The magnitudes of the radii of curvature are 35 cm and 45 cm for the two faces o
ID: 1401817 • Letter: T
Question
The magnitudes of the radii of curvature are 35 cm and 45 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.
(a) For a very distant object, locate the image formed by violet light. (Please answer to at least one decimal place. Give the image distance.)
______________ cm
(b) For a very distant object, locate the image formed by red light. (Please answer to at least one decimal place. Give the image distance.)
_________________ cm
Explanation / Answer
a)
by using the formula
1 / f = (n - 1 ) ( 1 / R1 - 1 / R2 )
1 / f = ( 1.53 - 1 ) * ( 1 / -35 - 1 / 45)
f = -37.14 cm
R1 is negative because the center of curvature of the first surface is on the virtual image side
the image formed on the focal length 37.14 cm
b)
use the same formula
1 / f = ( n -1) * ( 1 / R1 - 1 / R2)
1 / f = (1.51 - 1 ) * ( 1 / -35 - 1 / 45)
f = -38.6 cm
image formed by the red light on 38.6 cm
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