Question

a glass sheet 1.20 um thick is suspended in air. in reflected light, there are gaps in the visible spectrum at 564 nm and 630 nm. calculate the minimum value of the index

Answer 1

To find the minimum value of the index of refraction for a glass sheet, we can use the equation for destructive interference and solve for the minimum thickness. Once we have the minimum thickness, we can substitute it into the equation to find the minimum value of the index of refraction.

Step-by-step explanation:

To determine the minimum value of the index of refraction (n) for a glass sheet, we need to find the minimum thickness that causes destructive interference for two specific wavelengths. In this case, the wavelengths are 564 nm and 630 nm.

Using the equation for destructive interference, Δ = (2t/nλ), where Δ is the path length difference, t is the thickness of the sheet, n is the index of refraction, and λ is the wavelength.

By setting Δ = λ/2, we can solve for the minimum thickness and then substitute it into the equation to find the minimum value of the index of refraction.

Answer 2

the minimum value of the index of refraction for a glass sheet, we can use the concept of interference. The gaps in the visible spectrum at 564 nm and 630 nm in reflected light occur due to destructive interference when light reflects off the front and back surfaces of the glass sheet. By solving the equation involving path difference and wavelength, we can find the minimum value of the index of refraction.

Step-by-step explanation:

To calculate the minimum value of the index of refraction for a glass sheet, we can use the concept of interference. The gaps in the visible spectrum at 564 nm and 630 nm in reflected light occur due to destructive interference when light reflects off the front and back surfaces of the glass sheet. For destructive interference to occur, the path difference between the two reflected beams must be equal to an odd multiple of half the wavelength. The path difference can be calculated using the formula:

Path Difference = 2nt

where n is the index of refraction of the glass sheet and t is the thickness of the glass sheet. By equating the calculated path difference to an odd multiple of half the wavelength, we can solve for the minimum value of the index of refraction.

For example, at 564 nm:

Path Difference = 2nt = (2n)(1.20 um) = (2n)(1.20 × 10^-6 m)

Equating this to an odd multiple of half the wavelength:

(2n)(1.20 × 10^-6 m) = (2m + 1)(564 × 10^-9 m)

Where m is an integer. By solving this equation, we can find the minimum value of the index of refraction for the glass sheet.