8\,\mu m\) is introduced in front of a slit. If central maxima shifts to the position of 4th minima, then find refractive index of glass slab \((\lambda = 500\,\text{nm})\):

• A. \( \mu = 1.11 \)
• B. \( \mu = 1.22 \)
• C. \( \mu = 1.32 \)
• D. \( \mu = 2.2 \)

Answer 1

The Correct Option is B

Step 1: Understanding the Concept:
Introducing a transparent slab of thickness \(t\) and refractive index \(\mu\) in one path of a YDSE setup introduces an extra optical path length. This causes the entire fringe pattern to shift. The shift corresponds to the path difference generated by the slab.
Step 2: Key Formula or Approach:
Path difference introduced by the slab: \(\Delta x = (\mu - 1)t\) Path difference condition for the \(n^{\text{th}}\) minima: \(\Delta x = \left(n - \frac{1}{2}\right)\lambda\) Equating the two gives the condition for the central maxima shifting to the \(n^{\text{th}}\) minima.
Step 3: Detailed Explanation:
For the \(4^{\text{th}}\) minima, \(n = 4\): \[ \Delta x = \left(4 - \frac{1}{2}\right)\lambda = \frac{7\lambda}{2} \] Equate this to the path difference caused by the slab: \[ (\mu - 1)t = \frac{7\lambda}{2} \] Rearrange to solve for \(\mu - 1\): \[ \mu - 1 = \frac{7\lambda}{2t} \] Given \(\lambda = 500 \times 10^{-9} \, \text{m}\) and \(t = 8 \times 10^{-6} \, \text{m}\): \[ \mu - 1 = \frac{7 \times 500 \times 10^{-9}}{2 \times 8 \times 10^{-6}} = \frac{3500 \times 10^{-9}}{16 \times 10^{-6}} \] \[ \mu - 1 = \frac{3.5 \times 10^{-6}}{16 \times 10^{-6}} = \frac{3.5}{16} = 0.21875 \] \[ \mu = 1 + 0.21875 \approx 1.22 \]
Step 4: Final Answer:
The refractive index of the glass slab is \(1.22\).