Question

how far from -41.0-mm-focal-length lens must an object be placed if its image is to be magnified 3.50× and be virtual? follow the sign conventions hint: consider real objects and virtual objects (formed by some other piece of optics). express your answer to three significant figures and include the appropriate units. d_o1 = -52.7 mm submit previous answers all attempts used; correct answer displayed part b how far from -41.0-mm-focal-length lens must an object be placed if its image is to be magnified 3.50× and be real? follow the sign conventions hint: consider real objects and virtual objects (formed by some other piece of optics). express your answer to three significant figures and include the appropriate units. d_o2 = value units

Explanation:

Step1: Recall Lens Formula and Magnification

The lens formula is $\frac{1}{f} = \frac{1}{d_o} + \frac{1}{d_i}$, and magnification $m = -\frac{d_i}{d_o}$. For real image, $m$ is negative (since real images are inverted, magnification sign convention: negative for inverted, positive for upright). Given $m = - 3.50$ (real image, inverted), $f = - 41.0\ \text{mm}$ (diverging lens, focal length negative).

From $m = -\frac{d_i}{d_o}$, we get $d_i = - m d_o = - (-3.50) d_o = 3.50 d_o$? Wait, no: Wait, magnification for real image (inverted) so $m = -3.50$. So $m = -\frac{d_i}{d_o} \implies -3.50 = -\frac{d_i}{d_o} \implies d_i = 3.50 d_o$? Wait, no, sign conventions: for diverging lens, image is virtual for real objects, but here we need real image, so maybe object is virtual? Wait, the hint says consider real and virtual objects. Wait, let's re-examine.

Wait, focal length $f = -41.0\ \text{mm}$ (diverging lens). For a real image, $d_i$ must be positive? No, diverging lens always forms virtual images for real objects. So to get real image, the object must be virtual (formed by another optics), so $d_o$ is negative (virtual object).

Magnification $m = \frac{h_i}{h_o} = -\frac{d_i}{d_o}$. For real image, $h_i$ is inverted, so $m$ is negative? Wait, no: if object is virtual ($d_o < 0$), and image is real ($d_i > 0$), then $m = -\frac{d_i}{d_o} = -\frac{+}{-} = +$? Wait, maybe I messed up sign conventions. Let's use the standard sign convention: distances measured from lens, object distance $d_o$: positive for real objects (on left of lens), negative for virtual objects (on right of lens). Image distance $d_i$: positive for real images (on right of lens), negative for virtual images (on left of lens). Focal length $f$: positive for converging, negative for diverging.

Magnification $m = \frac{h_i}{h_o} = -\frac{d_i}{d_o}$.

We need real image, so $d_i > 0$. Magnification $m = 3.50$? Wait, no, the problem says "magnified 3.50× and be real". Real images from diverging lens: only possible if object is virtual (so $d_o < 0$), and image is real ($d_i > 0$).

So $m = -\frac{d_i}{d_o}$. Since $d_o < 0$ (virtual object) and $d_i > 0$ (real image), $m = -\frac{+}{-} = +$, so magnification is positive 3.50? Wait, the problem says "magnified 3.50×", maybe magnitude. Wait, the first part (Part A) had virtual image, magnification positive (upright), here real image, so inverted, so magnification should be negative? Wait, the problem says "magnified 3.50×", maybe absolute value. Wait, let's check Part A: in Part A, virtual image, magnification positive (upright), so $m = +3.50$. For Part B, real image, so inverted, $m = -3.50$? Wait, the problem statement: "magnified 3.50× and be real" – maybe the magnification is -3.50 (inverted, real).

So let's proceed with $m = -3.50$ (real image, inverted), $f = -41.0\ \text{mm}$.

From lens formula: $\frac{1}{f} = \frac{1}{d_o} + \frac{1}{d_i}$.

From magnification: $m = -\frac{d_i}{d_o} \implies d_i = - m d_o = - (-3.50) d_o = 3.50 d_o$? Wait, no: $m = -3.50 = -\frac{d_i}{d_o} \implies d_i = 3.50 d_o$.

