25.3 The Law of Refraction

f = 25.0mm di = 31.0mm do = 129mm m = -di/do m = -31.0mm/129mm m = -0.240

A camera lens used for taking close-up photographs has a focal length of 25.0 mm. The farthest it can be placed from the film is 31.0 mm. (a) What is the closest object (in mm) that can be photographed? 129mm (b) What is the magnification of this closest object? ✕

f = 25.0mm di = 31.0mm 1/f = 1/di + 1/do 1/f - 1/di = 1/do 1/(25.0mm) - 1/(31.0mm) = 1/do 1/do = 0.00774/mm do = 129mm

A camera lens used for taking close-up photographs has a focal length of 25.0 mm. The farthest it can be placed from the film is 31.0 mm. (a) What is the closest object (in mm) that can be photographed? mm

f = 15.5cm do = 12.5cm 1/f = 1/di + 1/do 1/f -1/do = 1/di 1/15.5 - 1/12.5 = 1/di 1/di = -1/(0.01548cm) di = -64.6cm = -0.646m

A doctor examines a mole with a 15.5 cm focal length magnifying glass held 12.5 cm from the mole. (a) Where is the image? (Enter the image distance in meters. Include the sign of the value in your answer.) m

f = 15.5cm do = 12.5cm = 0.125m di = -64.6cm = -0.646m m = -di/do m = +0.646m/0.125m m = 5.17

A doctor examines a mole with a 15.5 cm focal length magnifying glass held 12.5 cm from the mole. (a) Where is the image? (Enter the image distance in meters. Include the sign of the value in your answer.) m (b) What is its magnification?

f = 15.5cm do = 12.5cm = 0.125m di = -64.6cm = -0.646m m = 5.17 image diameter = 4.95mm diameter x magnification = new diameter 4.95mm x 5.17 = magnified diameter magnified diameter = 25.6mm

A doctor examines a mole with a 15.5 cm focal length magnifying glass held 12.5 cm from the mole. (a) Where is the image? (Enter the image distance in meters. Include the sign of the value in your answer.) m (b) What is its magnification? (c) How big in millimeters is the image of a 4.95 mm diameter mole? mm

theta for real diver triangle = 90° - 35.3° = 54.7° opp = 2.0m tan(theta) = opp/adj adj = opp/tan(theta) adj = 2.0m/tan(54.7°) adj real diver triangle = 1.416m theta imaginary diver triangle = 39.62° adj imaginary diver triangle = 1.416m tan(theta) = opp/adj adj = opp/tan(theta) adj x tan(theta) = opp 1.416m x tan(39.62°) = opp opp = 1.17m

A scuba diver training in a pool looks at his instructor as shown in the figure below. The angle between the ray in the water and the perpendicular to the water is 35.3°. (a) Using information in the figure above, find the height (in m) of the instructor's head above the water, noting that you will first have to calculate the angle of incidence. 1.66m (b) Find the apparent depth (in m) of the diver's head below water as seen by the instructor. m

sintheta1 = 35.3° sintheta2 = ? incidence of water = n1 = 1.333 incidence of air = n2 = 1.000 n1 x sintheta1 = n2 x sintheta2 asin(n1 x sintheta1 /n2) = theta2 asin(1.333 x sin35.3° /1.000) = theta2 theta2 = 50.38° theta incidence = 90° - theta2 theta incidence = 90° - 50.38° = 39.62° adj = 2.0m tantheta = opp/adj tantheta x adj = opp tan39.62° x 2.0m = opp opp = 1.66m

A scuba diver training in a pool looks at his instructor as shown in the figure below. The angle between the ray in the water and the perpendicular to the water is 35.3°. (a) Using information in the figure above, find the height (in m) of the instructor's head above the water, noting that you will first have to calculate the angle of incidence. m

sin(theta1) = 38.8° sin(theta2) = ?° refractive index water = n1 = 1.333 refractive index air = n2 = 1.000 n1 x sintheta1 = n2 x sintheta2 asin(n1 x sintheta1 /n2) = theta2 asin(1.333sin38.8° /1.000) = theta2 theta2 = 56.6°

