JEE Advanced · 2016 · 22 May 2016 · Paper 1 · Geometrical Optics

A plano-convex lens is made of a material of refractive index

n

. When a small object is placed 30 cm 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?
(A) The refractive index of the lens is 2.5
(B) The radius of curvature of the convex surface is 45 cm
(C) The faint image is erect and real
(D) The focal length of the lens is 20 cm

(A)The refractive index of the lens is 2.5✓ Correct
(B)The radius of curvature of the convex surface is 45 cm
(C)The faint image is erect and real
(D)The focal length of the lens is 20 cm✓ Correct

Answer: (A), (D)

Solution

For refraction (lens action): With magnification

m = - 2

and

u = - 30

cm, we have

v = - 2 u = 60

cm.

Using the lens equation:

\frac{1}{v} - \frac{1}{u} = \frac{1}{f}

\frac{1}{60} - \frac{1}{- 30} = \frac{1}{f}

, giving

f = 20

cm. (D) is correct.

For reflection (mirror action): With

u = - 30

cm and

v = 10

cm (virtual image):

\frac{1}{v} + \frac{1}{u} = \frac{2}{R}

\frac{1}{10} + \frac{1}{- 30} = \frac{2}{R}

, giving

R = 30

cm. (B) is incorrect.

For the lens:

(n - 1) (\frac{1}{R}) = \frac{1}{f}

(n - 1) (\frac{1}{30}) = \frac{1}{20}

, giving

n = 2.5

. (A) is correct.

The faint reflected image is erect and virtual (not real). (C) is incorrect.

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JEE Advanced · 2016 · 22 May 2016 · Paper 1 · Geometrical Optics

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