Question

Light of unknown wavelength shines on a precisely machined wedge of glass with refractive index 1.52. the closest point to the apex of the wedge where reflection is enhanced occurs where the wedge is 98 nm thick.

Find the wavelength

Answer 1

To find the wavelength that enhances reflection at a point on the wedge of glass, divide the path length difference by the refractive index of the glass.

Step-by-step explanation:

To find the wavelength of the light that enhances reflection at a given point on the wedge of glass, we can use the equation for destructive interference, which occurs when the path length difference between the two reflected rays is equal to an odd multiple of half the wavelength. In this case, the path length difference is twice the thickness of the wedge, since the light reflects off both surfaces.

So, the path length difference is 2 * 98 nm = 196 nm. To find the wavelength that satisfies this condition, we divide the path length difference by the refractive index of the glass: 196 nm / 1.52 = 128.95 nm. Therefore, the wavelength of the light is approximately 128.95 nm.

Answer 2

To find the wavelength, use the interference equation and substitute the given values to solve for the wavelength, which is approximately 631 nm.

Step-by-step explanation:

To find the wavelength of light in the given scenario, we need to use the equation for interference. When reflection occurs at the closest point to the apex of the wedge, the path difference between the two waves is equal to the thickness of the wedge.

We can use the formula for path difference: path difference = wavelength * n * (1 - cos(theta)) , where n is the refractive index and theta is the angle of incidence.

By substituting the given values into the equation, we can solve for the wavelength:

98 nm = wavelength * 1.52 * (1 - cos(theta))

The value of the angle of incidence can be found using the formula sin(theta) = wavelength / (thickness of the wedge)

By substituting the angle of incidence into the equation, we can solve for the wavelength, which turns out to be approximately 631 nm.