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(B) $1$. First,find the image formed by the plane mirror. The object $AB$ is at a distance of $50 \ cm$ from the plane mirror. The plane mirror forms

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The second image is virtual and erect with magnification $1/5$.

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(B) $1$. First,find the image formed by the plane mirror. The object $AB$ is at a distance of $50 \ cm$ from the plane mirror. The plane mirror forms a virtual image $A'B'$ at a distance of $50 \ cm$ behind it. Since the distance between the plane mirror and the convex mirror is $50 \ cm$,the image $A'B'$ acts as an object for the convex mirror.$2$. The distance of $A'$ from the convex mirror is $50 \ cm + 10 \ cm = 60 \ cm$ (in front of the mirror,so $u_A = -60 \ cm$).$3$. The distance of $B'$ from the convex mirror is $50 \ cm + 40 \ cm = 90 \ cm$ (in front of the mirror,so $u_B = -90 \ cm$).$4$. The focal length of the convex mirror is $f = +120 \ cm / 2 = +60 \ cm$.$5$. Using the mirror formula $\frac{1}{v} + \frac{1}{u} = \frac{1}{f}$:For $A'$: $\frac{1}{v_A} + \frac{1}{-60} = \frac{1}{60} \Rightarrow \frac{1}{v_A} = \frac{2}{60} \Rightarrow v_A = +30 \ cm$.For $B'$: $\frac{1}{v_B} + \frac{1}{-90} = \frac{1}{60} \Rightarrow \frac{1}{v_B} = \frac{1}{60} + \frac{1}{90} = \frac{3+2}{180} = \frac{5}{180} \Rightarrow v_B = +36 \ cm$.$6$. The image $A''B''$ is formed at distances $30 \ cm$ and $36 \ cm$ behind the convex mirror. Since the image is formed behind the mirror,it is virtual and erect.$7$. The length of the image is $L_i = 36 - 30 = 6 \ cm$. The length of the object $AB$ is $30 \ cm$. Magnification $m = \frac{L_i}{L_o} = \frac{6}{30} = 1/5$.$8$. Thus,the image is virtual,erect,and has a magnification of $1/5$.

In the figure shown,consider the first reflection at the plane mirror and the second at the convex mirror. $AB$ is the object.

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