Now substitute into lens formula:

$\frac{1}{-41.0} = \frac{1}{d_o} + \frac{1}{3.50 d_o}$

Combine terms: $\frac{1}{d_o} (1 + \frac{1}{3.50}) = \frac{1}{d_o} (\frac{3.50 + 1}{3.50}) = \frac{4.50}{3.50 d_o}$

So:

$\frac{4.50}{3.50 d_o} = \frac{1}{-41.0}$

Solve for $d_o$:

$d_o = \frac{4.50 \times (-41.0)}{3.50} = \frac{-184.5}{3.50} \approx -52.7\ \text{mm}$? Wait, no, that can't be. Wait, maybe I messed up the magnification sign. Let's try $m = +3.50$ (real image, upright? No, real images are inverted). Wait, maybe the object is virtual, so $d_o < 0$, and image is real ($d_i > 0$), so $m = -\frac{d_i}{d_o} = -\frac{+}{-} = +$, so magnification is positive (upright real image? But real images are usually inverted. Maybe the sign convention for magnification: if $m$ is positive, image is upright; negative, inverted. So real image can be upright if object is virtual?

Wait, let's start over.

Given: $f = -41.0\ \text{mm}$ (diverging lens).

We need image to be real: $d_i > 0$.

Magnification $|m| = 3.50$, so $m = \pm 3.50$. For real image, if $d_i > 0$ and $d_o < 0$ (virtual object), then $m = -\frac{d_i}{d_o} = -\frac{+}{-} = +$, so $m = +3.50$ (upright real image).

So $d_i = - m d_o = - 3.50 d_o$? Wait, no: $m = -\frac{d_i}{d_o} \implies d_i = - m d_o$. If $m = +3.50$, then $d_i = -3.50 d_o$.

Now lens formula: $\frac{1}{f} = \frac{1}{d_o} + \frac{1}{d_i}$.

Substitute $d_i = -3.50 d_o$ and $f = -41.0$:

$\frac{1}{-41.0} = \frac{1}{d_o} + \frac{1}{-3.50 d_o}$

Simplify right side: $\frac{1}{d_o} (1 - \frac{1}{3.50}) = \frac{1}{d_o} (\frac{3.50 - 1}{3.50}) = \frac{2.50}{3.50 d_o}$

So:

$\frac{2.50}{3.50 d_o} = \frac{1}{-41.0}$

Solve for $d_o$:

$d_o = \frac{2.50 \times (-41.0)}{3.50} = \frac{-102.5}{3.50} \approx -29.3\ \text{mm}$? No, that's not matching. Wait, maybe I got the magnification sign wrong. Let's consider that for real image, $d_i > 0$, and for diverging lens, to get $d_i > 0$, $d_o$ must be negative (virtual object), and $m = -\frac{d_i}{d_o} = -\frac{+}{-} = +$, so $m = 3.50$ (upright). Then:

$\frac{1}{f} = \frac{1}{d_o} + \frac{1}{d_i}$

$d_i = - m d_o = -3.50 d_o$ (since $m = -\frac{d_i}{d_o} \implies d_i = - m d_o$)

Wait, no: $m = -\frac{d_i}{d_o} \implies d_i = - m d_o$. If $m = 3.50$, then $d_i = -3.50 d_o$.

Now substitute into lens formula:

$\frac{1}{-41.0} = \frac{1}{d_o} + \frac{1}{-3.50 d_o} = \frac{1}{d_o} (1 - \frac{1}{3.50}) = \frac{2.50}{3.50 d_o}$

So:

$\frac{2.50}{3.50 d_o} = \frac{1}{-41.0}$

$d_o = \frac{2.50 \times (-41.0)}{3.50} = \frac{-102.5}{3.50} \approx -29.3\ \text{mm}$? No, that's not right. Wait, maybe the object is real, so $d_o > 0$, but for diverging lens, image is virtual ($d_i < 0$), so to get real image, object must be virtual ($d_o < 0$), and image is real ($d_i > 0$).

Wait, let's use the formula for magnification and lens formula correctly.

Let $m = -3.50$ (real image, inverted), so $d_i = - m d_o = 3.50 d_o$ (since $m = -3.50$).

Lens formula: $\frac{1}{f} = \frac{1}{d_o} + \frac{1}{d_i}$

Substitute $d_i = 3.50 d_o$ and $f = -41.0$:

$\frac{1}{-41.0} = \frac{1}{d_o} + \frac{1}{3.50 d_o} = \frac{3.50 + 1}{3.50 d_o} = \frac{4.50}{3.50 d_o}$

So:

$\frac{4.50}{3.50 d_o} = \frac{1}{-41.0}$

$d_o = \frac{4.50 \times (-41.0)}{3.50} = \frac{-184.5}{3.50} \approx -52.7\ \text{mm}$? But that's the same as Part A? No, that can't be. Wait, maybe I made a mistake in the sign of $d_i$. For real image, $d_i$ should be positive, but if $d_o$ is negative (virtual object), and $d_i = 3.50 d_o$, then $d_i$ would be negative (since $d_o$ is negative), which is virtual image. So that's wrong.

