JEE Main 2025 Apr 2 Shift 1, Physics Q4: Thin Lens and Mirrors

Image of A (on axis). Using the lens equation with sign convention ($u=-30\text{cm}$, $f=+20\text{cm}$):

$$\frac{1}{v}-\frac{1}{u}=\frac{1}{f}\Rightarrow \frac{1}{v}=\frac{1}{20}+\frac{1}{-30}=\frac{3-2}{60}=\frac{1}{60}$$

So $v=+60\text{cm}$, image of A is at A' (60 cm to the right of the lens). Linear magnification:

$$m=\frac{v}{u}=\frac{60}{-30}=-2$$

Longitudinal vs transverse magnification. For a small object close to A:

Transverse magnification (perpendicular to axis): $m_{\perp }=m=-2$.

Longitudinal magnification (along axis): $m_{\parallel }=m^2=4$, with a sign reversal of orientation.

The object endpoint B is offset by $du=1\text{cm}$ along the axis and $dy=2\text{cm}$ perpendicular. The image endpoint B' is therefore offset from A' by:

$dv=m^{2}du=4\times 1=4\text{cm}$ (along axis, with inversion)

$d{y}^{\prime }=mdy=-2\times 2=-4\text{cm}$ (perpendicular, inverted)

The slope of the image relative to the principal axis is $\frac{|d{y}^{\prime }|}{|dv|}=\frac{4}{4}=1$, but with the inversion the image makes an angle of $-45°$ with the axis.

$\therefore$ Option (2) is correct.