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Rom4ik [11]
4 years ago
15

5. You're examining some of the tiny printing on one of the newer twenty-dollar bills. A 1.5 mm tall letter appears 3 mm tall an

d upright when viewed through a converging lens. If the lens is held 4 cm from the bill, what is the focal length of the lens
Physics
2 answers:
kow [346]4 years ago
5 0

Answer:

80mm or 8cm

Explanation:

According to the lens formula,

1/f = 1/u+1/v

If the object distance u = 4cm = 40mm

Object height = 1.5mm

Image height = 3mm

First, we need to get the image distance (v) using the magnification formula Magnification = image distance/object distance = Image height/object height

v/40=3/1.5

1.5v = 120

v = 120/1.5

v = 80mm

The image distance is 80mm

To get the focal length, we will substitute the image distance and the object distance in the mirror formula to have;

1/f = 1/40+1/-80

Note that the image formed by the lens is an upright image (virtual), therefore the image distance will be negative.

Also the focal length of the converging lens is positive. Our formula will become;

1/f = 1/40-1/80

1/f = 2-1/80

1/f = 1/80

f = 80mm

The focal length of the lens 80mm or 8cm

Paraphin [41]4 years ago
4 0

Answer: f = 4cm

Explanation: The lateral magnification formulae is given as

m = height of image /height of object or

m = - (image distance) /(object distance)

Image height (Hi) = 3mm, object height (Hob) = 1.5mm, object distance (u) =?, image distance (v) = - 4cm... The reason we have a negative image distance is because the image is upright and an upright image (aka virtual image) has negative image distance because the the refracting ray is on the opposite side of the image.

m = Hi/Hob

m = 3/1.5 = 2.

Since m =2, and v = - 4cm, we can get 'u' as follows

m = - v/u

2 = - (-4)/u

2 = 4/u

2u = 4, u = 4/2, u = 2cm.

Using the lens formulae

1/u + 1/v = 1/f

1/2 - 1/4 = 1/f

0.5 - 0.25 = 1/f

0.25 = 1/f

f = 1/ 0.25

f = 4cm

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Answer:

Explanation:

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