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(C) The distance between the two mirrors is $d = 6\,cm + 15\,cm = 21\,cm$. Let $x_1 = 6\,cm$ (distance from $M_1$) and $x_2 = 15\,cm$ (distance from $

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$78\,cm$ from $M_1$

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(C) The distance between the two mirrors is $d = 6\,cm + 15\,cm = 21\,cm$. Let $x_1 = 6\,cm$ (distance from $M_1$) and $x_2 = 15\,cm$ (distance from $M_2$).Images formed by $M_1$ are at distances:$I_1 = x_1 = 6\,cm$$I_3 = x_1 + 2d = 6 + 42 = 48\,cm$$I_5 = x_1 + 4d = 6 + 84 = 90\,cm$Images formed by $M_2$ are at distances from $M_2$:$I_2 = x_2 = 15\,cm$$I_4 = x_2 + 2d = 15 + 42 = 57\,cm$To find the position of the $4^{th}$ image formed by $M_1$,we look at the sequence of images. The images formed by $M_1$ are at $6\,cm, 48\,cm, 90\,cm, \dots$ from $M_1$. The images formed by $M_2$ are at $15\,cm, 57\,cm, 99\,cm, \dots$ from $M_2$. Converting the $M_2$ image distances to be relative to $M_1$:$I_2$ is at $15 + 21 = 36\,cm$ from $M_1$.$I_4$ is at $57 + 21 = 78\,cm$ from $M_1$.Ordering the images by distance from $M_1$: $6\,cm (M_1), 36\,cm (M_2), 48\,cm (M_1), 78\,cm (M_2), 90\,cm (M_1), \dots$The $4^{th}$ image is at $78\,cm$ from $M_1$.

An object $O$ is placed between two parallel plane mirrors $M_1$ and $M_2$. The distance of the object from $M_1$ is $6\,cm$ and from $M_2$ is $15\,cm$. The position of the $4^{th}$ image formed by mirror $M_1$ is:

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