A thin glass plate of thickness is `(2500)/3lamda` (`lamda` is wavelength of light used) and refractive index `mu=1.5` is inserted between one of the slits and the screen in Young's double slit experiment. At a point on the screen equidistant from the slits, the ratio of the intensities before and after the introduction of the glass plate is :

Options

A. `2:1`
B. `1:4`
C. `4:1`
D. `4:3`

Answer

To solve the problem, we will follow these steps:

### Step 1: Understand the Initial Conditions
In Young's double slit experiment, when no glass plate is inserted, the intensity at a point on the screen equidistant from the slits is given by:
\[ I = 4I_0 \cos^2\left(\frac{\phi}{2}\right) \]
where \( I_0 \) is the intensity due to each slit, and \( \phi \) is the phase difference between the two waves arriving at that point.

### Step 2: Calculate the Initial Intensity
Since the point is equidistant from both slits, the initial phase difference \( \phi \) is 0. Thus:
\[ I = 4I_0 \cos^2\left(0\right) = 4I_0 \cdot 1 = 4I_0 \]

### Step 3: Introduce the Glass Plate
When a thin glass plate of thickness \( t = \frac{2500}{3} \lambda \) and refractive index \( \mu = 1.5 \) is inserted in front of one of the slits, it introduces a phase shift.

### Step 4: Calculate the Path Difference
The path difference \( \Delta x \) introduced by the glass plate is given by:
\[ \Delta x = (\mu - 1) \cdot t \]
Substituting the values:
\[ \Delta x = (1.5 - 1) \cdot \frac{2500}{3} \lambda = 0.5 \cdot \frac{2500}{3} \lambda = \frac{1250}{3} \lambda \]

### Step 5: Calculate the Phase Difference
The phase difference \( \Delta \phi \) corresponding to the path difference \( \Delta x \) is given by:
\[ \Delta \phi = \frac{2\pi}{\lambda} \Delta x = \frac{2\pi}{\lambda} \cdot \frac{1250}{3} \lambda = \frac{2500\pi}{3} \]

### Step 6: Calculate the New Intensity
Now, the intensity after the introduction of the glass plate \( I' \) is given by:
\[ I' = 4I_0 \cos^2\left(\frac{\Delta \phi}{2}\right) \]
Substituting the value of \( \Delta \phi \):
\[ I' = 4I_0 \cos^2\left(\frac{2500\pi}{6}\right) \]
Since \( \cos\left(\frac{2500\pi}{6}\right) \) can be simplified (as \( \frac{2500}{6} = 416.67 \) which is not a standard angle, we can find its equivalent angle in the unit circle):
\[ \cos\left(\frac{2500\pi}{6}\right) = \cos\left(\frac{2500 \mod 12 \cdot \pi}{6}\right) = \cos\left(\frac{4\pi}{6}\right) = \cos\left(\frac{2\pi}{3}\right) = -\frac{1}{2} \]
Thus:
\[ I' = 4I_0 \left(-\frac{1}{2}\right)^2 = 4I_0 \cdot \frac{1}{4} = I_0 \]

### Step 7: Calculate the Ratio of Intensities
Finally, we find the ratio of the intensities before and after the introduction of the glass plate:
\[ \frac{I}{I'} = \frac{4I_0}{I_0} = 4 \]

### Final Answer
The ratio of the intensities before and after the introduction of the glass plate is:
\[ \boxed{4} \]