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=+12.0cm and R2=+28.0cm. The index of refraction is 1.60.

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.

Compute the size of the image.

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

Find the size of the final image.

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

Explanation / Answer

1. we know that 1/f = (n-1){1/R1 - 1/R2} = 0.60{1/12 - 1/28}

f = 35.0 cm

we also know that

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

1/45 + 1/s' = 1/35

s' = 158 cm......Ans.

h' = -h(s'/s) =-0.50 cm)*(158/45) =-1.76 cm.........Ans.

2. Adding a second identical lens 315 cm to the right of the first meansthat the first lens’s image becomes an object for the second, a distance of 157cm from that second lens.

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

1/157 + 1/s' = 1/35

s' = 45 cm

h' =-1.76*(-45/157) = 0.5 cm

and the image is erect.

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