In Y.D.S.E., slits are at a distance of 0.5 mm and light of wavelength λ is used, if interference pattern is formed at a distance 1.2 m away and third dark fringe is at a distance of 1.2 cm from the central fringe, calculate the distance between central fringe and third dark frings if whole set-up is immersed in a liquid of refractive index 1.33.

Answer

To solve the problem step by step, we will use the formula for the position of dark fringes in Young's Double Slit Experiment (Y.D.S.E) and account for the change in wavelength due to the refractive index of the medium.

### Step 1: Understand the given data
- Distance between the slits (d) = 0.5 mm = 0.5 × 10^-3 m
- Distance from the slits to the screen (D) = 1.2 m
- Distance of the third dark fringe from the central fringe (y3) = 1.2 cm = 0.012 m
- Refractive index of the liquid (n) = 1.33

### Step 2: Use the formula for the position of dark fringes
The position of the nth dark fringe in Y.D.S.E is given by the formula:
\[ y_n = \frac{(n - 1/2) \lambda D}{d} \]
For the third dark fringe (n = 3):
\[ y_3 = \frac{(3 - 1/2) \lambda D}{d} = \frac{(5/2) \lambda D}{d} \]

### Step 3: Rearranging the formula to find λ
From the above equation, we can express λ as:
\[ \lambda = \frac{2 d y_3}{5 D} \]

### Step 4: Substitute the known values
Substituting the known values into the equation:
- \( d = 0.5 \times 10^{-3} \) m
- \( y_3 = 0.012 \) m
- \( D = 1.2 \) m

\[ \lambda = \frac{2 \times (0.5 \times 10^{-3}) \times (0.012)}{5 \times 1.2} \]

### Step 5: Calculate λ
Calculating the above expression:
\[ \lambda = \frac{2 \times 0.5 \times 10^{-3} \times 0.012}{6} \]
\[ \lambda = \frac{0.000012}{6} \]
\[ \lambda = 0.000002 \, \text{m} = 2 \times 10^{-6} \, \text{m} = 2000 \, \text{nm} \]

### Step 6: Calculate the new wavelength in the liquid
When the setup is immersed in a liquid with refractive index n, the new wavelength (λ') becomes:
\[ \lambda' = \frac{\lambda}{n} \]
Substituting the values:
\[ \lambda' = \frac{2000 \, \text{nm}}{1.33} \]

### Step 7: Calculate λ'
Calculating the new wavelength:
\[ \lambda' = 1503.77 \, \text{nm} \]

### Step 8: Calculate new position of the third dark fringe
Now we can find the new position of the third dark fringe (y3') using the same formula:
\[ y_3' = \frac{(3 - 1/2) \lambda' D}{d} = \frac{(5/2) \lambda' D}{d} \]

### Step 9: Substitute the new wavelength
Substituting the new wavelength:
\[ y_3' = \frac{(5/2) \times 1503.77 \times 10^{-9} \times 1.2}{0.5 \times 10^{-3}} \]

### Step 10: Calculate y3'
Calculating the above expression:
\[ y_3' = \frac{(5/2) \times 1503.77 \times 10^{-9} \times 1.2}{0.5 \times 10^{-3}} \]
\[ y_3' = \frac{(5 \times 1503.77 \times 1.2)}{2 \times 0.5} \times 10^{-6} \]
\[ y_3' = \frac{(5 \times 1503.77 \times 1.2)}{1} \times 10^{-6} \]

### Step 11: Final Calculation
Calculating the final distance:
\[ y_3' = 9.025 \times 10^{-6} \, \text{m} = 0.009 \, \text{m} = 0.9 \, \text{cm} \]

### Conclusion
The distance between the central fringe and the third dark fringe when the setup is immersed in a liquid of refractive index 1.33 is **0.9 cm**.