The figure (Figure 1) shows a small plant near a thin lens. The ray shown is one

Question

The figure (Figure 1) shows a small plant near a thin lens. The ray shown is one of the principal rays for the lens. Each square is 1.8 cm along the horizontal direction, but the vertical direction is not to the same scale. Use information from the diagram to answer the following questions:

Part A

What is the focal length of the lens? Express your answer using two significant figures.

Part C

Calculate where the image should be. Express your answer using two significant figures.

Explanation / Answer

A) Since the refracted ray is parallel to the optical axis, the incident ray comes from the focus direction, so the focus is where the incident ray prolongation touches the optical axis. This poit is 9 squares from the center of the lens, so focal length f is 9x1.80 = 16.20 cm

C) The plant is placed 5 squares from the center, so the object distance is
o = 5x1.80 = 9 cm
According to the thin lenses equation
1/f = 1/o + 1/i
whence
i = of/(o - f) = 9 x 16.20 / ( 9 - 16.20) = -20.25 cm
It is a virtual image (i < 0), and is placed therefore at the left of the lens at a distance of 20.25/1.80 = 11.25 squares from the center

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