A thin film of refractive index 1.5 and thickness `4 xx 10^(-5)` cm is illuminated by light normal to the surface. What wavelength within the visible spectrum will be intensified in the reflected beam?

Options

A. `4800 Å`
B. `5800 Å`
C. `6000 Å`
D. `6800 Å`

Explanation

To solve the problem of finding the wavelength that will be intensified in the reflected beam from a thin film of refractive index 1.5 and thickness \(4 \times 10^{-5}\) cm, we can follow these steps:

### Step 1: Understand the Condition for Bright Fringes
The condition for constructive interference (bright fringes) in thin films is given by the formula:
\[ 2nd = (m + \frac{1}{2}) \lambda \]
where:
- \(n\) is the refractive index of the film,
- \(d\) is the thickness of the film,
- \(m\) is the order of the fringe (an integer),
- \(\lambda\) is the wavelength of light in vacuum.

### Step 2: Rearrange the Formula
We need to rearrange the formula to solve for \(\lambda\):
\[ \lambda = \frac{2nd}{m + \frac{1}{2}} \]

### Step 3: Substitute the Known Values
Given:
- \(n = 1.5\)
- \(d = 4 \times 10^{-5}\) cm = \(4 \times 10^{-7}\) m (since \(1 \text{ cm} = 10^{-2} \text{ m}\))

Now substituting these values into the equation:
\[ \lambda = \frac{2 \times 1.5 \times 4 \times 10^{-7}}{m + \frac{1}{2}} \]
\[ \lambda = \frac{12 \times 10^{-7}}{m + \frac{1}{2}} \]

### Step 4: Determine the Value of \(m\)
To find the wavelength that lies within the visible spectrum (approximately 400 nm to 700 nm), we need to find an integer \(m\) such that \(\lambda\) falls within this range.

### Step 5: Calculate for Different Values of \(m\)
1. For \(m = 2\):
\[ \lambda = \frac{12 \times 10^{-7}}{2 + \frac{1}{2}} = \frac{12 \times 10^{-7}}{2.5} = \frac{12 \times 10^{-7}}{2.5} = 4.8 \times 10^{-7} \text{ m} = 480 \text{ nm} \]

2. Check if \(m = 1\) or \(m = 3\) gives a wavelength in the visible range:
- For \(m = 1\):
\[ \lambda = \frac{12 \times 10^{-7}}{1 + \frac{1}{2}} = \frac{12 \times 10^{-7}}{1.5} = 8 \times 10^{-7} \text{ m} = 800 \text{ nm} \quad (\text{not visible}) \]
- For \(m = 3\):
\[ \lambda = \frac{12 \times 10^{-7}}{3 + \frac{1}{2}} = \frac{12 \times 10^{-7}}{3.5} \approx 3.43 \times 10^{-7} \text{ m} = 343 \text{ nm} \quad (\text{not visible}) \]

### Step 6: Conclusion
The only value of \(m\) that gives a wavelength within the visible spectrum is \(m = 2\), which results in:
\[ \lambda = 480 \text{ nm} \]

Thus, the wavelength that will be intensified in the reflected beam is **480 nm**.