JEE Main · 2024 · 29 Jan 2024 · Shift 2 · Geometrical Optics

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If the distance between object and its two times magnified virtual image produced by a curved mirror is 15 cm, the focal length of the mirror must be :

(A)15 cm
(B)-12 cm
(C)-10 cm✓ Correct
(D) $10/3$ cm

Answer: (C)

Solution

Let the object sit at distance

u_{0}

in front of the mirror and the virtual, magnified image sit at distance

v_{0}

behind the mirror. Since a virtual image is erect, the magnification is positive:

m = \frac{v _{0}}{u _{0}} = 2

Because the object and image lie on opposite sides of the mirror's pole, the straight-line distance between them is the sum of their distances from the pole:

u_{0} + v_{0} = 15

Substituting

v_{0} = 2 u_{0}

u_{0} + 2 u_{0} = 15 ⟹ u_{0} = 5 cm, v_{0} = 10 cm

Applying the sign convention (object distance negative, virtual image distance positive):

u = - 5

cm,

v = + 10

cm.

\frac{1}{f} = \frac{1}{v} + \frac{1}{u} = \frac{1}{10} - \frac{1}{5} = - \frac{1}{10}

f = - 10 cm

The mirror is concave, with focal length

- 10

cm.

JEE Main · 2024 · 29 Jan 2024 · Shift 2 · Single correct option

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JEE Main · 2024 · 29 Jan 2024 · Shift 2 · Geometrical Optics

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