A scuba diver training in a pool looks at his instructor as shown in the figure below. What angle does the ray from the instructor's face make with the perpendicular to the water at the point where the ray enters? The angle between the ray in the water and the perpendicular to the water is 38.8°. °

θc = sin^-1(n2/n1) n1 = 1.69 n2 = 1.45 θc = sin^-1(1.45/1.69) θc = 59.1°

An optical fiber uses one glass clad with another glass. What is the critical angle? (Assume the glass in the fiber has an index of refraction of 1.69, and the glass in the cladding has an index of refraction of 1.45.) °

n1 air = 1.00 n2 olive oil = 1.47 n3 water = 1.333 θ2 olive oil = 31.4° θ1 air = 50.0° θ3 water = ?° θ2 water = ?° n2 x sinθ2 = n3 x sinθ3 1.47 x sin31.4° = 1.333 x sinθ3 asin(1.47 x sin31.4° /1.333) = θ3 θ3 = 35.1°

As shown in the figure, a light beam travels from air, through olive oil, and then into water. If the angle of refraction θ2 for the light in the olive oil is 31.4°, determine the angle of incidence θ1 in air and the angle of refraction θ3 in water. The index of refraction for olive oil is 1.47. θ1 = 50.0° θ3 = °

n1 air = 1.00 n2 olive oil = 1.47 n3 water = 1.333 θ2 olive oil = 31.4° θ1 air = ?° θ1 air = ?° n1 x sinθ1 = n2 x sinθ2 1.00 x sinθ1 = 1.47 x sin31.4° θ1 = asin(1.47 x sin31.4° /1) θ1 = 50.0°

As shown in the figure, a light beam travels from air, through olive oil, and then into water. If the angle of refraction θ2 for the light in the olive oil is 31.4°, determine the angle of incidence θ1 in air and the angle of refraction θ3 in water. The index of refraction for olive oil is 1.47. θ1 = °

θc = sin^-1(n2/n1) n1 = 1.501 n2 = 1.434 θc = sin^-1(1.434/1.501) θc = 72.82 degrees

At what minimum angle will you get total internal reflection of light traveling in benzene and reflected from fluorite? °

speed of light in medium = v = 1.559e8m/s speed of light in vacuum = c = 2.998e8m/s nv = c n = c/v n = 2.998e8m/s /1.559e8m/s n = 1.923e8m/s

Calculate the index of refraction for a medium in which the speed of light is 1.559 ✕ 10^8 m/s. (The speed of light in vacuum is 2.998 ✕ 10^8 m/s. Enter your answer to at least three decimal places.)

n zircon = 1.923

Calculate the index of refraction for a medium in which the speed of light is 1.559 ✕ 10^8 m/s. (The speed of light in vacuum is 2.998 ✕ 10^8 m/s. Enter your answer to at least three decimal places.) n = 1.923 Identify the most likely substance based on this table. ---Select--- benzene carbon disulfide carbon tetrachloride ethanol glycerine water, fresh diamond fluorite glass, crown glass, flint ice at 20°C polystyrene Plexiglas quartz, crystalline quartz, fused sodium chloride zircon

wavelength in air = 632.8nm n zircon = 1.923 c = 3.00e8m/s v = 1.56e8m/s wavelength in zircon = 329nm speed = wavelength x frequency 1.56e8m/s = 329nm x frequency 1.56e8m/s /329nm = frequency 1.56e8m/s /329e-9m = frequency frequency = 0.00474e17/s frequency = 4.74e14hz

