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= 12.0 cm and R2 = 28.0 cm . The index of refraction of the lens material is 1.60.

A) Compute the position and size of the image of an object in the form of an arrow 5.00 mm tall, perpendicular to the lens axis and 45.0 cm to the left of the lens.

B) A second converging lens with the same focal length is placed 3.15 m to the right of the first. Find the position and size of the final image.

C) Is the final image erect or inverted with respect to the original object?

Explanation / Answer

lens makers equation
R1= +ve........>>>R2 = +Ve

1/F = (n-1)*((1/R1)-(1/R2)

1/F = (1.6-1)*((1/12) - (1/28))

F = 35 cm
-------------------------------------
A) S1 = 45 cm

1/S1 + 1/S1' = 1/F

1/45 + 1/S1' = 1/35

S1' = 157.5 cm .....position

hi /ho = -S1' / S1

h i = -17.5 mm.........inverted
----------------------------------------------------

B) the image of first lens will act as object

S2 = 315 - 157.5 = 157.5 cm

1/S2 + 1/S2' = 1/F2

1/157.5 + 1/S2 = 1/35

S2' = 45 cm

--------------------------------------------------------------------------
C) total magnification M = (-S2'/S2)*(S1'/S1) = +1

M = +1 the final image is erect ........

Related Questions

Navigate

• Browse (All)
• Browse T
• Subjects