Ah! Here's the mistake: For real image, $d_i > 0$, so $d_i$ must be positive. So $d_i = +|d_i|$. Then $m = -\frac{d_i}{d_o}$. If $d_o$ is negative (virtual object), then $-\frac{d_i}{d_o} = -\frac{+}{-} = +$, so $m$ is positive (upright real image). So $m = +3.50$, $d_i = +|d_i|$, $d_o = -|d_o|$.

Then $m = -\frac{d_i}{d_o} \implies 3.50 = -\frac{d_i}{-|d_o|} = \frac{d_i}{|d_o|} \implies d_i = 3.50 |d_o|$.

Lens formula: $\frac{1}{f} = \frac{1}{d_o} + \frac{1}{d_i} = \frac{1}{-|d_o|} + \frac{1}{3.50 |d_o|} = \frac{-3.50 + 1}{3.50 |d_o|} = \frac{-2.50}{3.50 |d_o|}$

Set equal to $\frac{1}{f} = \frac{1}{-41.0}$:

$\frac{-2.50}{3.50 |d_o|} = \frac{1}{-41.0}$

Multiply both sides by -1:

$\frac{2.50}{3.50 |d_o|} = \frac{1}{41.0}$

Solve for $|d_o|$:

$|d_o| = \frac{2.50 \times 41.0}{3.50} = \frac{102.5}{3.50} \approx 29.3\ \text{mm}$

But $d_o$ is negative (virtual object), so $d_o = -29.3\ \text{mm}$? No, that's not right. Wait, maybe the object is real, so $d_o > 0$, but for diverging lens, image is virtual ($d_i < 0$), so to get real image, we need a converging lens? No, the lens is diverging ($f < 0$). Diverging lenses can only form virtual images for real objects. To form real images, the object must be virtual (created by another lens), so $d_o < 0$.

Wait, let's use the formula correctly with signs:

Let $d_o = x$ (can be positive or negative), $d_i = y$ (positive for real, negative for virtual), $f = -41.0\ \text{mm}$, $m = 3.50$ (real image, so $y > 0$, and $m = -\frac{y}{x} = 3.50$ (since $x < 0$ (virtual object), $-\frac{y}{x} = -\frac{+}{-} = +$).

So $-\frac{y}{x} = 3.50 \implies y = -3.50 x$

Lens formula: $\frac{1}{-41.0} = \frac{1}{x} + \frac{1}{y} = \frac{1}{x} + \frac{1}{-3.50 x} = \frac{-3.50 + 1}{-3.50 x} = \frac{-2.50}{-3.50 x} = \frac{2.50}{3.50 x}$

So:

$\frac{2.50}{3.50 x} = \frac{1}{-41.0}$

$x = \frac{2.50 \times (-41.0)}{3.50} = \frac{-102.5}{3.50} \approx -29.3\ \text{mm}$

Wait, now $x = d_o \approx -29.3\ \text{mm}$, $y = -3.50 x = -3.50 \times (-29.3) \approx 102.6\ \text{mm}$ (positive, real image). Let's check lens formula:

$\frac{1}{-41.0} + \frac{1}{102.6} \approx -0.02439 + 0.009746 \approx -0.01464$

$\frac{1}{-29.3} \approx -0.03413$, no, that's not matching. Wait, I messed up the algebra.

Wait, from $y = -3.50 x$, then $\frac{1}{y} = \frac{1}{-3.50 x}$

So $\frac{1}{x} + \frac{1}{y} = \frac{1}{x} - \frac{1}{3.50 x} = \frac{3.50 - 1}{3.50 x} = \frac{2.50}{3.50 x}$

Set equal to $\frac{1}{f} = \frac{1}{-41.0}$:

$\frac{2.50}{3.50 x} = \frac{1}{-41.0}$

Cross-multiplying: $2.50 \times (-41.0) = 3.50 x$

$x = \frac{2.50 \times (-41.0)}{3.50} = \frac{-102.5}{3.50} \approx -29.3\ \text{mm}$

Now check lens formula with $x = -29.3\ \text{mm}$, $y = -3.50 x = 102.55\ \text{mm}$:

