Electromagnetic waves > Properties of emission and reflection of electromagnetic waves > Problem 12.1.20

Statement

$12.1.20.$

A layer of photoemulsion is applied on a mirror metal substrate. When light
falls normally at a distance of 10−5mm from the metal surface, the emulsion
becomes blackened. Explain this effect. Determine the wavelength of light
incident on a metal surface. At what distance from the substrate surface will
the second layer of blackened emulsion be located?

Solution

Light incident normally on the metallic mirror interferes with the reflected light, generating a standing wave. The electric field has a node at the metal surface $(E = 0) $ and antinodes (intensity maxima) at distances $\lambda/4, 3\lambda/4, 5\lambda/4, \dots $from the surface. The photographic emulsion darkens at those antinodes.

Wavelength

The first darkening (first antinode) occurs at $d_1 = 10^{-5}\ \text{cm}$. Since $d_1 = \lambda/4$:

$\boxed{\lambda = 4 \times 10^{-5}\ \text{cm} = 400\ \text{nm}}$.

Distance between darkened layers

Consecutive antinodes are separated by half a wavelength:

$\boxed{x = \frac{\lambda}{2} = 2 \times 10^{-5}\ \text{cm}}.$

Answer

$\boxed{\lambda = 4 \times 10^{-5}\ \text{cm} = 400\ \text{nm}}$.

$\boxed{x = \frac{\lambda}{2} = 2 \times 10^{-5}\ \text{cm}}.$