JEE MAIN - Physics (2025 - 2nd April Morning Shift - No. 12)
A slanted object $A B$ is placed on one side of convex lens as shown in the diagram. The image is formed on the opposite side. Angle made by the image with principal axis is :
_2nd_April_Morning_Shift_en_12_1.png)
$-\alpha$
$-\frac{\alpha}{2}$
$-45^{\circ}$
$+45^{\circ}$
Explanation
_2nd_April_Morning_Shift_en_12_2.png)
Location of image of A :-
$$\begin{aligned} & \frac{1}{v}-\frac{1}{u}=\frac{1}{f} \Rightarrow \frac{1}{v}-\frac{1}{-30}=\frac{1}{20} \Rightarrow \frac{1}{v}=\frac{1}{60} \Rightarrow v=60 \mathrm{~cm} \\ & \therefore m=2 \end{aligned}$$
Since size of object is small wrt the location hence
$$\begin{aligned} & d v=m^2 d u \Rightarrow d v=4 \times 1=4 \mathrm{~cm} \\ & h_i=m h_0 \Rightarrow h_i(d y)=2 \times 2=4 \mathrm{~cm} \end{aligned}$$
$\therefore$ Angle made with principle axis $=-45^{\circ}$
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