$\frac{1}{-29.3} + \frac{1}{102.55} \approx -0.03413 + 0.00975 \approx -0.02438$

$\frac{1}{-41.0} \approx -0.02439$, which matches (considering rounding errors). So $d_o = -29.3\ \text{mm}$

Answer:

Step1: Recall Lens Formula and Magnification

The lens formula is $\frac{1}{f} = \frac{1}{d_o} + \frac{1}{d_i}$, and magnification $m = -\frac{d_i}{d_o}$. For real image, $m$ is negative (since real images are inverted, magnification sign convention: negative for inverted, positive for upright). Given $m = - 3.50$ (real image, inverted), $f = - 41.0\ \text{mm}$ (diverging lens, focal length negative).

From $m = -\frac{d_i}{d_o}$, we get $d_i = - m d_o = - (-3.50) d_o = 3.50 d_o$? Wait, no: Wait, magnification for real image (inverted) so $m = -3.50$. So $m = -\frac{d_i}{d_o} \implies -3.50 = -\frac{d_i}{d_o} \implies d_i = 3.50 d_o$? Wait, no, sign conventions: for diverging lens, image is virtual for real objects, but here we need real image, so maybe object is virtual? Wait, the hint says consider real and virtual objects. Wait, let's re-examine.

Wait, focal length $f = -41.0\ \text{mm}$ (diverging lens). For a real image, $d_i$ must be positive? No, diverging lens always forms virtual images for real objects. So to get real image, the object must be virtual (formed by another optics), so $d_o$ is negative (virtual object).

Magnification $m = \frac{h_i}{h_o} = -\frac{d_i}{d_o}$. For real image, $h_i$ is inverted, so $m$ is negative? Wait, no: if object is virtual ($d_o < 0$), and image is real ($d_i > 0$), then $m = -\frac{d_i}{d_o} = -\frac{+}{-} = +$? Wait, maybe I messed up sign conventions. Let's use the standard sign convention: distances measured from lens, object distance $d_o$: positive for real objects (on left of lens), negative for virtual objects (on right of lens). Image distance $d_i$: positive for real images (on right of lens), negative for virtual images (on left of lens). Focal length $f$: positive for converging, negative for diverging.

Magnification $m = \frac{h_i}{h_o} = -\frac{d_i}{d_o}$.

We need real image, so $d_i > 0$. Magnification $m = 3.50$? Wait, no, the problem says "magnified 3.50× and be real". Real images from diverging lens: only possible if object is virtual (so $d_o < 0$), and image is real ($d_i > 0$).

So $m = -\frac{d_i}{d_o}$. Since $d_o < 0$ (virtual object) and $d_i > 0$ (real image), $m = -\frac{+}{-} = +$, so magnification is positive 3.50? Wait, the problem says "magnified 3.50×", maybe magnitude. Wait, the first part (Part A) had virtual image, magnification positive (upright), here real image, so inverted, so magnification should be negative? Wait, the problem says "magnified 3.50×", maybe absolute value. Wait, let's check Part A: in Part A, virtual image, magnification positive (upright), so $m = +3.50$. For Part B, real image, so inverted, $m = -3.50$? Wait, the problem statement: "magnified 3.50× and be real" – maybe the magnification is -3.50 (inverted, real).

So let's proceed with $m = -3.50$ (real image, inverted), $f = -41.0\ \text{mm}$.

From lens formula: $\frac{1}{f} = \frac{1}{d_o} + \frac{1}{d_i}$.

From magnification: $m = -\frac{d_i}{d_o} \implies d_i = - m d_o = - (-3.50) d_o = 3.50 d_o$? Wait, no: $m = -3.50 = -\frac{d_i}{d_o} \implies d_i = 3.50 d_o$.

Now substitute into lens formula:

$\frac{1}{-41.0} = \frac{1}{d_o} + \frac{1}{3.50 d_o}$

Combine terms: $\frac{1}{d_o} (1 + \frac{1}{3.50}) = \frac{1}{d_o} (\frac{3.50 + 1}{3.50}) = \frac{4.50}{3.50 d_o}$

So:

$\frac{4.50}{3.50 d_o} = \frac{1}{-41.0}$

Solve for $d_o$:

$d_o = \frac{4.50 \times (-41.0)}{3.50} = \frac{-184.5}{3.50} \approx -52.7\ \text{mm}$? Wait, no, that can't be. Wait, maybe I messed up the magnification sign. Let's try $m = +3.50$ (real image, upright? No, real images are inverted). Wait, maybe the object is virtual, so $d_o < 0$, and image is real ($d_i > 0$), so $m = -\frac{d_i}{d_o} = -\frac{+}{-} = +$, so magnification is positive (upright real image? But real images are usually inverted. Maybe the sign convention for magnification: if $m$ is positive, image is upright; negative, inverted. So real image can be upright if object is virtual?

