Light of wavelength `lambda` in vacuum strikes a lens that is made of glass with index of refraction 1.6. The lens has been coated with a film of thickness t and index of refraction 1.3. For which one of the following conditions will there be no reflection?

Options

A. `2t=(lambda)/(2)`
B. `2t=(1)/(2)((lambda)/(1.3))`
C. `2t=(lambda)/(1.33)`
D. `2t=(1)/(2)((lambda)/(1.6))`

Answer

To solve the problem, we need to determine the condition under which there will be no reflection when light strikes a lens coated with a film. This involves understanding the principles of interference and the conditions for destructive interference.

### Step-by-Step Solution:

1. **Understanding Reflection and Interference**:
- When light reflects off a medium with a higher refractive index, it undergoes a phase change of 180 degrees (or λ/2). In our case, light reflects off the glass lens (n = 1.6) and the film (n = 1.3).
- For no reflection, we want to achieve destructive interference, which means we need to set up a condition where the reflected waves cancel each other out.

2. **Condition for Destructive Interference**:
- The condition for destructive interference is given by:
\[ 2t = \left( m + \frac{1}{2} \right) \frac{\lambda}{n} \]
where \( m \) is an integer (0, 1, 2,...), \( t \) is the thickness of the film, \( \lambda \) is the wavelength of light in vacuum, and \( n \) is the refractive index of the film.

3. **Substituting Values**:
- In our case, the refractive index of the film is \( n = 1.3 \).
- For the first minimum (where \( m = 0 \)):
\[ 2t = \left( 0 + \frac{1}{2} \right) \frac{\lambda}{1.3} \]
Simplifying gives:
\[ 2t = \frac{\lambda}{2 \cdot 1.3} \]

4. **Solving for Thickness \( t \)**:
- Rearranging the equation:
\[ t = \frac{\lambda}{4 \cdot 1.3} \]

5. **Conclusion**:
- The condition for no reflection (destructive interference) occurs when the thickness \( t \) of the film is:
\[ t = \frac{\lambda}{4 \cdot 1.3} \]