Ufa State Aviation Technical University EXAMINATION TICKET NO. 38 A1. To enlighten the lens (n,=1.5), a thin film whose refractive index is n_(2)= 1.28 is applied to the surface of the lens. Light from lambda =0.55 um normally falls on the lens. At what minimum film thickness the reflected rays are maximally attenuated __ jim 4. 0.5 1. 0.4 2. 0,3 3. 0.2

To solve this problem, we need to use the formula for thin film interference:<br /><br />\[ 2 \cdot n_2 \cdot t = m \cdot \lambda \]<br /><br />where:<br />- \( n_2 \) is the refractive index of the thin film,<br />- \( t \) is the thickness of the film,<br />- \( m \) is an integer (1, 2, 3,...),<br />- \( \lambda \) is the wavelength of the light.<br /><br />Given:<br />- \( n_2 = 1.28 \)<br />- \( \lambda = 0.55 \) um<br /><br />We need to find the minimum film thickness for which the reflected rays are maximally attenuated. This occurs when \( m = 1 \).<br /><br />Substituting the given values into the formula:<br /><br />\[ 2 \cdot 1.28 \cdot t = 1 \cdot 0.55 \]<br /><br />Solving for \( t \):<br /><br />\[ t = \frac{0.55}{2 \cdot 1.28} \]<br /><br />\[ t = \frac{0.55}{2.56} \]<br /><br />\[ t \approx 0.215 \text{ um} \]<br /><br />Therefore, the minimum film thickness at which the reflected rays are maximally attenuated is approximately 0.215 um. The closest answer from the given options is:<br /><br />3. 0.2