Identify apply snells law the water ice and ice air interfaces

Figure 33.2

33-1

respect to the normal, not the surface.

(a) SET UP: The incident, reflected and refracted rays are shown in Figure 33.7.

Figure 33.7

EVALUATE: The light is bent toward the normal when the light enters the material of larger refractive index.

SET UP: Let the light initially be in the material with refractive index

gives n a sin θ a = n b sin θ b .

EVALUATE: The final direction of travel depends on the angle of incidence in the first slab and the refractive

boundary is the same on both sides of the boundary, so the frequency does not change. The wavelength and speed

without referring them to the values in vacuum (or air).

the two angles labeled φ′are equal because of the law of reflection. The two angles labeled θare equal because the

lines forming one angle are perpendicular to the lines forming the other angle.

EVALUATE: The ray refracts into a material of smaller n, so it is bent away from the normal.

Figure 33.17

(b) Using the same procedure as in part (a), we have 1.53 sin θc = 1.333 sin 90° and θc = 60.6°.

EVALUATE: Since the refractive index of water is closer to the refractive index of glass than the refractive index of air is, the critical angle for glass-to-water is greater than for glass-to-air.

Figure 33.22

33.25. IDENTIFY: When unpolarized light passes through a polarizer the intensity is reduced by a factor of 1 2and the

transmitted light is polarized along the axis of the polarizer. When polarized light of intensity I max is incident on a

the intensity is ( I 0 /2)(cos60 ) 2 = 0.125 I 0 , and the light is polarized along the axis of the second polarizer. At

point C the intensity is (0.125 )(cos30 ) 0 2 = 0.0938 I 0 .

Figure 33.26

33.27. IDENTIFYand SET UP: Reflected beam completely linearly polarized implies that the angle of incidence equals

Note: φ = 180.0° −θ r−θ b and θ r = θ a . Thus

φ = 180.0° −54.5° −35.5° = 90.0 ;° the

EVALUATE: When the incident light is polarized along the axis of the first filter, φ must be larger to achieve the

same overall reduction in intensity than when the incident light is unpolarized.

SET UP: For the second filter, φ = 62.0°−25.0°= 37.0° .

EXECUTE: After the first filter the intensity is 1

33.32. IDENTIFY: After passing through the first filter the light is linearly polarized along the filter axis. After the

second filter, I = I max(cos ) 2 , where φ is the angle between the axes of the two filters.

EVALUATE: As φ increases toward 90° the axes of the two filters are closer to being perpendicular to each other

and the transmitted intensity decreases.

Figure 33.33a

light is linearly polarized along the axis of the first polarizer.

Figure 33.33b

0I . Then repeat the calculation with only the first and third polarizers.

SET UP: The intensity of the scattered light is proportional to 1/λ4, we can write it as I = (constant)/λ4.

EXECUTE:(a) Since I is proportional to 1/λ4, we have I = (constant)/λ4. Taking the ratio of the intensity of the

EVALUATE: In the scattered light, the intensity of the short-wavelength violet light is about 7 times as great as that of the red light, so this scattered light will have a blue-violet color.

Figure 33.36

SET UP: The definition of the index of refraction is n = c/v, where v is the speed of light in the jelly.

EXECUTE: First get the length L of the tube using air. In the air, we have L = ct = (3.00 × 108 m/s)(8.72 ns) = 2.616 m.

SET UP: The ray is sketched in Figure 33.41.

EXECUTE: For glass→air at point A, Snell's law gives (1.38)sin θ crit = (1.00)sin90°and θ crit = 46.4°.

Figure 33.41

33.42. IDENTIFY: As the light crosses the glass-air interface along AB, it is refracted and obeys Snell’s law.

EVALUATE: The lower ray is deflected more than the upper ray because that wavelength has a slightly greater

index of refraction than the upper ray.

EXECUTE: In the circularly polarized light the two perpendicular polarization components are 90° out of phase.

