Snell's Law Worksheet

KEY Snell’s Law 1.) What is the speed of light in a clear plastic whose index of refraction is 1.40? Answer – =  1.40 =  . ×   = 2.14 × 10     2.) The speed of light in a clear liquid is 2.3 × 10 . What is its index of refraction?  Answer – =  =  . ×   = 1.30 . ×  3.) A beam of light strikes the surface of a block of glass ( = 1.50) and produces a refracted angle of 10.0 . What is the incident angle? Answer – 1.0003 sin  = 1.50 sin 10   =    = 15.1° 4.) What is the wavelength of light in water ( = 1.33) if its wavelength in air is 5.30 × 10%& '? Answer – the change in optical densities of the mediums is equal to the change in wavelength that will occur. So . . . the change in optical density is ) = 3.986 × 10%& ' . . 0.754 × 5.30 × 10%& = 0.754 ) = 3.99 × 10%& ' 5.) Monochromatic liquid (light of one color) has a wavelength of 6.0 × 10%& ' in air and 5.0 × 10%& ' in a clear liquid. What is the index of refraction of the clear liquid? Answer – -. × ./ 0. × ./ = 1.2 1.2 × 1.0003 ) = 1.200 ) = 1.2 6.) Monochromatic light has a wavelength of 5.75 × 10%& ' in air and 4.32 × 10%& ' in a clear liquid. If a ray of light enters this clear liquid at an angle of incidence of 25.0°, what is the angle of refraction? Answer – 0.&0× ./ 1. × ./ = 1.3310 1.3310 × 1.0003   =    = 18.512°  = 1.3314 1.0003 sin 25.0 = 1.3314 sin   = 18.5° 1 KEY 7.) Monochromatic light has a wavelength of 5.20 × 10 ' in air and 3.91 × 10 ' in a clear liquid. What is %& %& the speed of light in the clear liquid? Answer – 0. × ./ .2 × ./ =   = 1.32992 1.32992 × 1.0003 1.3310 = . ×   = 2.25509 × 10   = 1.3303    = 2.26 × 10   8.) What is the index of refraction of a substance if the angle of incidence of this substance is 53.0° and the angle of refraction in this substance is 41.0°? Answer –   =   1.0003 sin 53 =  sin 41  = 1.21768  = 1.22 9.) A ray of light strikes the surface of water ( = 1.33) at an angle of 60.0° from the water surface. What is the angle of refraction? ***Don’t forget to measure angle from the normal*** Answer – 1.0003 sin 30 = 1.33 sin    =    = 22.0893°  = 22.1° 10.) What is the critical angle for an air-glass interface if the index of refraction of glass is 1.50? Answer –   =   1.50 sin  = 1.0003 sin 90  = 41.8257°  = 41.8° 11.) What is the critical angle for a water-lucite interface if the index of refraction of water is 1.33 and of Lucite is 1.51? Answer –   =   1.51 sin  = 1.33 sin 90  = 61.738°  = 61.7° 12.) The critical angle for a certain liquid-air interface is 48.8°. What is the index of refraction of the liquid? Answer –   =    sin 48.8 = 1.0003 sin 90  = 1.3333  = 1.33 13.) What is the critical angle of a substance whose index of refraction is 1.81? Answer –   =   1.81 sin  = 1.0003 sin 90  = 33.5491°  = 33.5° 2 KEY 14.) What is the index of refraction of a substance whose critical angle is 42.0°? Answer –  sin 42.0 = 1.0003 sin 90   =    = 1.4949  = 1.50 15.) The speed of light in a clear liquid is three quarters the speed of light in air. What is the critical angle of the liquid? Answer – =   1.003 = . ×  6 = 2.99910 × 10 345 789: = 0.75 × 2.99910 × 10 .222 ×  .12 ×    789: = 2.2493 × 10   = 1.333333 1.3333 sin  = 1.0003 sin 90   =    = 48.6099°  = 48.6° 16.) A ray of light travels from air into water and then into glass ( = 1.50) as shown below. Find the angle of the refraction in the glass. air 35.0o water glass Diagram is not drawn to scale Answer –   =   1.0003 sin 55.0 = 1.33 sin   = 38.03105°   =   1.33 sin 38.03 = 1.50 sin   = 33.11111°  = 33.1° 17.) A ray of light travels from glass ( = 1.50) into water and then into air as shown below. Find the angle that the light leaves the water-air interface. glass 54.8o water air Diagram is not drawn to scale Answer –   =   1.50 sin 35.2 = 1.33 sin   = 40.5500°   =   1.33 sin 40.55 = 1.0003 sin   = 59.8130°  = 59.8° 3 KEY 18.) A ray of light strikes a side of an equilateral lucite prism ( = 1.50) at an angle of 36° as shown below. Find the angle that the light leaves the prism. *** triangle angles are 60°*** Answer – 36o   =   ? 1.0003 sin 36 = 1.50 sin   = 23.0774°   =   1.50 sin 36.9225 = 1.0003 sin   = 64.26778° 36.9° Diagram is not drawn to scale  = 64° 19.) A ray of light strikes a side of lucite ( = 1.50) prism at 40° as shown below. Find the angle that the light leaves the prism. 70o Answer –   =   ? o 1.0003 sin 50 = 1.50 sin   = 30.72043° 40   =   Diagram is not drawn to scale 1.50 sin 39.2795 = 1.0003 sin   = 71.69033°  = 72° 20.) A ray of light reflects from a mirror onto the surface of a clear liquid as shown in the diagram. Determine the index of refraction of the liquid. 29.0° Answer – air liquid 65.0° Diagram is not drawn to scale   =   1.0003 sin 61.0 =  sin 25.0  = 2.07015  = 2.07 4 KEY 21.) A ray of light travels through a clear liquid into a clear plastic as shown in the diagram. Find the index 4.0 cm of refraction of the plastic compared to the liquid. tan  = 1 0 &   = 38.6598° 5.0 cm tan  = liquid  = 74.0546° plastic 2.0 cm Answer –   =   7.0 cm Diagram is not drawn to scale  sin 38.65 =  sin 74.05 =5 =4 =5 = 0.649692 =4 = 0.65 22.) What is the frequency of light in diamond ( = 2.42) if the frequency in air is 6.20 × 101 >?? Answer – frequency never changes due to medium change. Therefore the frequency in diamond is the same as in air (6.20 × 101 >?)? 23.) Monochromatic light of a wavelength of 6.22 × 10 ' enters lucite ( = 1.51). What is the frequency of the light in the Lucite? Answer – =  Change nm to metres. and  = )@  so . . .  =  AB 1.51 = 6.22 × 10 ' × . ×    = = 6.22 × 10%& ' @ = 3.1941 × 101 >? -.× ./ × B @ = 3.19 × 101 >? 24.) Monochromatic light of a wavelength of 4.00 × 10%& ' enters water. What is the period of the light in water? Answer – . . = 0.754 × 4.00 × 10%& = 0.754  1.33 =   = A C . ×  ) = 3.016 × 10%& '  = 2.25564 × 10  2.25564 × 10 = . -× ./   D = 1.33709 × 10%0  C D = 1.34 × 10%0  25.) The period of a light wave in air is 1.70 × 10%0 . What is its wavelength in water? Answer –  = . . A C = 0.7521 3.0 × 10 = A .& × .EF ) = 5.10 × 10%& ' 0.7521 × 5.10 × 10%& ) = 3.8357 × 10%& ' ) = 3.84 × 10%& ' 5