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(D) For a concave mirror,the image position coincides with the object when the object is placed at the centre of curvature. Thus,the radius of curvatu

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$3/2$

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(D) For a concave mirror,the image position coincides with the object when the object is placed at the centre of curvature. Thus,the radius of curvature $R = 0.5\, m$.When the mirror is placed in a liquid,the light from the object (pin) travels through the liquid to reach the mirror. The apparent depth of the object as seen by the mirror must be equal to the radius of curvature $R = 0.5\, m$ for the image to coincide with the object.The pin is placed at a distance of $0.4\, m$ from the mirror. This distance consists of $0.2\, m$ of air and $0.2\, m$ of liquid.The apparent distance $d_{app}$ of the object from the mirror is given by the sum of the real distance in air and the apparent distance in the liquid:$d_{app} = 0.2\, m + \frac{0.2\, m}{\mu} = 0.5\, m$$\frac{0.2}{\mu} = 0.5 - 0.2 = 0.3$$\mu = \frac{0.2}{0.3} = \frac{2}{3}$Wait,re-evaluating the physics: The light travels from the pin through the air and then through the liquid to the mirror. The distance in air is $0.2\, m$ and the distance in liquid is $0.2\, m$. The apparent distance of the object from the mirror is $d_{app} = 0.2 + \mu(0.2) = 0.5\, m$.$0.2\mu = 0.5 - 0.2 = 0.3$$\mu = \frac{0.3}{0.2} = 1.5 = 3/2$.

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