# Focal Glass

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/**A light beam is directed from the air into a glass prism in the form of an equilateral triangle.?**

*The refractive index of glass is 1.47.* I am having trouble answering this part of the problem: Calculate the angle from a line normal to the surface of the right side of the prisim in which the output beam prism. Also, if the lens is placed under water, the focal length of the lens, decrease or remain the same as in the air? Explain

Ok, so I've left out some information, but we can work through it anyway. First draw the prism (triangle) and show the beam bending, as enters and exits the prism. Lets call the angle of incidence for the input beam Ai (which is the angle between the ray and the normal to the face of the prism through which the input beam), and call the angle of refraction of the beam entering Ar. Similarly for the beam as it exits, we can call the incident angle (inside the prism) and Bi Br refractive angle can be seen in the geometry of the problem that there is a triangle between the tip of the prism and point of entry and exit for the beam of light … Thus Therefore, we have the relation between Ar and Bi angles of a triangle sum to 180 degrees (90 – Ar) + 60 + (90 – Bi) = 180 Ar + Bi = 60. Now we call the index of refraction of glass. Snell's Law we have that: Ai sin = n * sin and sin Ar Br = n * sin know that Bi Bi = 60 – Ar, by taking what we find Bi: Bi = 60 – Ar = 60 – arcsin (sin (Ai) / n) and Br (the angle at which the output beam) Br = arcsin (n * sin Bi) —— ——– ———– The complete equation for the problem above is: Br = sin (n2/n1) * sin (60 – arcsin ((N1/N2) * sin Ai)) … where N1 = 1 is the index of refraction of air and N2 = 1.47 is the index of refraction of glass. If the prism immersed in water and then '(N1/N2) "would increase since n1 becomes the index of refraction of water, which is greater than that of air. we can see from the equation, as "(N1/N2) 'increases, so does' arcsin ((N1/N2) * Sinai) 'also' (n2/n1) 'decreases so that "sin (Br)' and thus' Br decline. The result is that the angle at which the beam leaves the prism is less if the prism is immersed in water. In a lens then we see that the focal length is increased by placing it under water. The larger the difference in refractive index between the lens and the substance is in the lens, the more powerful the (the shorter the focal length).

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/**Powers Art Glass Lampwork Focal Bead**

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