The figure ( Figure 1 ) shows a small plant near a thin lens. The ray shown is o
ID: 2264417 • Letter: T
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 2.4cm along the horizontal direction, but the vertical direction is not to the same scale. Use information from the diagram to answer the following questions:
A)Using only the ray shown, decide what type of lens this is.?
b)What is the focal length of the lens?
C)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 2.4cm along the horizontal direction, but the vertical direction is not to the same scale. Use information from the diagram to answer the following questions: Using only the ray shown, decide what type of lens this is.? What is the focal length of the lens? Calculate where the image should be.?Explanation / Answer
type of lens this is converging
First of all the lens-maker's equation is correctly written as:1/f = 1/u + 1/v,
And the appropriate sign convention is worked out for each specific case in question.
cases:
convex or concave lens, image distance is negative only for virtual images
(projected images) that is images formed on the same side of the lens.
plane or spherical mirror, image distance: v is negative, since image is virtual
Now, 1/u + 1/v => combine fractions by using the common denominator: uv
1/u + 1/v = (v + u)/uv = 1/f
f = uv/(u+v)
the image is at where the ray is meeting the principle axis
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