In Young's double slit experiment with light of wavelength `lambda=600 nm`, intensity of central fringe is `I_(0)`. Now one of the slit is covered by glass plate of refractive index 1.4 and thickness t=`5mum`. The new intensity at the same point on screen will be :

Options

A. `(I_(0))/(4)`
B. `(3I_(0))/(4)`
C. `I_(0)`
D. `(I_(0))/(2)`

Answer

To solve the problem, we will follow these steps:

### Step 1: Understand the Initial Conditions
In Young's double slit experiment, the intensity of the central fringe is given as \( I_0 \) when both slits are open without any obstruction.

### Step 2: Determine the Optical Path Difference
When one of the slits is covered by a glass plate of refractive index \( \mu = 1.4 \) and thickness \( t = 5 \, \mu m = 5 \times 10^{-6} \, m \), we need to calculate the optical path difference introduced by the glass plate.

The optical path difference (OPD) due to the glass plate is given by:
\[ \text{OPD} = (\mu - 1) \cdot t \]
Substituting the values:
\[ \text{OPD} = (1.4 - 1) \cdot (5 \times 10^{-6}) = 0.4 \cdot (5 \times 10^{-6}) = 2 \times 10^{-6} \, m \]

### Step 3: Calculate the Phase Difference
The phase difference \( \Delta \phi \) corresponding to the optical path difference can be calculated using the formula:
\[ \Delta \phi = \frac{2\pi}{\lambda} \cdot \text{OPD} \]
Where \( \lambda = 600 \, nm = 600 \times 10^{-9} \, m \).

Substituting the values:
\[ \Delta \phi = \frac{2\pi}{600 \times 10^{-9}} \cdot (2 \times 10^{-6}) \]
Calculating this gives:
\[ \Delta \phi = \frac{4\pi \times 10^{-6}}{600 \times 10^{-9}} = \frac{4\pi}{0.6} = \frac{20\pi}{3} \]

### Step 4: Calculate the New Intensity
The intensity at the central fringe when both slits are open is given by:
\[ I = I_0 \cos^2\left(\frac{\Delta \phi}{2}\right) \]
Substituting the phase difference:
\[ I = I_0 \cos^2\left(\frac{20\pi}{6}\right) = I_0 \cos^2\left(\frac{10\pi}{3}\right) \]
Since \( \frac{10\pi}{3} \) is equivalent to \( \frac{4\pi}{3} \) (as \( 10\pi/3 - 6\pi/3 = 4\pi/3 \)), we have:
\[ I = I_0 \cos^2\left(\frac{4\pi}{3}\right) \]
Calculating \( \cos\left(\frac{4\pi}{3}\right) \):
\[ \cos\left(\frac{4\pi}{3}\right) = -\frac{1}{2} \quad \text{(since it is in the third quadrant)} \]
Thus,
\[ I = I_0 \left(-\frac{1}{2}\right)^2 = I_0 \cdot \frac{1}{4} \]

### Final Answer
The new intensity at the same point on the screen will be:
\[ I = \frac{I_0}{4} \]

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