Where must an object be placed to have a
magnification of 3.74 in front of a converging
lens of focal length 14.1 cm?
Let do be the object distance and di be the image distance.
di/do = 3.74, the magnification.
1/do + 1/di = 1/do + 1/(3.74 do)
= 4.74/(3.74 do)
= 1.267/do = 1/14.1
Solve for do.
Three identical small Styrofoam balls of mass
1.53 g are suspended from a fixed point by
three nonconducting threads, each with a
length of 56.4 cm and negligible mass. At
equilibrium the three balls form an equilateral triangle with sides of 31 cm.
What is the common charge carried by
each ball? The acceleration of gravity is
9.8 m/s
2
and the value of Coulomb’s constant
is 8.98755 × 10
9
N · m2
/C
2
.
Answer in units of µC
To find the position where an object must be placed in front of a converging lens to have a particular magnification, you can use the lens formula:
1/f = 1/v - 1/u
where:
f = focal length of the lens (given as 14.1 cm)
v = image distance (unknown)
u = object distance (unknown)
The magnification, m, is given by:
m = -v/u
Given that the magnification, m, is 3.74, and the focal length, f, is 14.1 cm, we can rearrange the formula as follows:
3.74 = -v/u
Now, let's solve for u. Rearrange the equation as:
u = -v/3.74
Substituting the value of the focal length into the lens formula, we get:
1/14.1 = 1/v - 1/u
Substituting the expression for u, we have:
1/14.1 = 1/v - 3.74/v
Combining the fractions on the right side, we get:
1/14.1 = (1 - 3.74)/v
Simplifying further:
1/14.1 = -2.74/v
Cross-multiplying:
v = -14.1 / 2.74
Therefore, the image distance, v, is approximately -5.15 cm (negative sign indicates that the image is formed on the same side as the object).
Since the image distance, v, is negative, the image forms on the same side as the object. So, the object should be placed approximately 5.15 cm in front of the converging lens to have a magnification of 3.74.