A slanted object AB is placed on one side of | vedclass

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(B) For point $A$,the object distance is $u = -30 \ cm$ and the focal length is $f = +20 \ cm$. Using the lens formula $\frac{1}{v} - \frac{1}{u} = \f

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$-45^{\circ}$

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(B) For point $A$,the object distance is $u = -30 \ cm$ and the focal length is $f = +20 \ cm$. Using the lens formula $\frac{1}{v} - \frac{1}{u} = \frac{1}{f}$:$\frac{1}{v} - \frac{1}{-30} = \frac{1}{20} \Rightarrow \frac{1}{v} = \frac{1}{20} - \frac{1}{30} = \frac{3-2}{60} = \frac{1}{60}$.Thus,$v = 60 \ cm$. The magnification is $m = \frac{v}{u} = \frac{60}{-30} = -2$.For a small object,the longitudinal magnification is $m_L = \frac{dv}{du} = m^2 = (-2)^2 = 4$. The change in image position is $dv = m^2 du = 4 	imes 1 \ cm = 4 \ cm$.The transverse magnification is $m = \frac{h_i}{h_o} = -2$,so the height of the image is $h_i = m 	imes h_o = -2 	imes 2 \ cm = -4 \ cm$.The image is inverted and real. The angle $	heta$ made by the image with the principal axis is given by $	an 	heta = \frac{|h_i|}{|dv|} = \frac{4}{4} = 1$. Since the image is inverted,the angle is $-45^{\circ}$.

$A$ slanted object $AB$ is placed on one side of a convex lens as shown in the diagram. The image is formed on the opposite side. The angle made by the image with the principal axis is

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