A particle travels in a circle of radius 2.0 cm . A concave mirror having a focal length of 20 cm is positioned so that its principal axis passes through the centre of the circle and is perpendicular to its plane. The

Correct answer: D. 4.0 cm Using the mirror formula, (1)/(v) + (1)/(u) = (1)/(f) Given u = -30 cm and f = -20 cm (1)/(v) = (1)/(-20) - (1)/(-30) = (-3 + 2)/(60) = -(1)/(60) v = -60 cm The transverse magnification is given by m = -(v)/(u) m = -(-60)/(-30) = -2 The radius of the circle traced by the image is the transverse height of the image. R_ image = |m| × R_ object R_ image = |-2| × 2.0 cm = 4.0 cm

Concepts Of Physics MCQ Edition [Volume 1]

Exam: Concepts Of Physics MCQ Edition [Volume 1]
Subject: Physics
Chapter: Geometrical Optics

Concepts Of Physics MCQ Edition [Volume 1] · Physics · Geometrical Optics

A particle travels in a circle of radius 2.0 cm . A concave mirror...

A particle travels in a circle of radius

2.0 \text{ cm}

. A concave mirror having a focal length of

20 \text{ cm}

is positioned so that its principal axis passes through the centre of the circle and is perpendicular to its plane. The distance from the pole of the mirror to the centre of the circle is

30 \text{ cm}

. Determine the radius of the circle traced by the image.

Options

A $1.0 \text{ cm}$
B $8.0 \text{ cm}$
C $6.0 \text{ cm}$
D $4.0 \text{ cm}$

Correct answer

D. $4.0 \text{ cm}$

Step-by-step solution

Using the mirror formula,

\dfrac{1}{v} + \dfrac{1}{u} = \dfrac{1}{f}

Given

u = -30 \text{ cm}

and

f = -20 \text{ cm}

\dfrac{1}{v} = \dfrac{1}{-20} - \dfrac{1}{-30} = \dfrac{-3 + 2}{60} = -\dfrac{1}{60}

v = -60 \text{ cm}

The transverse magnification is given by

m = -\dfrac{v}{u}

m = -\dfrac{-60}{-30} = -2

The radius of the circle traced by the image is the transverse height of the image.

R_{image} = |m| \times R_{object}

R_{image} = |-2| \times 2.0 \text{ cm} = 4.0 \text{ cm}

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