Question

A very thin film in reflected white light appears
A. Colored
B. White
C. Black
D. Red

Answer 1

We are given a very thin film. When this film is reflected in light we know the equation for the path difference. Since the thickness of the film is very less we can neglect the term containing thickness in the equation. Thus we obtain the path difference of the wave. Compare the path difference with the path difference of destructive and constructive interference to know the color of the light.
Formula used:
Path difference,
$\Delta x=2\mu t\cos r+\dfrac{\lambda }{2}$

Complete answer:
In the question we have a very thin film.
We need to find the color of this thin film when it is reflected in white light.
We know the path difference of the reflection of thin film is given by the equation,
$\Delta x=2\mu t\cos r+\dfrac{\lambda }{2}$, were ‘$\Delta x$’ is the path difference, ‘$\mu$’ is the refractive index, ‘t’ is the thickness of the film, ‘r’ is the angle of reflection and ‘$\lambda$’ is the wavelength.
In the question it is given that the thickness of the film is very less.
Therefore when comparing to ‘$\lambda$’ the value of ‘t’ is very less,
$t << \lambda$
Since ‘t’ is very less we can neglect that part in the equation.
Therefore we get path difference of the reflection of thin film as,
$\Delta x=\dfrac{\lambda }{2}$
We know that the path difference of destructive interference is given as $\dfrac{\lambda }{2}$.
Comparing the path difference of the reflection of thin film with the path difference of destructive interference, we can see that both are equal.
Therefore for a very thin film the path difference is the same as that of destructive interference.
Due to destructive interference, the color of a very thin film when reflected appears to be black.

Hence the correct answer is option C.

Note:
We know that for a very thin film the distance traveled inside the film will be insignificant.
Since the film is very thin there will be phase change at one of the surfaces and due to this the two reflected waves will be almost out of phase with each other. Thus the waves interfere destructively and hence the color appears black.