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(A) The shift produced by a slab of thickness $t$ and refractive index $\mu_{slab}$ in a surrounding medium of refractive index $\mu_{surrounding}$ is

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On the object

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(A) The shift produced by a slab of thickness $t$ and refractive index $\mu_{slab}$ in a surrounding medium of refractive index $\mu_{surrounding}$ is given by $\Delta x = t(1 - \frac{\mu_{surrounding}}{\mu_{slab}})$.The total shift produced by the three slabs is:$\Delta x_{total} = 45(1 - \frac{4/3}{1.5}) + 24(1 - \frac{4/3}{1}) + 54(1 - \frac{4/3}{1.5})$$= 45(1 - \frac{4/3}{3/2}) + 24(1 - 4/3) + 54(1 - \frac{4/3}{3/2})$$= 45(1 - 8/9) + 24(-1/3) + 54(1 - 8/9)$$= 45(1/9) - 8 + 54(1/9)$$= 5 - 8 + 6 = 3 \, cm$.The real distance of the object from the mirror is $d_{real} = 20 + 45 + 24 + 54 + 10 = 153 \, cm$.The apparent distance of the object from the mirror is $d_{apparent} = d_{real} - \Delta x_{total} = 153 - 3 = 150 \, cm$.Since the apparent distance is equal to the radius of curvature $(R = 150 \, cm)$,the object appears to be at the center of curvature. Therefore,the final image is formed at the object itself.

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