A slab of glass,of thickness 6, cm and refrac | vedclass

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(C) When an object is placed in front of a concave mirror such that the reflected image coincides with the object,the rays must strike the mirror norm

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$42$

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(C) When an object is placed in front of a concave mirror such that the reflected image coincides with the object,the rays must strike the mirror normally. This means the rays must appear to come from the center of curvature $C$ of the mirror after passing through the glass slab.Let the distance of the object from the mirror be $u$. The glass slab causes a normal shift $\Delta x$ in the position of the object,given by $\Delta x = t \left( 1 - \frac{1}{\mu} \right)$,where $t = 6\, cm$ and $\mu = 1.5$.$\Delta x = 6 \left( 1 - \frac{1}{1.5} \right) = 6 \left( 1 - \frac{2}{3} \right) = 6 \left( \frac{1}{3} \right) = 2\, cm$.The apparent position of the object must coincide with the center of curvature $C$ of the mirror. The distance of the center of curvature from the mirror is equal to the radius of curvature $R = 40\, cm$.Therefore,the apparent distance of the object from the mirror is $40\, cm$.The actual distance of the object from the mirror is $u = R + \Delta x = 40\, cm + 2\, cm = 42\, cm$.

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