Before starting this problem, check out the section on the thin-lens formula and

Question

Before starting this problem, check out the section on the thin-lens formula and the magnification formula. Check out the Summary of Sign Conventions for Lenses in the text.
A 3.90-cm tall pencil is placed 12.6 cm in front of a lens. An image is formed 5.00 cm on the opposite side of the lens from the object. What is the focal length of the lens?

Is the lens converging or diverging?

Diverging
Converging

A 5.90-cm tall pencil is placed 13.3 cm in front of a lens. An image is formed 4.80 cm on the same side of the lens as the object. What is the focal length of the lens?

Is the lens converging or diverging?

Converging
Diverging

A 3.80-cm tall pencil is placed 11.0 cm in front of a lens. The focal length of the lens is +29.7 cm. What is the image distance (with proper sign)?

Is the image real or virtual?

Virtual
Real      

Explanation / Answer

s = 12.6   s' = 5

1/s + 1/s' = 1/f

(1/12.6)+(1/5) = 1/f

f = +3.59 cm

converging lens

------------------

s = 13.3                 s' = -4.8

1/s + 1/s' = 1/f

f = (s*s')/(s+s')

f = -(13.3*4.8)/(13.3-4.8) = -7.51 cm

diverging

------------------

s = 11

1/s + 1/s' = 1/f

(1/11) + (1/s') = 1/29.7 = -17.47 cm

s' = negative

the image is virtual

Related Questions

Navigate

• Browse (All)
• Browse B
• Subjects

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.