A very thin sheet of plastic having refractive index `mu=1.5` covers one slit of a double slit apparatus illuminated by 700 nm light . The central point on the screen , instead of being maximum , is dark . What is the possible thickness of the plastic in (nm) ?

Options

A. 700 nm
B. 1400 nm
C. 350nm
D. 2100 nm

Answer

To solve the problem, we need to determine the thickness of the plastic sheet that causes a dark fringe at the central point on the screen in a double slit experiment. This occurs when the additional path difference introduced by the plastic sheet is equal to half the wavelength of the light used.

### Step-by-step Solution:

1. **Identify the Given Values:**
- Refractive index of plastic, \( \mu = 1.5 \)
- Wavelength of light, \( \lambda = 700 \, \text{nm} \)

2. **Understand the Condition for Dark Fringe:**
- A dark fringe occurs when the additional path difference caused by the plastic sheet is equal to \( \frac{\lambda}{2} \).

3. **Calculate the Additional Path Difference:**
- The additional path difference \( \Delta x \) due to the plastic sheet can be expressed as:
\[ \Delta x = (\mu - 1) t \]
where \( t \) is the thickness of the plastic sheet.

4. **Set Up the Equation:**
- Since we want the additional path difference to be \( \frac{\lambda}{2} \):
\[ (\mu - 1) t = \frac{\lambda}{2} \]

5. **Substitute the Values:**
- Substitute \( \mu = 1.5 \) and \( \lambda = 700 \, \text{nm} \):
\[ (1.5 - 1) t = \frac{700 \, \text{nm}}{2} \]
- This simplifies to:
\[ 0.5 t = 350 \, \text{nm} \]

6. **Solve for Thickness \( t \):**
- Rearranging gives:
\[ t = \frac{350 \, \text{nm}}{0.5} = 700 \, \text{nm} \]

### Final Answer:
The possible thickness of the plastic sheet is \( t = 700 \, \text{nm} \).

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