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麻烦给解下物理题.Asmallbubbleisformedatthecentreofaglasscubeofside4cmandrefractiveindex1.5.Identicalcircularopaquediscsaregluedtothesixcubefacessothatthebubblecannotbeovservedfromtheoutsideofthecube.W
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麻烦给解下物理题.
A small bubble is formed at the centre of a glass cube of side 4 cm and refractive index 1.5.Identical circular opaque discs are glued to the six cube faces so that the bubble cannot be ovserved from the outside of the cube.What is the minimum radius of the discs?
A.2.00cm B.1.89cm C.1.79cm D.1.67cm
A small bubble is formed at the centre of a glass cube of side 4 cm and refractive index 1.5.Identical circular opaque discs are glued to the six cube faces so that the bubble cannot be ovserved from the outside of the cube.What is the minimum radius of the discs?
A.2.00cm B.1.89cm C.1.79cm D.1.67cm
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答案和解析
the answer is C.
Actually the topic is very simple,you see ,which saw bubbles,glass cube center,by optical path can instead,but think eye is reversible view,and the largest incident angle is refracted 90 degrees,41.81 degree angle is out,by the laws of refraction can (and symbols to save ) again just across the center of the light refraction for critical incident point at this time round,41.81 radius of tangent length value multiplied by cube half(as 2),then 1.7888 is the answer.Maybe you could draw a picture.
答案是 C.
其实题目很简单,看到小气泡,即看到玻璃立方体中心,由光路可逆,可认为眼睛为光源,最大视线入射角是90度,折射后角度为41.81度,(由折射定律可得,符号难打,我省了)再折射光线刚好穿过中心的入射点为临界情况,此时圆半径为41.81的正切值乘以立方体棱长一半(即2),答案为1.7888.画图就出来了.
Actually the topic is very simple,you see ,which saw bubbles,glass cube center,by optical path can instead,but think eye is reversible view,and the largest incident angle is refracted 90 degrees,41.81 degree angle is out,by the laws of refraction can (and symbols to save ) again just across the center of the light refraction for critical incident point at this time round,41.81 radius of tangent length value multiplied by cube half(as 2),then 1.7888 is the answer.Maybe you could draw a picture.
答案是 C.
其实题目很简单,看到小气泡,即看到玻璃立方体中心,由光路可逆,可认为眼睛为光源,最大视线入射角是90度,折射后角度为41.81度,(由折射定律可得,符号难打,我省了)再折射光线刚好穿过中心的入射点为临界情况,此时圆半径为41.81的正切值乘以立方体棱长一半(即2),答案为1.7888.画图就出来了.
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