A magnifying glass uses a converging lens with a focal length of 15.5 cm. It produces a virtual and upright image that is 3.1 times larger than the object. How far is the object from the lens?
To solve this problem, we can use the lens formula:
1/f = 1/v - 1/u
where:
- f is the focal length of the lens,
- v is the image distance, and
- u is the object distance.
Given that the focal length (f) is 15.5 cm and the magnification (M) is 3.1, we can also use the magnification formula:
M = -v/u
Since the image is virtual and upright, the magnification (M) should be positive.
Let's solve for the object distance (u) using the magnification formula:
M = -v/u
3.1 = -v/u (since M is positive)
-3.1u = v
Now, we substitute this value of v into the lens formula:
1/f = 1/v - 1/u
1/15.5 = 1/(-3.1u) - 1/u
To simplify this equation, let's find a common denominator:
1/15.5 = (-1 + 3.1) / (3.1u)
Multiplying both sides of the equation by 15.5 and simplifying:
1 = (2.1) / (3.1u)
Cross-multiplying:
3.1u = 2.1
Now, solve for u:
u = 2.1 / 3.1
u ≈ 0.6774 cm
Therefore, the object is approximately 0.6774 cm away from the lens.
To find the distance of the object from the lens, we can use the lens formula:
1/f = 1/v - 1/u
Where:
f is the focal length of the lens
v is the image distance
u is the object distance
In this case, the lens has a focal length of 15.5 cm and it produces a virtual image. For a converging lens, when the image is virtual and upright, the image distance (v) is positive.
Since the image produced is 3.1 times larger than the object, the magnification (M) can be calculated using the formula:
M = v/u
Substituting these values into the lens formula, we have:
1/15.5 = 1/v - 1/u
Since the image is virtual and upright, the magnification (M) is positive:
3.1 = v/u
Rearranging the equation, we get:
u = v/3.1
Substituting this into the lens formula:
1/15.5 = 1/v - 1/(v/3.1)
Simplifying the equation, we have:
1/15.5 = 1/v - 3.1/v
Combining like terms:
1/15.5 = (v - 3.1v)/v
Simplifying further:
1/15.5 = -2.1v/v
Multiplying both sides by v:
v/15.5 = -2.1
Dividing both sides by -2.1:
v = -15.5/2.1
The negative sign indicates that the image is virtual. However, distance cannot be negative. Since we are looking for the object distance, which is always positive, we can ignore the negative sign.
Therefore, the object distance (u) is approximately:
u = abs(-15.5/2.1) ≈ 7.38 cm
So, the object is approximately 7.38 cm away from the lens.