The double concave lens in the figure above is fabricated of glass having index

Question

The double concave lens in the figure above is fabricated of glass having index of refraction n -1.68. The magnitudes of the radii of curvature of the left and right surfaces of the lens are R 19 cm and |RrI-14 cm, respectively. The focal length of this lens can be found using the lens maker's formula: 1/f- (n 1)(1/R- 1/Rr). The following sign convention has to be followed: the surface with the center of curvature located to the left of the lens will have a negative radius, while the surface with the center of curvature located to the right of the lens will have a positive radius 1) Calculate the power P of the lens in I"diopters diopters Submit Help 2) An upright object of height H 0.6 cm is placed at a distance of 20 cm to the left of the lens. Calculate the image distance. cm Submit Help 3) Calculate the image height. cm Submit Help

Explanation / Answer

a)1/f = (n-1)(1/R + 1/R)

1/f = (1.68-1)[1/19 + 1/14] = 0.68(0.05263 + 0.07143) = 0.68*0.12406 = 0.08436

f = 1/0.08188 = 11.862 cm
diopter = 1/f = 0.08436

b) b)
1/f = 1/o + 1/i
0.08436 = 1/20 + 1/i,....=> i=1/(0.08436 - 0.05)s=1/0.03436=29.412cm

c) M = i/o = Hi/Ho = 29.41/20= 1.47, => Hi = 1.47*0.6 = 0.88cm

d)By using P = 1/s + 1/s2'
s2' = 8.466cm

there may be calculation error but concept is right

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