How far apart are an object and an image formed by a 75 cm focal length converging lens if the image is 2.5x larger than the object and is real?

To solve this problem, we can use the lens formula:

1/f = 1/d₀ + 1/dᵢ

Where:
f = focal length of the lens (given as 75 cm)
d₀ = distance of the object from the lens (unknown)
dᵢ = distance of the image from the lens (unknown)

Given that the image is 2.5 times larger than the object, we can say:

hᵢ/h₀ = 2.5

Where:
h₀ = height of the object
hᵢ = height of the image

Now, let's solve the problem step-by-step:

Step 1: Determine the height relationship

Since the image is 2.5 times larger than the object, we can say:

hᵢ = 2.5 * h₀

Step 2: Calculate the distances

Using the magnification equation:

magnification (m) = hᵢ/h₀ = -dᵢ/d₀

Since we know that the image is real, magnification (m) will be positive. Therefore, we can rewrite the equation as:

m = hᵢ/h₀ = dᵢ/d₀

Substitute the given magnification ratio:

2.5 = dᵢ/d₀

Step 3: Substitute into the lens formula

Substituting the above equation into the lens formula:

1/f = 1/d₀ + 1/dᵢ

we get:

1/75 = 1/d₀ + 1/(2.5 * d₀)

Step 4: Simplify the equation

Multiply through by the common denominator:

(1/75) * (2.5 * d₀) = (1/d₀) * (2.5 * d₀) + (1/(2.5 * d₀)) * (2.5 * d₀)

0.0333 = 2.5 + 1

0.0333 = 3.5

This is not a true statement, so there might be an error in the problem statement or calculation. Please double-check the given information or recheck the calculations.

To determine the distance between the object and the image, you can use the lens formula, which is given by:

1/f = 1/d₀ + 1/dᵢ

Where:
- f is the focal length of the lens,
- d₀ is the distance of the object from the lens,
- dᵢ is the distance of the image from the lens.

In this case, the focal length (f) of the converging lens is given as 75 cm. We need to find the distance (d₀) of the object from the lens.

We are also given that the image formed is real and 2.5 times larger than the object. This indicates that the magnification (m) is positive and equals 2.5:

m = -dᵢ/d₀

Since the image is magnified, the distance (dᵢ) of the image from the lens will be positive.

To solve for d₀, we can substitute the given values into the magnification formula:

2.5 = -dᵢ/d₀

Multiplying both sides by d₀, we get:

2.5d₀ = -dᵢ

Now, we can substitute the values into the lens formula:

1/75 = 1/d₀ + 1/(-dᵢ) [Note the negative sign.]

Rearranging the equation, we have:

1/d₀ - 1/(-dᵢ) = 1/75

To simplify further, we can combine the fractions:

(-dᵢ - d₀)/(-dᵢ * d₀) = 1/75

Multiplying both sides by -75 * (-dᵢ * d₀), we get:

(-dᵢ - d₀) = (-dᵢ * d₀) / 75

Expanding the equation, we have:

-dᵢ - d₀ = -dᵢ * d₀ / 75

Multiplying both sides by 75 to eliminate the fractions, we get:

-75dᵢ - 75d₀ = -dᵢ * d₀

Rearranging the equation, we have:

d₀dᵢ - 75d₀ - 75dᵢ = 0

Now, we have a quadratic equation in terms of d₀ and dᵢ. We can solve it using the quadratic formula:

d₀ = (-(-75) ± √((-75)² - 4 * 1 * (-75dᵢ)) / (2 * 1)

Simplifying this, we get two solutions for d₀:

d₀₁ = (75 + √(5625 + 300dᵢ)) / 2
d₀₂ = (75 - √(5625 + 300dᵢ)) / 2

Since the image is real, the distance (d₀) of the object from the lens should be positive. So, we discard the second solution:

d₀ = (75 + √(5625 + 300dᵢ)) / 2

Now, plugging in the given magnification value (m = 2.5), we can solve for dᵢ:

2.5 = -dᵢ/d₀

Multiplying both sides by d₀, we get:

2.5d₀ = -dᵢ

Substituting the equation for d₀, we have:

2.5((75 + √(5625 + 300dᵢ)) / 2) = -dᵢ

Simplifying, we get:

(75 + √(5625 + 300dᵢ)) = -2dᵢ

Squaring both sides to eliminate the square root, we have:

(75 + √(5625 + 300dᵢ))^2 = 4dᵢ²

Expanding and rearranging the equation, we have:

5625 + 150√(5625 + 300dᵢ) + (5625 + 300dᵢ) = 4dᵢ²

Simplifying this equation, we get:

900dᵢ + 150√(5625 + 300dᵢ) = 0

We can solve this equation numerically using numerical methods like the Newton-Raphson method or the bisection method. Since the equation is quadratic, it will have two solutions for dᵢ. To find the valid solution, we will select the negative value of dᵢ (since it is on the object's side) and substitute it back into the lens formula:

1/f = 1/d₀ + 1/dᵢ

Simplifying, we can solve for the distance (d₀) between the object and the lens.

From the magnification equation, you can relate M to di/do

Use that to solve. I will be happy to critique your thinking.