To ensure almost `100%` transmittivity, photographic lenses are often coated with a thin layer of dielectric material, like `MgF_(2)(mu=1.38)` . The minimum thickness of the film to be used so that at the centre of visible spectrum `(lambda = 5500 Å)` there is maximum transmission.

Options

A. 5000 Ã…
B. 2000 Ã…
C. 1000 Ã…
D. 3000 Ã…

Explanation

To find the minimum thickness of the film (MgF₂) for maximum transmission at the center of the visible spectrum (λ = 5500 Å), we can follow these steps:

### Step 1: Understand the condition for maximum transmission
For maximum transmission through a thin film, the condition is given by the formula:
\[ 2 \mu t \cos r = n \lambda \]
where:
- \( \mu \) = refractive index of the film (for MgF₂, \( \mu = 1.38 \))
- \( t \) = thickness of the film
- \( r \) = angle of refraction (which is approximately 0 for normal incidence)
- \( n \) = order of interference (for the first maximum, \( n = 1 \))
- \( \lambda \) = wavelength of light (given as \( 5500 \, \text{Å} \))

### Step 2: Simplify the equation for normal incidence
At normal incidence, the angle of refraction \( r \) is 0 degrees, thus \( \cos r = 1 \). The equation simplifies to:
\[ 2 \mu t = n \lambda \]

### Step 3: Substitute the values
Substituting \( n = 1 \), \( \lambda = 5500 \, \text{Å} \), and \( \mu = 1.38 \):
\[ 2 \times 1.38 \times t = 1 \times 5500 \]

### Step 4: Solve for thickness \( t \)
Rearranging the equation to find \( t \):
\[ t = \frac{5500}{2 \times 1.38} \]

Calculating the denominator:
\[ 2 \times 1.38 = 2.76 \]

Now substituting back:
\[ t = \frac{5500}{2.76} \]
\[ t \approx 1992.75 \, \text{Å} \]

### Step 5: Round to the nearest significant figure
To find the minimum thickness, we can round this to:
\[ t \approx 1993 \, \text{Å} \]

### Final Answer
The minimum thickness of the film to ensure maximum transmission at the center of the visible spectrum is approximately **1993 Å**.

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