10) PLEASE BE CLEAR ON YOUR RESPONSE AND ACCURACY, THANKS. **Transmission through Thin Layers** In the figure, light is incident perpendicularly on a thin layer of material 2 that lies between (thicker) materials 1 and 3. (The rays are tilted only for clarity.) Part of the light ends up in material 3 as ray \(r_3\) (the light does not reflect inside material 2) and \(r_4\) (the light reflects twice inside material 2). The waves of \(r_3\) and \(r_4\) interfere, and here we consider the type of interference to be either maximum (max) or minimum (min). The table below provides the indices of refraction \(n_1\), \(n_2\), and \(n_3\), the type of interference, and the thin layer thickness \(L\) in nanometers. Give the wavelength that is in the visible range.**Diagram Explanation:**- A vertical representation of three layers is depicted with: - \(n_1\) for material 1 on the left, - \(n_2\) for the thin layer of material 2 in the middle, with thickness \(L\) indicated, - \(n_3\) for material 3 on the right.- The incident light ray \(i\) travels through these layers: - It forms ray \(r_3\) which exits the material without reflection inside layer 2. - Ray \(r_4\) reflects twice within material 2 and exits.**Table of Values:**| \(n_1\) | \(n_2\) | \(n_3\) | Type | \(L\) (nm) | \(\lambda\) (nm) ||---------|---------|---------|------|------------|------------------|| 1.77 | 1.66 | 1.57 | max | 411 | |**Input Fields:** - **Number**: (With information tooltip)- **Units**: (Dropdown selection) This setup provides data to calculate the wavelength of light that remains within the visible spectrum after interference.

Given: Refractive Index of 1, n1=1.77Refractive Index of 2, n2=1.66Refractive Index of 3, n3=1.57Length, L=411 nm=411×10-9 m As the specified value of the refractive index, the refractive index of 1 is greater than the refractive index of 2, and the refractive index of 2 is greater than the refractive index of 3, and the type is maximum. Material 1 and material 3 are thicker than material 2; the relation between the refractive index is specified. There will be no phase shift due to the relation between the refractive index. The expression of the constructive interference is as follows: 2n2L=m+12λ ..... (1) Also, write it as follows: λ=2n2Lm+12 .... (2) Here, the value of m=0,1,2,3.... For m=2, Substitute the value in equation (2). λ=21.66411×10-9m2+12λ=1364.52×10-92.5mλ=545.808×10-9m The wavelength in the visible range is 545.808×10-9m.

Science

Advanced Physics

Transmission through thin layers. In the figure, light is incident perpendicularly on a thin layer of material 2 that lies between (thicker) materials 1 and 3. (The rays are tilted only for clarity.) Part of the light ends up in material 3 as ray r3 (the light does not reflect inside material 2) and ra (the light reflects twice inside material 2). The waves of r3 and ra interfere, and here we consider the type of interference to be either maximum (max) or minimum (min). The table below provides the indexes of refraction ng, n2, and ng, the type of interference, and the thinlayer thickness Lin nanometers. Give the wavelength that is in the visible range. n1 n2 n3 Type L A 1.77 1.66 1.57 max 411 Number Units

Transmission through thin layers. In the figure, light is incident perpendicularly on a thin layer of material 2 that lies between (thicker) materials 1 and 3. (The rays are tilted only for clarity.) Part of the light ends up in material 3 as ray r3 (the light does not reflect inside material 2) and ra (the light reflects twice inside material 2). The waves of r3 and ra interfere, and here we consider the type of interference to be either maximum (max) or minimum (min). The table below provides the indexes of refraction ng, n2, and ng, the type of interference, and the thinlayer thickness Lin nanometers. Give the wavelength that is in the visible range. n1 n2 n3 Type L A 1.77 1.66 1.57 max 411 Number Units