Determine the speed, wavelength, and frequency of light from a helium-neon laser as it travels through zircon. The wavelength of the light from the laser is 632.8 nm in air and the index of refraction of zircon is 1.923. speed = 1.56e8m/s wavelength = 329nm frequency = ?Hz

wavelength in air = 632.8nm n zircon = 1.923 c = 3.00e8m/s v = 1.56e8m/s speed = wavelength x frequency speed in air = wavelength in air x frequency speed in zircon = wavelength in zircon x frequency speed in zircon/speed in air = wavelength in zircon x frequency /(wavelength in air x frequency) speed in zircon/speed in air = wavelength in zircon /wavelength in air 1.56e8m/s /3.00e8m/s = wavelength in zircon /632.8nm 632.8nm x 1.56e8m/s /3.00e8m/s = wavelength in zircon wavelength in zircon = 329nm

Determine the speed, wavelength, and frequency of light from a helium-neon laser as it travels through zircon. The wavelength of the light from the laser is 632.8 nm in air and the index of refraction of zircon is 1.923. speed = 1.56e8m/s wavelength = ?nm

wavelength = 632.8nm in air n zircon = 1.923 c = 3.00e8m/s nv = c 1.923 x v = 3.00e8m/s v = 3.00e8m/s /1.923 v = 1.56e8m/s

Determine the speed, wavelength, and frequency of light from a helium-neon laser as it travels through zircon. The wavelength of the light from the laser is 632.8 nm in air and the index of refraction of zircon is 1.923. speed = ?m/s

di = 20.0cm do = 38.0cm 1/do + 1/di = 1/f solved for f f = di x do /(di + do) f = 20cm x 38cm /(20+38)cm f = 13.1cm

If a converging lens forms a real, inverted image 20.0 cm to the right of the lens when the object is placed 38.0 cm to the left of a lens, determine the focal length of the lens. cm

f = 150mm = 150e-3m do = 18.0km = 18.0e3m di = 0.150m height image = ?cm height cliff = 1200m m = -di/do m = -0.150m/18.0e3m m = -8.33e-6 height object x magnification = height magnified object 1200m x -8.33e-6 = height magnified object height magnified object = 0.0100m height image = -1.00cm

Suppose a 150 mm focal length telephoto lens is being used to photograph mountains 18.0 km away. (a) Where (in m) is the image? m (b) What is the height (in cm) of the image of a 1200 m high cliff on one of the mountains? (Include the sign of the value in your answer.) cm

f = 150mm = 150e-3m do = 18.0km = 18.0e3m di = ?m 1/f = 1/do - 1/di 1/f - 1/do = 1/di 1/(150e-3m) - 1/(180e3m) = 1/di 1/di = 6.667/m di = 0.150m

Suppose a 150 mm focal length telephoto lens is being used to photograph mountains 18.0 km away. (a) Where (in m) is the image? m (b) What is the height (in cm) of the image of a 1200 m high cliff on one of the mountains? (Include the sign of the value in your answer.) cm

n1 air = 1.000 n2 crown glass = 1.52 Δx = ? theta incident = 39.5° thickness of n2 = 2.40cm = 24.0mm n1 x sintheta1 = n2 x sintheta2 1 x sin39.5° = 1.52 x sintheta2 asin(sin39.5°/1.52) = theta2 theta2 = 24.74° solving for undisplaced x theta1 = 39.5° adj = 24.0mm tantheta = opp/adj tan39.5° = opp/24.0mm tan39.5° x 24.0mm = opp opp for undisplaced ray = 19.78mm solving for displaced x theta2 = 24.74° adj = 24.0mm tantheta = opp/adj tan24.74° = opp/24.0mm tan24.74° x 24.0mm = opp opp for displaced ray = 11.06mm displacement distance = 19.78-11.06 = 8.72mm

Suppose the figure below represents a ray of light going from air through crown glass into water, such as going into a fish tank. Calculate the amount the ray is displaced by the glass (Δx in mm), given that the incident angle is 39.5° and the glass is 2.40 cm thick. mm

µ(g) sin i = µ(a) sin r At critical angle (i) , angle of refraction (r) = 90 µ(air) = 1 and i = 45˚ sin i = 1/µ µ = 1/sin 45 = 1.41

The figure below shows a 45° - 90° - 45° prism with a ray of light entering and exiting on the long side of the prism after undergoing two total internal reflections. This arrangement is an optically useful method of reversing the direction of travel of the light. Determine the minimum index of refraction of the prism in order for this to occur.

