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 1.6cm along the horizontal direction, but the vertical direction is not to the same scale. Use information from the diagram to answer the following questions.

What is the focal length of the lens?

Calculate where the image should be.

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 1.6cm along the horizontal direction, but the vertical direction is not to the same scale. Use information from the diagram to answer the following questions. What is the focal length of the lens? Calculate where the image should be.

Explanation / Answer

focal length of the lense = -5*1.6

= -8 cm (The lense is a divering lense) <<<<<<<<<<<----------Answer

Form the figure,
object distance, u = 8*1.6

= 12.8 cm

let v is the image distance

now use the equation

1/u + 1/v = 1/f

1/v = 1/f - 1/u

1/v = 1/(-8) - 1/12.8

v = -4.923 cm <<<<<<<<<<<----------Answer

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