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Figure P23.28 shows a curved surface separating a material with index of refraction n1 from a material with index n2. The surface forms an image I of object O. The ray shown in red passes through the surface along a radial line. Its angles of incidence and refraction are both zero, so its direction does not change at the surface. For the ray shown in blue, the direction changes according to n1 sin θ1 = n2 sin θ2. For paraxial rays, we assume θ1 and θ2 are small, so we may write n1 tan θ1 – n2 tan θ2. The magnification is defined as M = h′/h. Prove that the magnification is given by M = –n1q/n2p.
Figure P23.28
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Chapter 23 Solutions
Bundle: College Physics, Loose-Leaf Version, 10th, + WebAssign Printed Access Card for Serway/Vuille's College Physics, 10th Edition, Multi-Term
- A Lucite slab (n = 1.485) 5.00 cm in thickness forms the bottom of an ornamental fish pond that is 40.0 cm deep. If the pond is completely filled with water, what is the apparent thickness of the Lucite plate when viewed from directly above the pond?arrow_forwardThe drawing shows a rectangular block of glass (n = 1.52) surrounded by a liquid with n = 1.69. A ray of light is incident on the glass at point A with a 30.0° angle of incidence. At what angle does the ray leave the glass at point B? A 30.0arrow_forwardA block of crown glass is immersed in water as in the figure below. A light ray is incident on the top face at an angle of θ1= 41° with the normal and exits the block at point P. Find the angle of refraction θ2 of the light ray leaving the block at P. 80.2° 41° 43.3° 68.9°arrow_forward
- You send a beam of light from a material with index of refraction 1.19 into an unknown material. In order to help identify this material, you determine its index of refraction by measuring the angles of incidence and refraction for which you find the values 40.7° and 36.5°, respectively. What is the index of refraction n of the unknown material? n =arrow_forwardIn the figure below light begins in material 1 with index of refraction ng = 1.21 and makes an angle with the normal (dotted) line of 01= 34° as it strikes material 2. The light then refracts into material 2 with index of refraction n2 = 1.78 and unknown angle of refraction 02. Lastly, the light enters material 3 with unknown index of refraction ng and makes an angle 03= 26° with the normal line. What is the index of refraction of the bottom material n3? NOTE: Please enter your answer with three significant figures. Do not use scientific notation. Do not enter units (your answer is unitless).arrow_forwardChoose the correct statement regarding light traveling in air and glass mediums. Assume that the angle of incidence is not perpendicular to the surface. Refractive index of air is nair=1.00029; refractive index of glass is nglass=1.517. For light traveling from glass to air, the ray becomes bent toward the normal. O Light travels at a slower speed in air than in glass. For light traveling from air to glass, the ray becomes bent away from the normal. O For light traveling from air to glass, the incidence angle is larger than the refraction angle. O For light traveling from glass to air, the refraction angle is smaller than the incidence angle. Submit Answer Tries 0/2 Post Discussion Send Feedbaclarrow_forward
- Figure P23.28 shows a curved surface separating a material with index of refraction n1 from a material with index n2 . The surface forms an image I of object o. The ray shown in red passes through the surface along a radial line. Its angles of incidence and refraction are both zero, so its direction does not change at the surface. For the ray shown in blue, the direction changes according to n1 sin θ1 = n2 sin02 . For paraxial rays, we assume θ1 and θ2 are small, so we may write n1 tan θ1 = n2 tan θ2. The magnification is defined as M =h′/h. Prove that the magnification is given by M = −n1 q/n2p.arrow_forwardYou have a slab of diamond that is attached to a slab of sapphire. A laser beam starts off in the diamond (index of refraction =2.42) and then exits into sapphire (index of refraction =1.77). The beam makes an angle of 30 degrees with the normal in the diamond. What is the maximum angle of incidence that the laser will refract into the sapphire?arrow_forwardChoose the correct statement regarding light traveling in air and glass mediums. Assume that the angle of incidence is not perpendicular to the surface. Refractive index of air is nair=1.00029; refractive index of glass is nglass=1.517. For light traveling from glass to air, the ray becomes bent toward the normal. For light traveling from air to glass, the ray becomes bent away from the normal. For light traveling from air to glass, the refraction angle is smaller than the incidence angle. For light traveling from air to glass, the refraction angle is larger than the incidence angle. For light traveling from air to glass, the incidence angle is smaller than the refraction angle.arrow_forward
- A horizontal light ray reflects upward from a plane mirror that makes an acute angle φ with the horizontal. The reflected ray makes an angle α = 76° with the incident ray. Refer to the figure. Find the angle θ, in degrees. Find the angle φ, in degrees.arrow_forwardA fish that is d=d= 2.9 m below the surface looks up and sees a child fishing from the shore. What angle of incidence (θ1) does the ray from the person’s face make with the perpendicular to the water at the point where the ray enters? The angle of refraction (θ2) between the ray in the water and the perpendicular to the water is 35.8°. θ1 = 51.1 What is the height of the person’s head above the water? Assume the person is standing L= 7.4 m away from the point where the incident ray intersects the water. h = marrow_forwardThe drawing shows a rectangular block of glass (n = 1.52) surrounded by a liquid with n = 1.56. A ray of light is incident on the glass at point A with a 30.0° angle of incidence. At what angle does the ray leave the glass at point B? 30.0° Number i A B eTextbook and Media Unitsarrow_forward
- Physics for Scientists and Engineers: Foundations...PhysicsISBN:9781133939146Author:Katz, Debora M.Publisher:Cengage Learning
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