A light beam of wavelength `600nm` in air passes through film `1(n_1=1.2)` of thickness `1.0 mu m`,then through film `2`(air) of thickness `1.5mu m`, and finally through film `3(n_3=1.8)` of `thickness `1.0mu m`.
(a) Which film does the light cross in the least time, and what is that least time?
(b) What are the total number of wavelength (at any instant) across all three films together?

Explanation

To solve the problem step by step, we will break it down into two parts as per the question.

### Part (a): Which film does the light cross in the least time, and what is that least time?

1. **Identify the parameters:**
- Wavelength in air, \( \lambda_0 = 600 \, \text{nm} = 600 \times 10^{-9} \, \text{m} \)
- Film 1: \( n_1 = 1.2 \), thickness \( L_1 = 1.0 \, \mu m = 1.0 \times 10^{-6} \, \text{m} \)
- Film 2: \( n_2 = 1.0 \) (air), thickness \( L_2 = 1.5 \, \mu m = 1.5 \times 10^{-6} \, \text{m} \)
- Film 3: \( n_3 = 1.8 \), thickness \( L_3 = 1.0 \, \mu m = 1.0 \times 10^{-6} \, \text{m} \)

2. **Calculate the time taken to cross each film:**
The time \( t \) taken to cross a film is given by the formula:
\[ t = \frac{L}{v} \]
where \( v = \frac{c}{n} \) (speed of light in the medium).

Thus, the time can be rewritten as:
\[ t = \frac{nL}{c} \]

3. **Calculate time for each film:**
- For Film 1:
\[ t_1 = \frac{n_1 L_1}{c} = \frac{1.2 \times 1.0 \times 10^{-6}}{3 \times 10^8} = \frac{1.2 \times 10^{-6}}{3 \times 10^8} = 4 \times 10^{-15} \, \text{s} \]

- For Film 2:
\[ t_2 = \frac{n_2 L_2}{c} = \frac{1.0 \times 1.5 \times 10^{-6}}{3 \times 10^8} = \frac{1.5 \times 10^{-6}}{3 \times 10^8} = 5 \times 10^{-15} \, \text{s} \]

- For Film 3:
\[ t_3 = \frac{n_3 L_3}{c} = \frac{1.8 \times 1.0 \times 10^{-6}}{3 \times 10^8} = \frac{1.8 \times 10^{-6}}{3 \times 10^8} = 6 \times 10^{-15} \, \text{s} \]

4. **Determine the least time:**
Comparing \( t_1, t_2, \) and \( t_3 \):
- \( t_1 = 4 \times 10^{-15} \, \text{s} \)
- \( t_2 = 5 \times 10^{-15} \, \text{s} \)
- \( t_3 = 6 \times 10^{-15} \, \text{s} \)

The least time is \( t_1 = 4 \times 10^{-15} \, \text{s} \) for Film 1.

### Part (b): What are the total number of wavelengths (at any instant) across all three films together?

1. **Calculate the wavelength in each medium:**
The wavelength in a medium is given by:
\[ \lambda = \frac{\lambda_0}{n} \]
- For Film 1:
\[ \lambda_1 = \frac{600 \times 10^{-9}}{1.2} = 500 \times 10^{-9} \, \text{m} \]
- For Film 2:
\[ \lambda_2 = \frac{600 \times 10^{-9}}{1.0} = 600 \times 10^{-9} \, \text{m} \]
- For Film 3:
\[ \lambda_3 = \frac{600 \times 10^{-9}}{1.8} = 333.33 \times 10^{-9} \, \text{m} \]

2. **Calculate the total distance traveled through all films:**
\[ L_{\text{total}} = L_1 + L_2 + L_3 = 1.0 \times 10^{-6} + 1.5 \times 10^{-6} + 1.0 \times 10^{-6} = 3.5 \times 10^{-6} \, \text{m} \]

3. **Calculate the total number of wavelengths:**
The total number of wavelengths across all films is given by:
\[ N = \frac{L_{\text{total}}}{\lambda_1} + \frac{L_{\text{total}}}{\lambda_2} + \frac{L_{\text{total}}}{\lambda_3} \]
\[ N = \frac{3.5 \times 10^{-6}}{500 \times 10^{-9}} + \frac{3.5 \times 10^{-6}}{600 \times 10^{-9}} + \frac{3.5 \times 10^{-6}}{333.33 \times 10^{-9}} \]
\[ N = 7 + 5.83 + 10.5 \approx 23.33 \]

### Final Answers:
(a) The film that the light crosses in the least time is Film 1, with a least time of \( 4 \times 10^{-15} \, \text{s} \).

(b) The total number of wavelengths across all three films together is approximately \( 23.33 \).