The magnitudes of the radii of curvature are 33.1 cm and 43.3 cm for the two fac
ID: 2244028 • Letter: T
Question
The magnitudes of the radii of curvature are 33.1 cm and 43.3 cm for the two faces of a biconcave lens. The glass has index of refraction 1.53 for violet light and 1.51 for red light.
The magnitudes of the radii of curvature are 33.1 cm and 43.3 cm for the two faces of a biconcave lens. The glass has index of refraction 1.53 for violet light and 1.51 for red light. For a very distant object, locate the image formed by violet light. (Give your answer to two decimal places. Indicate the position of the image relative to the lens with the sign of your answer.) Your response is within 10% of the correct value. This may be due to roundoff error, or you could have a mistake in your calculation. Carry out all intermediate results to at least four-digit accuracy to minimize roundoff error. For a very distant object, locate the image formed by red light. (Give your answer to two decimal places.Indicate the position of the image relative to the lens with the sign of your answer.)Explanation / Answer
find focal length first using: 1/f = (n-1)(1/R1-1/R2) so plug and chug
a)
[(1.53-1)(1/-33.1 - 1/43.3)]^-1 (R1 is negative because of the convex surface) and you get -35.39cm (to two decimal places) when p = (undefined) q=f so then your answer to a is going to be -35.39cm.
b)using the same formula but replacing 1.53 with 1.51 you get
q = -36.78cm
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