The quarter-wave plate shifts the relative phase by 90° and then the two components are either in phase or

wavelength in the material.

EVALUATE: Without the glass plate the number of wavelengths between the source and screen is

edge of the halo.

SET UP: The ray incident at the critical angle is sketched in Figure 33.45.

the liquid.

After the liquid is poured in, θ is the same and the refracted ray passes through the center of the bottom of the a

glass, as shown in Figure 33.47b.

bn the refractive index of the liquid:

EXECUTE: Apply Snell's law to the air→liquid refraction. (1.00)sin(42.5 ) = (1.63)sin θ b and θ = b 24.5°.

θ b = φ and φ = θ′a , so θ a′ = θ b = 24.5°. Snell's law applied to the liquid→air refraction gives

Figure 33.49

SET UP: The path of the ray is sketched in Figure 33.50. The ray enters the prism at normal incidence so is not

SET UP: The paths of the two rays are sketched in Figure 33.51.

EXECUTE:(a) In the water, θ = S R= (1.09 m)/(1.10 m) = 0.991 rad = 56.77°. This is the critical angle. So, using the refractive index for water from Table 33.1, we get n = (1.333) sin 56.77° = 1.12
(b) (i) The laser beam stays in the water all the time, so

The total time is 4.09 ns + (9.78 ns)/2 = 8.98 ns
EVALUATE: The gas must be under considerable pressure to have a refractive index as high as 1.12.

SET UP: The path of a ray from the sun is sketched in Figure 33.55.

(a = vacuum of space, refractive, index 1.0; b = atmosphere, refractive index n)

EVALUATE: The calculated δ is about the same as the angular radius of the sun.

(b) Taking the derivative with respect to x of the time and setting it to zero yields

SET UP: Refer to the angles and distances defined in the figure that accompanies the problem.

EVALUATE: The lateral displacement in part (d) is large, of the same order as the thickness of the plate.

surface of the film at an angle of incidence α . Snell’s law applied to the refraction as ray B leaves the film gives

they emerge from the film.

(c) If two colors have different indices of refraction for the glass, then the deflection angles for them will differ:

is 90.00° – 58.15° = 31.85°. Applying Snell’s law at the water surface gives

The Nature and Propagation of Light 33-19

(b) Multiplying Eq. (1) by cos β and Eq. (2) by cosα yields:

(d) Expanding the left-hand side, we have:

	x																2	(sin			2		β +				cos		2	β )			+	y	2	(sin				2	α +					cos				2	α )		−		2		xy		(sin α sin						β +			cos α cos
	=																x		2	+		y		2	−	2		xy	(sin α sin							β +				cos α cos											β )			=		x	2	+		y	2	−	2	xy		cos( α−
	The right-hand side can be rewritten:																													a	2	(sin α cos							β−					sin					β cos )							2	=		a	2	sin ( 2 α−
	x	2	+	y	2	−	2	xy		cos( α−					β )			=		a	2		sin ( 2 α−							β ). Or,						x		2	+		y		2	−			2		xy		cos δ					=		a	2	sin		2	δ , where						δ	=	α−
	EVALUATE:(e)											δ =			0:		x		2	+		y		2	−	2		xy	=		(	x	−	y		)	2	=	0			⇒			x			=		y		, which is a straight diagonal line
	δ =			π	:	x	2	+	y	2	−	2 xy				=		a		2	, which is an ellipse
33.66.				4														2
	δ =			π 2x			2	+	y	2	=	a	2	,which is a circle. This pattern repeats for the remaining phase differences.

larger for red than for violet, so red in the rainbow is higher above the horizon. 33.67. IDENTIFY: Follow similar steps to Challenge Problem 33.66.
SET UP: Refer to Figure 33.20e in the textbook.

EVALUATE: The angles we calculated agree with those given in Figure 37.20e in the textbook.The color that appears higher above the horizon is violet. The colors appear in reverse order in a secondary rainbow compared to a primary rainbow.