Question

Young's double slit experiment is conducted with monochromatic light of wavelength \(5000\,\text{\AA}\), with slit separation of \(3\,\text{mm}\) and observer at \(20\,\text{cm}\) away from the slits. If a \(1\,\text{mm}\) transparent plate is placed in front of one of the slits, the fringes shift by \(6\,\text{mm}\). The refractive index of the transparent plate is

• A. \(1.08\)
• B. \(1.09\)
• C. \(1.1\)
• D. \(1.2\)

Hint:

In YDSE, the fringe shift due to a thin transparent plate is \[ S=\frac{D}{d}(\mu-1)t \] where \(t\) is the thickness of the plate.

Answer 1

The Correct Option is B

Step 1: Recall the fringe shift formula.
When a transparent slab of thickness $t$ and refractive index $\mu$ is placed in front of one slit in YDSE, the entire fringe pattern shifts by: \[ S = \frac{D}{d}(\mu - 1)t \] where $D$ is the distance from slits to screen and $d$ is the slit separation.
Step 2: Identify the given quantities.
From the problem: $S = 6\,\text{mm}$, $D = 20\,\text{cm} = 200\,\text{mm}$, $d = 3\,\text{mm}$, $t = 1\,\text{mm}$. Note that the wavelength $\lambda = 5000\,\text{\AA}$ is not needed for this calculation since fringe shift depends on $\mu$ only.
Step 3: Substitute the known values into the formula.
\[ 6 = \frac{200}{3} \times (\mu - 1) \times 1 \] \[ 6 = \frac{200(\mu - 1)}{3} \]
Step 4: Solve for $(\mu - 1)$.
\[ \mu - 1 = \frac{6 \times 3}{200} = \frac{18}{200} = 0.09 \]
Step 5: Find the refractive index $\mu$.
\[ \mu = 1 + 0.09 = 1.09 \]
Step 6: Verify the answer makes physical sense.
A refractive index of $1.09$ is greater than $1$ (as expected for any transparent medium denser than air) and is a reasonable value for a thin transparent slab. Therefore, the refractive index of the transparent plate is: \[ \boxed{\mu = 1.09} \]