Question

5. For all setups, use a focal length of 2cm and place a real object to the left of the lens. For each case a-f:

i. Draw the ray diagram using principal rays.

ii. Determine whether the image is real or virtual, and where your eye would need to be to see the image.

iii. Estimate the image magnification from your drawing.

iv. Use the thin lens equation to check that the image location corresponds to your drawing.

v. Calculate the magnification and check it against your estimate.

a. converging lens with object 6cm from lens (i.e. farther from the lens than the focal point) b. converging lens with object 1cm from lens (i.e. closer to the lens than the focal point) c. converging lens with object at the focal point d. diverging lens with object 6cm from lens (i.e. farther from the lens than the focal point) e. diverging lens with object 1cm from lens (i.e. closer to the lens than the focal point) f. diverging lens with the object at the focal point

Answer 1

(a)

image is real

image magnification from drawing = 0.5

(1/v) - (1/u) = 1/f   ; 1/v = (1/2) -(1/6) = 1/3 or v = 3 cm

v is lens-to-image distance, u is lens-to-object distance, f is focal length

magnification v/u = 3 /( -6) = -1/2

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(b)

image is virtual , eye is at right side of lens so that light rays are approaching eye to view image

image magnification from drawing = 2

(1/v) - (1/u) = 1/f   ; 1/v = (1/2) -(1) = -1/2 or v = -2 cm

magnification v/u = -2 /( -1) = 2

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image is formed at infinity

(1/v) - (1/u) = 1/f   ; 1/v = (1/2) -(1/2) = 0 or v = infinity

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image is virtual , eye is at right side of lens so that light rays are approaching eye to view image

image magnification from drawing = 0.25

(1/v) - (1/u) = 1/f   ; 1/v = (-1/2) -(1/6) = -2/3 or v = -1.5 cm

magnification v/u = -1.5 /( -6) = 0.25

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image is virtual, eye is at right side of lens so that light rays are approaching eye to view image

image magnification from drawing = 0.66

(1/v) - (1/u) = 1/f   ; 1/v = (-1/2) -(1) = -3/2 or v = -2/3 cm

magnification v/u = (-2/3) /( -1) = 2/3

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image is virtual, eye is at right side of lens so that light rays are approaching eye to view image

image magnification from drawing = 0.5

(1/v) - (1/u) = 1/f   ; 1/v = (-1/2) -(1/2) = -1 or v = -1 cm

magnification v/u = -1 /( -2) = 0.5