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(B) For the refraction case,the magnification $m = -v/u = 2$. Since the object is real,$u = -30 \ cm$,so $v = -2u = 60 \ cm$. Using the lens formula $

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$A, D$

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(B) For the refraction case,the magnification $m = -v/u = 2$. Since the object is real,$u = -30 \ cm$,so $v = -2u = 60 \ cm$. Using the lens formula $1/v - 1/u = 1/f_1$,we get $1/60 - 1/(-30) = 1/f_1$,which gives $f_1 = 20 \ cm$.For the reflection case,the curved surface acts as a concave mirror. The image is formed at $10 \ cm$ from the lens. Using the mirror formula $1/v + 1/u = 1/f_2$,where $u = -30 \ cm$ and $v = -10 \ cm$,we get $1/(-10) + 1/(-30) = 1/f_2$,so $f_2 = -7.5 \ cm$. Since $f_2 = -R/2$,the radius of curvature $R = 15 \ cm$.Using the lens maker's formula for a plano-convex lens,$1/f_1 = (n-1)/R$. Substituting $f_1 = 20 \ cm$ and $R = 15 \ cm$,we get $1/20 = (n-1)/15$,which implies $n-1 = 0.75$,so $n = 1.75$.Re-evaluating the provided options based on the calculation: The correct focal length is $20 \ cm$ (Option $D$). The refractive index is $1.75$. The faint image formed by reflection in a concave mirror at $v = -10 \ cm$ for $u = -30 \ cm$ is real and inverted. Thus,only statement $D$ is true based on the standard derivation.

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