A flat glass slab of thickness 6, cm and refr | vedclass

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(B) The glass slab causes a normal shift in the position of the object (the observer) as seen by the mirror. The shift is given by $\Delta t = t(1 - \

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$96$

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(B) The glass slab causes a normal shift in the position of the object (the observer) as seen by the mirror. The shift is given by $\Delta t = t(1 - \frac{1}{\mu}) = 6(1 - \frac{1}{1.5}) = 6(1 - \frac{2}{3}) = 6(\frac{1}{3}) = 2\, cm$.The observer is at a distance of $50\, cm$ from the mirror. Due to the slab,the mirror perceives the observer at a distance of $50 - 2 = 48\, cm$.The mirror forms an image at a distance of $48\, cm$ behind it. The light rays from this image must pass through the slab again to reach the observer,causing another shift of $2\, cm$ in the direction of the light.The total distance of the image from the observer is the sum of the distance from the observer to the mirror $(50\, cm)$ and the distance from the mirror to the image $(48\, cm)$,minus the effective shift caused by the slab on the return path.Distance $= 50 + 48 - 2 = 96\, cm$.

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