Suppose you are myopic (nearsighted). You can clearly focus on objects that are as far away as 52.5 cm away. You can clearly foc
1 answer:
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A. 6.67 cm
The focal length of the lens can be found by using the lens equation:
where we have
f = focal length
s = 12 cm is the distance of the object from the lens
s' = 15 cm is the distance of the image from the lens
Solving the equation for f, we find
B. Converging
According to sign convention for lenses, we have:
- Converging (convex) lenses have focal length with positive sign
- Diverging (concave) lenses have focal length with negative sign
In this case, the focal length of the lens is positive, so the lens is a converging lens.
C. -1.25
The magnification of the lens is given by
where
s' = 15 cm is the distance of the image from the lens
s = 12 cm is the distance of the object from the lens
Substituting into the equation, we find
D. Real and inverted
The magnification equation can be also rewritten as
where
y' is the size of the image
y is the size of the object
Re-arranging it, we have
Since in this case M is negative, it means that y' has opposite sign compared to y: this means that the image is inverted.
Also, the sign of s' tells us if the image is real of virtual. In fact:
- s' is positive: image is real
- s' is negative: image is virtual
In this case, s' is positive, so the image is real.
E. Virtual
In this case, the magnification is 5/9, so we have
which can be rewritten as
which means that s' has opposite sign than s: therefore, the image is virtual.
F. 12.0 cm
From the magnification equation, we can write
and then we can substitute it into the lens equation:
and we can solve for s:
G. -6.67 cm
Now the image distance can be directly found by using again the magnification equation:
And the sign of s' (negative) also tells us that the image is virtual.
H. -24.0 cm
In this case, the image is twice as tall as the object, so the magnification is
M = 2
and the distance of the image from the lens is
s' = -24 cm
The problem is asking us for the image distance: however, this is already given by the problem,
s' = -24 cm
so, this is the answer. And the fact that its sign is negative tells us that the image is virtual.
Answer:
p = 22.5 cm
Explanation:
For this exercise we must use the equation of the constructor
where f is the focal length, p and q are the distance to the object and the image respectively.
Let's start with the far point, the object is very far away (p = ∞) and the image must be formed at the far point of view of the person q = 45.0 cm
since the image is on the same side as the object according to the sign convention the distance is negative
f = -45.0 cm
now let's use the near point (q = 15.0 cm) at what distance the object should be
1 / p = 1 / -45 - 1 / -15
1 / p = -1/45 + 1/15
= 0.0444
p = 22.5 cm
this is the closest distance you can see an object clearly