Question

o Consider a thin double concave lens, with each face having a radius of curvature of 35 cm. The glass composing the lens has an index of refraction of 1.4 1.What is the focal length of this lens? 2.If an 1 cm object is placed at 1.5x the focal length, where is the image and how big is it, according to the thin lens equation? 3.Is the image real or virtual? Right side up? Hint: when sketching your ray diagram, for each ray, think carefully about the focal point location)

Answer 1

(a)

suppose the lens is in air

then

1/f = ( n1/n2 - 1) ( 1/r1 - 1/r2)

1/f = (1.4 - 1) ( 1/35 - (-1/35))

f = 43.75 cm

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(b)

do = 1.5 * f = 65.625 cm

then

1/di = 1/f - 1/do

1/di = 1/43.75 - 1/65.625

di = 131.25 cm

and

magnification, m = - 2

sl,

image size = 2 cm

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As m is negative, the ray is real, inverted and larger than the object.