A denser medium of refractive index 1.5 has a concave surface with respect to air of radius of curvature 12 cm. An object is situated in the denser medium at a distance of 9 cm from the pole locate the image due to refraction in air

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A real image at 8 cm

a virtual image at 8 cm

A real image at 4.8 cm

A virtual image at 4.8 cm

answer is D.

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Detailed Solution

μ2v-μ1u=μ2-μ1R;1v+1.59=1-1.512

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