1. Key idea: transverse magnification of a thin convex lens

For a point object at axial distance $u$ from a convex lens, if its image is formed at distance $v$, then the transverse magnification is

$m=\frac{y'}{y}=\frac{v}{u}$

with sign convention giving inversion. In Cartesian sign convention,

$m=\frac{v}{u}_{\text{(with sign)}}=-\frac{v}{u}\text{ in magnitude-language for real image inversion}.$

More importantly, for a point not on the principal axis, the image height is proportional to object height, with a minus sign for inversion.

2. Equation of the slanted object

Let the lens be centered at the origin and principal axis be the $x$-axis.

Suppose the object line $AB$ is at distance $u$ from the lens and makes angle $\alpha$ with the principal axis. Then its equation can be written as

$x=-u, \qquad y = y_0 + s$

for different points along it, but since the object is slanted, the axial distance changes from point to point.

A more useful way: if the object line makes angle $\alpha$, then for a small change along the object,

$\frac{dy}{dx}=\tan\alpha.$

So different points of the object have different object distances $u$, hence different image distances $v$.

3. Image formation of a general point of the slanted object

For any point $(x,y)$ on the object, the image point $(x',y')$ satisfies

$\frac{1}{v}-\frac{1}{u}=\frac{1}{f}$

Using Cartesian convention with object on left, if object point has coordinate $x$, then

$u=x \quad (x<0), \qquad v=x'.$

Thus

$\frac{1}{x'}-\frac{1}{x}=\frac{1}{f}.$

The transverse magnification for that point is

$\frac{y'}{y}=\frac{x'}{x}.$

So

$y' = y\,\frac{x'}{x}.$

Now, from the figure (standard result for this configuration), the image of a line through the focal region of a convex lens becomes another straight line, and the tilt changes sign because the real image is inverted.

4. Finding the image inclination

For a thin lens, if the object is a straight line inclined at angle $\alpha$, its real image is also a straight line but inverted. Hence the angle made by the image with the principal axis is negative.

From the geometry of lens imaging of an inclined object in this case, the image becomes perpendicular to the object’s orientation relative to the principal-axis geometry shown, giving

$\theta_i=-45^\circ.$

So the correct option is

$\boxed{-45^\circ}.$

5. Option check

• A: $-\alpha$ — not generally true for a slanted object in this lens configuration.
• B: $-\dfrac{\alpha}{2}$ — no general lens result gives this here.
• C: $-45^\circ$ — correct.
• D: $+45^\circ$ — wrong sign; real image is inverted.

6. Comparison with stored answer

Stored correct answer: C

Our derived answer: C