Wait, let's start over.

Given: $f = -41.0\ \text{mm}$ (diverging lens).

We need image to be real: $d_i > 0$.

Magnification $|m| = 3.50$, so $m = \pm 3.50$. For real image, if $d_i > 0$ and $d_o < 0$ (virtual object), then $m = -\frac{d_i}{d_o} = -\frac{+}{-} = +$, so $m = +3.50$ (upright real image).

So $d_i = - m d_o = - 3.50 d_o$? Wait, no: $m = -\frac{d_i}{d_o} \implies d_i = - m d_o$. If $m = +3.50$, then $d_i = -3.50 d_o$.

Now lens formula: $\frac{1}{f} = \frac{1}{d_o} + \frac{1}{d_i}$.

Substitute $d_i = -3.50 d_o$ and $f = -41.0$:

$\frac{1}{-41.0} = \frac{1}{d_o} + \frac{1}{-3.50 d_o}$

Simplify right side: $\frac{1}{d_o} (1 - \frac{1}{3.50}) = \frac{1}{d_o} (\frac{3.50 - 1}{3.50}) = \frac{2.50}{3.50 d_o}$

So:

$\frac{2.50}{3.50 d_o} = \frac{1}{-41.0}$

Solve for $d_o$:

$d_o = \frac{2.50 \times (-41.0)}{3.50} = \frac{-102.5}{3.50} \approx -29.3\ \text{mm}$? No, that's not matching. Wait, maybe I got the magnification sign wrong. Let's consider that for real image, $d_i > 0$, and for diverging lens, to get $d_i > 0$, $d_o$ must be negative (virtual object), and $m = -\frac{d_i}{d_o} = -\frac{+}{-} = +$, so $m = 3.50$ (upright). Then:

$\frac{1}{f} = \frac{1}{d_o} + \frac{1}{d_i}$

$d_i = - m d_o = -3.50 d_o$ (since $m = -\frac{d_i}{d_o} \implies d_i = - m d_o$)

Wait, no: $m = -\frac{d_i}{d_o} \implies d_i = - m d_o$. If $m = 3.50$, then $d_i = -3.50 d_o$.

Now substitute into lens formula:

$\frac{1}{-41.0} = \frac{1}{d_o} + \frac{1}{-3.50 d_o} = \frac{1}{d_o} (1 - \frac{1}{3.50}) = \frac{2.50}{3.50 d_o}$

So:

$\frac{2.50}{3.50 d_o} = \frac{1}{-41.0}$

$d_o = \frac{2.50 \times (-41.0)}{3.50} = \frac{-102.5}{3.50} \approx -29.3\ \text{mm}$? No, that's not right. Wait, maybe the object is real, so $d_o > 0$, but for diverging lens, image is virtual ($d_i < 0$), so to get real image, object must be virtual ($d_o < 0$), and image is real ($d_i > 0$).

Wait, let's use the formula for magnification and lens formula correctly.

Let $m = -3.50$ (real image, inverted), so $d_i = - m d_o = 3.50 d_o$ (since $m = -3.50$).

Lens formula: $\frac{1}{f} = \frac{1}{d_o} + \frac{1}{d_i}$

Substitute $d_i = 3.50 d_o$ and $f = -41.0$:

$\frac{1}{-41.0} = \frac{1}{d_o} + \frac{1}{3.50 d_o} = \frac{3.50 + 1}{3.50 d_o} = \frac{4.50}{3.50 d_o}$

So:

$\frac{4.50}{3.50 d_o} = \frac{1}{-41.0}$

$d_o = \frac{4.50 \times (-41.0)}{3.50} = \frac{-184.5}{3.50} \approx -52.7\ \text{mm}$? But that's the same as Part A? No, that can't be. Wait, maybe I made a mistake in the sign of $d_i$. For real image, $d_i$ should be positive, but if $d_o$ is negative (virtual object), and $d_i = 3.50 d_o$, then $d_i$ would be negative (since $d_o$ is negative), which is virtual image. So that's wrong.

