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

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(B) Let the object be placed at a distance $x$ from the concave mirror. The distance of the object from the plane mirror is $(22.5 - x)$. The plane mi

Vedclass · Accepted Answer

$15$

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(B) Let the object be placed at a distance $x$ from the concave mirror. The distance of the object from the plane mirror is $(22.5 - x)$. The plane mirror forms an image at a distance $(22.5 - x)$ behind it. The total distance of this image from the concave mirror is $v = -(22.5 + (22.5 - x)) = -(45 - x)$.Using the mirror formula for the concave mirror: $\frac{1}{v} + \frac{1}{u} = \frac{1}{f}$.Here,$u = -x$,$f = -10\;cm$,and $v = -(45 - x)$.Substituting these values: $\frac{1}{-(45 - x)} + \frac{1}{-x} = \frac{1}{-10}$.$\frac{1}{45 - x} + \frac{1}{x} = \frac{1}{10}$.$\frac{x + 45 - x}{x(45 - x)} = \frac{1}{10}$.$\frac{45}{45x - x^2} = \frac{1}{10}$.$450 = 45x - x^2$.$x^2 - 45x + 450 = 0$.$(x - 30)(x - 15) = 0$.So,$x = 30\;cm$ or $x = 15\;cm$.Since the distance between the mirrors is $22.5\;cm$,$x = 30\;cm$ is not possible. Therefore,the object must be at a distance of $15\;cm$ from the concave mirror.

$A$ plane mirror is placed $22.5\;cm$ in front of a concave mirror of focal length $10\;cm$. An object can be placed between the two mirrors so that the first images formed by both the mirrors coincide. The object distance from the concave mirror is (in $;cm$)

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