Let v1 be the speed of sound in the surrounding medium: then the speed of the sound in the liver is v2 = 1 - 0.085 = 0.915v1. Since the refraction index of a wave is inversely proportional to the speed of the wave, if n1 is the refraction index in the surrounding medium the refraction index in the liver is n2 = n1/0.915. Let θ be the refraction angle. According to Snell's law: sin θ / sin(55°) = n1/n2 = 0.915 sin θ = 0.915 x sin(55°) = 0.915 * 0.8192 = 0.750 θ = arcsin(0.750) = 48.55° The tumor depth is therefore d = 7.00 / tan(48.55°) = 7.00 / 1.132 = 6.18 cm

Ultrasonic diagnostics may be used to locate a tumor in an organ such as the liver. The figure shows a beam of ultrasonic waves that enter the surrounding medium, refract into the liver, reflect off the tumor, and refract back into the surrounding medium. If the speed of the wave as it travels through the liver is 8.8% less than when it is traveling through the surrounding medium, determine the depth of the tumor. cm

speed of light = c = 3.00e8m/s n crystalline quartz = 1.544 nv = c v = c/n v = 3.00e8m/s /1.544 v = 1.94e8m/s

What is the speed of light (in m/s) in water? 2.26e8m/s What is the speed of light (in m/s) in crystalline quartz? m/s

speed of light = c = 3.00e8m/s n fresh water at 20 degrees C = 1.33 nv = c v = c/n 3.00e8m/s /1.33 = 2.26e8m/s

What is the speed of light (in m/s) in water? m/s

f = 47.8cm m = -2.25 From the lens equations, we have the following relationships 1/do + 1/di = 1/f and m = − di/do, where do and di are respectively the object and image distance, f is the focal length of the lens, and m is the magnification for this particular object distance. Solve the second expression for the object distance, substitute the result into the first expression and solve for the image distance to obtain di = (1 − m)f. di = (1 + 2.25) x 47.8cm di = 155cm = 1.55m

You have a converging (convex) lens and a diverging (concave) lens and the magnitude of the focal length of each lens is 47.8 cm. (a) Determine which lens you should use in order to produce an image of a light bulb on a screen several meters away. converging diverging (b) Determine how far from the lens you should locate the screen in order to produce an image of the light bulb that is magnified by a factor of 2.25. m

(a) In order to project an image on the screen, the image must be real. Only converging (convex) lenses produce real images. As a result, you must use the converging lens.(b) converging

You have a converging (convex) lens and a diverging (concave) lens and the magnitude of the focal length of each lens is 47.8 cm. (a) Determine which lens you should use in order to produce an image of a light bulb on a screen several meters away. converging diverging (b) Determine how far from the lens you should locate the screen in order to produce an image of the light bulb that is magnified by a factor of 2.25. m

do = 15.4cm f = -12.0cm m1 = f/(f-do1) m2 = f/(f-do2) m2 = m1/2.90 m2 = f/(f-do2) m1/2.90 = f/(f-do2) (f-do2) = 2.90f /m1 -do2 = 2.90f /m1 - f??????????? (not sure how to get here from there) do2 = 2.90do1 - 1.90f do2 = (2.90 x 15.4cm) + ( 1.90 x 12.0cm)

You place an object 15.4 cm in front of a diverging lens which has a focal length with a magnitude of 12.0 cm. Determine how far in front of the lens the object should be placed in order to produce an image that is reduced by a factor of 2.90. cm