Ah! Here's the mistake: For real image, $d_i > 0$, so $d_i$ must be positive. So $d_i = +|d_i|$. Then $m = -\frac{d_i}{d_o}$. If $d_o$ is negative (virtual object), then $-\frac{d_i}{d_o} = -\frac{+}{-} = +$, so $m$ is positive (upright real image). So $m = +3.50$, $d_i = +|d_i|$, $d_o = -|d_o|$.

Then $m = -\frac{d_i}{d_o} \implies 3.50 = -\frac{d_i}{-|d_o|} = \frac{d_i}{|d_o|} \implies d_i = 3.50 |d_o|$.

Lens formula: $\frac{1}{f} = \frac{1}{d_o} + \frac{1}{d_i} = \frac{1}{-|d_o|} + \frac{1}{3.50 |d_o|} = \frac{-3.50 + 1}{3.50 |d_o|} = \frac{-2.50}{3.50 |d_o|}$

Set equal to $\frac{1}{f} = \frac{1}{-41.0}$:

$\frac{-2.50}{3.50 |d_o|} = \frac{1}{-41.0}$

Multiply both sides by -1:

$\frac{2.50}{3.50 |d_o|} = \frac{1}{41.0}$

Solve for $|d_o|$:

$|d_o| = \frac{2.50 \times 41.0}{3.50} = \frac{102.5}{3.50} \approx 29.3\ \text{mm}$

But $d_o$ is negative (virtual object), so $d_o = -29.3\ \text{mm}$? No, that's not right. Wait, maybe the object is real, so $d_o > 0$, but for diverging lens, image is virtual ($d_i < 0$), so to get real image, we need a converging lens? No, the lens is diverging ($f < 0$). Diverging lenses can only form virtual images for real objects. To form real images, the object must be virtual (created by another lens), so $d_o < 0$.

Wait, let's use the formula correctly with signs:

Let $d_o = x$ (can be positive or negative), $d_i = y$ (positive for real, negative for virtual), $f = -41.0\ \text{mm}$, $m = 3.50$ (real image, so $y > 0$, and $m = -\frac{y}{x} = 3.50$ (since $x < 0$ (virtual object), $-\frac{y}{x} = -\frac{+}{-} = +$).

So $-\frac{y}{x} = 3.50 \implies y = -3.50 x$

Lens formula: $\frac{1}{-41.0} = \frac{1}{x} + \frac{1}{y} = \frac{1}{x} + \frac{1}{-3.50 x} = \frac{-3.50 + 1}{-3.50 x} = \frac{-2.50}{-3.50 x} = \frac{2.50}{3.50 x}$

So:

$\frac{2.50}{3.50 x} = \frac{1}{-41.0}$

$x = \frac{2.50 \times (-41.0)}{3.50} = \frac{-102.5}{3.50} \approx -29.3\ \text{mm}$

Wait, now $x = d_o \approx -29.3\ \text{mm}$, $y = -3.50 x = -3.50 \times (-29.3) \approx 102.6\ \text{mm}$ (positive, real image). Let's check lens formula:

$\frac{1}{-41.0} + \frac{1}{102.6} \approx -0.02439 + 0.009746 \approx -0.01464$

$\frac{1}{-29.3} \approx -0.03413$, no, that's not matching. Wait, I messed up the algebra.

Wait, from $y = -3.50 x$, then $\frac{1}{y} = \frac{1}{-3.50 x}$

So $\frac{1}{x} + \frac{1}{y} = \frac{1}{x} - \frac{1}{3.50 x} = \frac{3.50 - 1}{3.50 x} = \frac{2.50}{3.50 x}$

Set equal to $\frac{1}{f} = \frac{1}{-41.0}$:

$\frac{2.50}{3.50 x} = \frac{1}{-41.0}$

Cross-multiplying: $2.50 \times (-41.0) = 3.50 x$

$x = \frac{2.50 \times (-41.0)}{3.50} = \frac{-102.5}{3.50} \approx -29.3\ \text{mm}$

Now check lens formula with $x = -29.3\ \text{mm}$, $y = -3.50 x = 102.55\ \text{mm}$:

$\frac{1}{-29.3} + \frac{1}{102.55} \approx -0.03413 + 0.00975 \approx -0.02438$

$\frac{1}{-41.0} \approx -0.02439$, which matches (considering rounding errors). So $d_o = -29.3\ \text{mm}$