A typical thin film of refractive index 1.33 and thickness 0.000015 m is illuminated by a point source of white light. light is made to fall(i) normally and at 60 deg. which colours will appear in the reflected light?

To determine which colors will appear in the reflected light from a thin film with a refractive index of 1.33 and a thickness of 0.000015 m, we need to consider the phenomenon of thin film interference. This occurs when light waves reflect off the top and bottom surfaces of the film, leading to constructive or destructive interference depending on the wavelength of the light and the angle of incidence.

Understanding Thin Film Interference

When light hits the thin film, some of it reflects off the top surface while some transmits into the film, reflects off the bottom surface, and then exits the film. The key factors influencing the interference pattern are:

• Wavelength of Light: Different colors (wavelengths) of light will interfere differently.
• Phase Change: A phase change of half a wavelength occurs when light reflects off a medium with a higher refractive index.
• Thickness of the Film: The thickness of the film determines the path length difference between the two reflected waves.

Case (i): Normal Incidence

When light is incident normally (at 0 degrees), the path difference between the two reflected rays is simply twice the thickness of the film. The condition for constructive interference is given by:

2nt = mλ

Where:

• n: Refractive index of the film (1.33)
• t: Thickness of the film (0.000015 m)
• m: Order of interference (an integer)
• λ: Wavelength of light in vacuum

Substituting the values, we can find the wavelengths that will constructively interfere:

First, we need to calculate the effective wavelength in the film:

λ' = λ/n

For white light, the wavelengths range from approximately 400 nm (violet) to 700 nm (red). Let's calculate the effective wavelengths for constructive interference:

Using the formula:

2(1.33)(0.000015) = m(λ/n)

Calculating the path difference:

2(1.33)(0.000015) = 0.0000399 m = 39.9 μm

Now we can find the wavelengths that satisfy the condition for different orders of m. For example, for m = 1, we can find:

λ = 39.9 μm / 1.33 ≈ 30 μm

Since this value is outside the visible spectrum, we can check for higher orders. The visible wavelengths will be those that fall within the 400 nm to 700 nm range. You can repeat this for m = 2, 3, etc., until you find wavelengths that fall within this range.

Case (ii): Incidence at 60 Degrees

When light strikes the film at an angle of 60 degrees, the situation becomes a bit more complex due to the change in path length caused by the angle of incidence. The effective path difference is given by:

2nt cos(θ) = mλ

Where θ is the angle of incidence. The cosine of 60 degrees is 0.5, so we can substitute this into our equation:

2(1.33)(0.000015)cos(60) = mλ

Calculating this gives:

2(1.33)(0.000015)(0.5) = 0.00001995 m = 19.95 μm

Again, we can find the effective wavelengths for different orders of m. The same process applies, but now we need to consider the angle's effect on the wavelength in the film:

λ' = λ/n cos(θ)

By repeating the calculations for various orders, you will find which colors appear in the reflected light for both cases.

Summary of Colors

In both cases, the colors that appear in the reflected light will depend on the specific wavelengths that satisfy the interference conditions. Typically, for thin films, you might see a range of colors from the visible spectrum, often producing a rainbow-like effect due to the varying wavelengths of light being reflected constructively or destructively.