The radii of curvature of the surfaces of a thin converging meniscus lens are R1

Question

The radii of curvature of the surfaces of a thin converging meniscus lens are R1=+12cm and R2=+24cm. The index of refraction is 1.60.
a) Compute the position of the image of an object in the form of an arrow 5.00 mm tall, perpendicular to the lens axis, 45.0 cm to the left of the lens.
s'= ?cm

b)Compute the size of the image. y=?mm

c)A second converging lens with the same focal length is placed 3.15 to the right of the first. Find the position of the final image.
s'=? cm to the right from the right lens

d)Find the size of the final image. y=? mm

e)Is the original object erect or inverted?

Explanation / Answer

Part A)

To find the focal length, use the Lens Makers Equation

1/f = (n-1)(1/R1 - 1/R2)

(1/f ) = (1.6 - 1)(1/12 - 1/24)

f = 40 cm

Then use the lens equation

1/f = 1/s + 1/s'

1/40 = 1/45 = 1/s'

s' = 360 cm

Part B)

h'/h = -s'/s

h'/5 = -(360)/45

h' = -32 mm    (The negative can be ignored, its sign convention indicating inverted - use 32 mm for height if you just need the magnitude)

Part C)

From the first part, the image is 360 cm to the right of the firsat lens. This image becomes a virtual object in a second lens 360 - 3.15 cm away = 356.85 (and it will be negative due to virtual object)

1/f = 1/s + 1/s'

1/40 = 1/-356.85 + 1/s'

s' = 35.97 cm to the right of the second lens

Part D)

h'/h = -s'/s

h'/-32 = -(35.97)/-(365.85)

h' = -3.15 mm (again the negative just indicates inverted and can be ignored for just magnitude)

Part E)

This part makes no sense. The original object is always erect. You can not have an inverted object

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