A plane mirror is placed 22.5; cm in front of | vedclass

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Question

(C) As shown in the figure,if the object is placed at a distance $x$ from the concave mirror,its distance from the plane mirror will be $(22.5 - x)$.

Vedclass · Accepted Answer

$-2$

Vedclass · Answer

(C) As shown in the figure,if the object is placed at a distance $x$ from the concave mirror,its distance from the plane mirror will be $(22.5 - x)$. The plane mirror will form an equal and erect image of the object at a distance $(22.5 - x)$ behind the plane mirror.According to the problem,the image formed by the concave mirror coincides with the image formed by the plane mirror. Therefore,for the concave mirror:$v = -[22.5 + (22.5 - x)] = -(45 - x)$Using the mirror formula $\frac{1}{v} + \frac{1}{u} = \frac{1}{f}$:$\frac{1}{-(45 - x)} + \frac{1}{-x} = \frac{1}{-10}$$\frac{45}{45x - x^2} = \frac{1}{10}$$x^2 - 45x + 450 = 0$$(x - 30)(x - 15) = 0$$x = 30\; cm$ or $x = 15\; cm$.Since the distance between the two mirrors is $22.5\; cm$,$x = 30\; cm$ is not possible. Thus,the object must be at a distance of $15\; cm$ from the concave mirror.$v = -(45 - 15) = -30\; cm$.Magnification $m = -\frac{v}{u} = -\frac{-30}{-15} = -2$.

$A$ plane mirror is placed $22.5\; cm$ in front of a concave mirror of focal length $10\; cm$. An object is placed between the two mirrors such that the images formed by both mirrors coincide. Find the magnification of the image produced by the concave mirror.

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