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Example 22.2 Angle of Refraction for Glass Goal Apply Snell's law to a slab of glass. Problem A light ray of wavelength 589 nm (produced by a sodium lamp) traveling through air is incident on smooth, flat slab of crown glass at an angle of ?? = 30° to the normal, as sketched in Figure 22.11. Find the angle of refraction, ??. Strategy Substitute quantities into Snell's law and solve for the unknown angle of refraction, ??. Incident ray Normal ?? Air Glass ?? Refracted ray Figure 22.11 Refraction of light by glass. Solution Solve Snell's law (Eq. 22.8) for sin ?? sin?? = (n?/n?)sin?? (1) From Table 22.1, find n? = 1.00 for air and n? = 1.52 for crown glass. Substitute these values into (1) and take the inverse sine of both sides. sin?? = (1.00 / 1.52)(sin30.0°) = 0.329 ?? = sin?¹(0.329) = ° Remarks Notice the light ray bends toward the normal when it enters a material of a higher index of refraction. If the ray left the material following the same path in reverse, it would bend away from the normal. Exercise 22.2 Hints: Getting Started | I'm Stuck If the light ray moves from inside the glass toward the glass-air interface at an angle of 30.0° to the normal, determine the angle of refraction. The ray bends ° away from the normal, as expected.

          Example 22.2 Angle of Refraction for Glass
Goal Apply Snell's law to a slab of glass.
Problem A light ray of wavelength 589 nm (produced by a sodium lamp) traveling through air is incident on smooth, flat slab of crown glass at an angle of ?? = 30° to the normal, as sketched in Figure 22.11. Find the angle of refraction, ??.
Strategy Substitute quantities into Snell's law and solve for the unknown angle of refraction, ??.
Incident ray
Normal
??
Air
Glass
??
Refracted ray
Figure 22.11 Refraction of light by glass.
Solution
Solve Snell's law (Eq. 22.8) for sin ??
sin?? = (n?/n?)sin?? (1)
From Table 22.1, find n? = 1.00 for air and n? = 1.52 for crown glass. Substitute these values into (1) and take the inverse sine of both sides.
sin?? = (1.00 / 1.52)(sin30.0°) = 0.329
?? = sin?¹(0.329) = °
Remarks Notice the light ray bends toward the normal when it enters a material of a higher index of refraction. If the ray left the material following the same path in reverse, it would bend away from the normal.
Exercise 22.2 Hints: Getting Started | I'm Stuck
If the light ray moves from inside the glass toward the glass-air interface at an angle of 30.0° to the normal, determine the angle of refraction.
The ray bends ° away from the normal, as expected.

$Example 22.2 Angle of Refraction for Glass Goal Apply Snell's law to a slab of glass. Problem A light ray of wavelength 589 nm (produced by a sodium lamp) traveling through air is incident on smooth, flat slab of crown glass at an angle of ?? = 30° to the normal, as sketched in Figure 22.11. Find the angle of refraction, ??. Strategy Substitute quantities into Snell's law and solve for the unknown angle of refraction, ??. Incident ray Normal ?? Air Glass ?? Refracted ray Figure 22.11 Refraction of light by glass. Solution Solve Snell's law (Eq. 22.8) for sin ?? sin?? = (n?/n?)sin?? (1) From Table 22.1, find n? = 1.00 for air and n? = 1.52 for crown glass. Substitute these values into (1) and take the inverse sine of both sides. sin?? = (1.00 / 1.52)(sin30.0°) = 0.329 ?? = sin?¹(0.329) = ° Remarks Notice the light ray bends toward the normal when it enters a material of a higher index of refraction. If the ray left the material following the same path in reverse, it would bend away from the normal. Exercise 22.2 Hints: Getting Started | I'm Stuck If the light ray moves from inside the glass toward the glass-air interface at an angle of 30.0° to the normal, determine the angle of refraction. The ray bends ° away from the normal, as expected.$

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