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A plano-convex lens is made of a material with refractive index n. A small object placed 30 cm in front of the curved (convex) surface forms an image that is twice the size of the object. Reflection from the convex surface also produces a faint image 10 cm from the lens. Which of the following statements are correct? (Multiple may be true.)

1. The refractive index of the lens is 2.5.
2. The radius of curvature of the convex surface is 45 cm.
3. The faint image is erect and real.
4. The focal length of the lens is 20 cm.

Correct answer: The refractive index of the lens is 2.5.

Solution

For the faint image (reflection from convex surface acting as convex mirror): mirror formula 1/v + 1/u = 1/f = 2/R. Object at u = -30 cm, image behind mirror so v = +10 cm (virtual, erect). 1/10 - 1/30 = 2/R => 2/30 = 2/R => R = 30 cm. Faint image: convex mirror gives virtual erect image, so statement C ('erect and real') is FALSE. For refraction (object in air, image in glass): n1=1, n2=n, u=-30. Magnification = (n1*v)/(n2*u)= v/(n*(-30)) = -2 (real, inverted, twice size) => v = 60n. Refraction formula: n/v - 1/u = (n-1)/R => n/(60n) - 1/(-30) = (n-1)/30 => 1/60 + 1/30 = (n-1)/30 => 3/60 = (n-1)/30 => 1/20 = (n-1)/30 => n-1 = 3/2 => n = 2.5. Statement A is TRUE. R = 30 cm, statement B (R=45 cm) is FALSE. Focal length of lens: 1/f = (n-1)(1/R - 1/inf) = (1.5)(1/30) = 1/20, so f = 20 cm. Statement D is TRUE.

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