Question

water interface (the surface of the lake), it is partly reflected back into the air and partly refracted or transmitted into the water. This explains why on the surface of a lake sometimes you see the reflection of the surrounding landscape and other times the underwater vegetation. These effects on light propagation occur because light travels at different speeds depending on the medium. The index of refraction of a material, denoted by n, gives an indication of the speed of light in the material. It is defined as the ratio of the speed of light c in vacuum to the speed v in the material, or n=cv. Figure 1 of 1The figure shows an incident ray falling to a horizontal interface. The ray makes an angle of theta subscript 1 with the normal to the interface. There are also reflected and refracted rays. The refracted ray continues below the interface and makes an angle theta subscript 2 with the normal. The reflected ray continues above the interface, on the other side of the normal from the incident ray.

Part A Part complete When light propagates from a material with a given index of refraction into a material with a smaller index of refraction, the speed of the light View Available Hint(s) increases. SubmitPrevious Answers Correct Part B Part complete What is the minimum value that the index of refraction can have? View Available Hint(s) What is the minimum value that the index of refraction can have? 0 +1 −1 between 0 and 1 SubmitPrevious Answers Correct The index of refraction of a material is always a positive number greater than 1 that tells us how fast the light travels in the material. The greater the index of refraction of a material, the more slowly light travels in the material. An example of reflection and refraction of light is shown in the figure. (Figure 1)An incident ray of light traveling in the upper material strikes the interface with the lower material. The reflected ray travels back in the upper material, while the refracted ray passes into the lower material. Experimental studies have shown that the incident, reflected, and refracted rays and the normal to the interface all lie in the same plane. Moreover, the angle that the reflected ray makes with the normal to the interface, called the angle of reflection, is always equal to the angle of incidence. (Both of these angles are measured between the light ray and the normal to the interface separating the two materials.) This is known as the law of reflection. The direction of propagation of the refracted ray, instead, is given by the angle that the refracted ray makes with the normal to the interface, which is called the angle of refraction. The angle of refraction depends on the angle of incidence and the indices of refraction of the two materials. In particular, if we let n1 be the index of refraction of the upper material and n2 the index of refraction of the lower material, then the angle of incidence, θ1, and the angle of refraction, θ2, satisfy the relation n1sinθ1=n2sinθ2. This is the law of refraction, also known as Snell's law. Part C Part complete Now consider a ray of light that propagates from water (n=1.33) to air (n=1). If the incident ray strikes the water-air interface at an angle θ1≠0, which of the following relations regarding the angle of refraction, θ2, is correct? View Available Hint(s) Now consider a ray of light that propagates from water () to air (). If the incident ray strikes the water-air interface at an angle , which of the following relations regarding the angle of refraction, , is correct? θ2>θ1 θ2<θ1 θ2=θ1 SubmitPrevious Answers Correct When light propagates from a certain material to another one that has a smaller index of refraction, that is, n1>n2, the speed of propagation of the light rays increases and the angle of refraction is always greater than the angle of incidence. This means that the rays are always bent away from the normal to the interface separating the two media. Part D Part complete Consider a ray of light that propagates from water (n=1.33) to glass (n=1.52). If the incident ray strikes the water-glass interface at an angle θ1≠0, which of the following relations regarding the angle of refraction θ2 is correct? View Available Hint(s) Consider a ray of light that propagates from water () to glass (). If the incident ray strikes the water-glass interface at an angle , which of the following relations regarding the angle of refraction  is correct? θ2>θ1 θ2<θ1 θ2=θ1 SubmitPrevious Answers Correct When light propagates from a certain material to another one that has a greater index of refraction, that is, n1<n2, the speed of propagation of the light rays decreases and the angle of refraction is always smaller than the angle of incidence. This means that the rays are always bent toward the normal to the interface separating the two media. Part E Part complete Consider a ray of light that propagates from air (n=1) to any one of the materials listed below. Assuming that the ray strikes the interface with any of the listed materials always at the same angle θ1, in which material will the direction of propagation of the ray change the most due to refraction? View Available Hint(s) Consider a ray of light that propagates from air () to any one of the materials listed below. Assuming that the ray strikes the interface with any of the listed materials always at the same angle , in which material will the direction of propagation of the ray change the most due to refraction? ice (n=1.309) water (n=1.333) turpentine (n=1.472) glass (n=1.523) diamond (n=2.417) SubmitPrevious Answers Correct The greater the change in index of refraction, the greater the change in the direction of propagation of light. To avoid or minimize undesired bending of the light rays, light should travel through materials with matching indices of refraction. Is light always both reflected and refracted at the interface separating two different materials? To answer this question, let's consider the case of light propagating from a certain material to another material with a smaller index of refraction (i.e., n1>n2). Part F Part complete In the case of n1>n2, if the incidence angle is increased, the angle of refraction View Available Hint(s) increases up to a maximum value of 90 degrees. SubmitPrevious Answers Correct Since the light is propagating into a material with a smaller index of refraction, the angle of refraction, θ2, is always greater than the angle of incidence, θ1. Therefore, as θ1 is increased, at some point θ2 will reach its maximum value of 90∘ and the refracted ray will travel along the interface. The angle of incidence for which θ2=90∘ is called the critical angle θcrit. For any angle of incidence greater than θcrit, no refraction occurs. The ray no longer passes into the second material. Instead, it is completely reflected back into the original material. This phenomenon is called total internal reflection and occurs only when light encounters an interface with a second material with a smaller index of refraction than the original material. Part G What is the critical angle θcrit for light propagating from a material with index of refraction of 1.50 to a material with index of refraction of 1.00? Express your answer in radians. View Available Hint(s) θcrit θ c r i t theta_crit = nothing   radians   Submit Provide Feedback Next

Correct. Followup. Since the light is propagating into a material with a smaller index of refraction, the angle of refraction, , is always greater than the angle of incidence, . Therefore, as is increased, at some point will reach its maximum value of 90 and the refracted ray will travel along the interface. The angle of incidence for which is called the critical angle . For any angle of incidence greater than , no refraction occurs. The ray no longer passes into the second material. Instead, it is completely reflected back into the original material. This phenomenon is called total internal reflection and occurs only when light encounters an interface with a second material with a smaller index of refraction than the original material. End of followup.

Answer 1

Part speed of light in a given medium is given by = e 4 U2 = 1 » Unia W₂02 o - af name then 10, >l₂ thus speed of light linereases ugut propagatello a smaller index of Part(B) Чтоцhon Minimum value 9 Index q verachon is Because R-1-(2) >1 always. Porto - Using > Wing Shell's bio 1. Masing-masing2 7 1:23 sing = ixsing2 dur Later 1 » 002 mwind - masing2 133.9941252512 M70204302 = 102201) Dart E) since, mar sina= n sing = |x C) = n ºn P- so the angle or 0 wil greatest for material having poi in greatest . Therefore correct option is Diamond ( 2-417) Part(F) Il non as the angle of inudence increase the angle or increases upto maximum angle of gou degrees Part (1) for oi Ocnita Oz 90° 1. AD XSI Berit - XSingoº si devets Sinil HBO) = yupo Onet - 0.7297 radian.