Question 5 of 295

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:

A$-\frac{\alpha}{2}$B$-45^{\circ}$C$+45^{\circ}$D$-\alpha$

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Answer: (B) $-45^{\circ}$

Location of image of $\mathrm{A}:-$
$\begin{aligned} & \frac{1}{\mathrm{v}}-\frac{1}{\mathrm{u}}=\frac{1}{\mathrm{f}} \Rightarrow \frac{1}{\mathrm{v}}-\frac{1}{-30}=\frac{1}{20} \Rightarrow \frac{1}{\mathrm{v}}=\frac{1}{60} \Rightarrow \mathrm{v}=60 \mathrm{~cm} \\ & \therefore \mathrm{~m}=2\end{aligned}$
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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