The double convex lens in the figure above is fabricated of glass with index of

Question

The double convex lens in the figure above is fabricated of glass with index of refraction n - 1.71. The magnitudes of the radii of curvature of the left and right surfaces of the lens are Rl-20 cm and RrI 15 cm, respectively. The focal length of this lens can be found using the lens maker's formula: 1 /f= (n-1)(1/R-IR). 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 ldiopters diopters SubHelp 2) An upright object of height H-0.5 cm is placed at a distance of 19 cm to the left of the lens. Calculate the image distance. (Give a numerical value and proper algebraic sign.) cm Submit Help 3) Calculate the magnitude of the image height. cm Submit Help

Explanation / Answer

1)

1/f = (n -1) (1/Rl - 1/Rr)

1/f = (1.71 - 1) (1/(20) - 1/(-15))

P = 100/f = 100 (1.71 - 1) (1/(20) - 1/(-15))

P = 8.28 dioptre

2)

f = focal length = 12.1 cm

do = object distance = 19 cm

di = image distance

Using the lens equation

1/di + 1/do = 1/f

1/di + 1/19 = 1/12.1

di = 33.32 cm

3)

using the magnification equation

hi/ho = - di/do

hi / 0.5 = - (33.32 / 19)

hi = - 0.88 cm

magnitude of image height = 0.88 cm

6)

f = focal length = 12.1 cm

do = object distance = 3 cm

di = image distance

Using the lens equation

1/di + 1/do = 1/f

1/di + 1/3 = 1/12.1

di = - 3.98 cm

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