A transparent rod 45.0 cm long and a refractive index of 1.68 is cut flat at the

Question

A transparent rod 45.0 cm long and a refractive index of 1.68 is cut flat at the right end and rounded to a hemispherical surface with a 16.0 cm radius at the left end. An object is placed on the axis of the rod 12.9 cm to the left of the vertex of the hemispherical end. A) What is the position of the final image, relative to the far end of the rod? (s' is negative if the final image is inside the rod.) B) What is its magnification?

Explanation / Answer

na/do + nb/di = (nb - na)/R Here na = 1.00 air and nb = 1.68 and R = 16 So 1/12.9 + 1.68/di = (1.68 - 1.0)/16 => 1.68/di = 0.68/16 - 1/12.9 = -0.035 So di = 1.68/(-0.035) = -48cm which is 48cm to the left of the vertex b) m = -1*di/nb*do = -1.0*(-48)/(1.68*12.9) = 2.21

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