The distance between an object and its image | vedclass

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Question

(A) Given magnification $M = -\frac{1}{3}$.Since $M = -\frac{v}{u}$,we have $-\frac{v}{u} = -\frac{1}{3}$,which implies $v = \frac{u}{3}$.The distance

Vedclass · Accepted Answer

$45$

Vedclass · Answer

(A) Given magnification $M = -\frac{1}{3}$.Since $M = -\frac{v}{u}$,we have $-\frac{v}{u} = -\frac{1}{3}$,which implies $v = \frac{u}{3}$.The distance between the object and the image is given by $|u - v| = 30 \ cm$. Since the image is real and inverted (magnification is negative),both object and image are on the same side of the mirror. Thus,$|u| - |v| = 30 \ cm$. Let $u = -u_0$ and $v = -v_0$,then $u_0 - v_0 = 30$.Substituting $v_0 = \frac{u_0}{3}$,we get $u_0 - \frac{u_0}{3} = 30 \Rightarrow \frac{2u_0}{3} = 30 \Rightarrow u_0 = 45 \ cm$.So,$u = -45 \ cm$ and $v = -15 \ cm$.Using the mirror formula $\frac{1}{f} = \frac{1}{v} + \frac{1}{u}$:$\frac{1}{f} = \frac{1}{-15} + \frac{1}{-45} = \frac{-3 - 1}{45} = -\frac{4}{45}$.Therefore,$f = -\frac{45}{4} \ cm$.The magnitude of the focal length is $\frac{45}{4} \ cm$. Comparing this with $\frac{x}{4} \ cm$,we get $x = 45$.

The distance between an object and its image (magnified by $-\frac{1}{3}$) is $30 \ cm$. The focal length of the mirror used is $\left(\frac{x}{4}\right) \ cm$,where the magnitude of the value of $x$ is . . . . . . .

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