A plano-covex lens is made of a material of refractive index n. When a small object is placed 30cm away in front of the curved surface of the lens, an image of double the size of the object is produced. Due to reflection from the convex surface of the lens, another faint image is observed at a distance of 10 cm away from the lens. Which of the following statement (s) is (are) true?

Options

A. refractive index of lens is `2.5`
B. radius of curvature of convex surface is `45 cm`
C. faint image is erect and real
D. focal length of the lens is `20 cm`

Answer

A, D

Explanation

As image is of double the size of object,
`v = 2 u = - 60 cm`.
Here, `u = - 30 cm`

As `(1)/(f_(1)) = (1)/(v) - (1)/(u) = (1)/(60) + (1)/(30) = (1)/(20) :. f_(1) = 20 cm`.
In 2nd case, convex surface of lens acts as a convex mirror forming a faint image at a distance of `10 cm`.
`u = - 30 cm, v = 10 cm`

using mirror formula
`(1)/(f_(2)) = (1)/(v) + (1)/(u) = (1)/(-30) + (1)/(10) = (2)/(30) = (1)/(15)`
`f_(2) = 15 cm, R = 2 f_(2) = 30 cm`
From `(1)/(f_(1)) = (mu - 1)((1)/(R_(1)) - (1)/(R_(2))) = (mu - 1)/(R )`
`(because R_(2) = 00 and R_(1) = R)`
`mu = 1 = (R )/(f_(1)) = (30 cm)/(20 cm) = 1.5`
`mu = 1.5 + 1 = 2.5`
Choice (a) and (b) are correct.