An object that is 3 cm tall is positioned at a distance of 7.5 cm from a convex mirror with a focal length of 6 cm . Determine the location, size, and nature of the image formed.

Correct answer: D. (10)/(3) cm behind the mirror, 1.33 cm , virtual and erect Using the mirror formula, (1)/(v) + (1)/(u) = (1)/(f) Given u = -7.5 cm and f = +6 cm for a convex mirror: (1)/(v) = (1)/(6) - (1)/(-7.5) = (1)/(6) + (1)/(7.5) = (1)/(6) + (2)/(15) (1)/(v) = (5 + 4)/(30) = (9)/(30) = (3)/(10) v = (10)/(3) cm Since v is positive, the image is formed (10)/(3) cm behind the mirror. Magnification m = -(v)/(u) = -((10)/(3))/(-7.5) = (10)/(3 × 7.5) = (10)/(22.5) = (4)/(9) Height of the image h_i = m × h_o = (4)/(9) × 3 cm = (4)/(3) cm 1.33 cm Since m is positive, the image is virtual and erect.

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

An object that is 3 cm tall is positioned at a distance of 7.5 cm...

An object that is

3 \text{ cm}

tall is positioned at a distance of

7.5 \text{ cm}

from a convex mirror with a focal length of

6 \text{ cm}

. Determine the location, size, and nature of the image formed.

Options

A $\dfrac{10}{3} \text{ cm}$ in front of the mirror, $1.33 \text{ cm}$ , real and inverted
B $30 \text{ cm}$ in front of the mirror, $12 \text{ cm}$ , real and inverted
C $30 \text{ cm}$ behind the mirror, $12 \text{ cm}$ , virtual and erect
D $\dfrac{10}{3} \text{ cm}$ behind the mirror, $1.33 \text{ cm}$ , virtual and erect

Correct answer

D. $\dfrac{10}{3} \text{ cm}$ behind the mirror, $1.33 \text{ cm}$ , virtual and erect

Step-by-step solution

Using the mirror formula,

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

Given

u = -7.5 \text{ cm}

and

f = +6 \text{ cm}

for a convex mirror:

\dfrac{1}{v} = \dfrac{1}{6} - \dfrac{1}{-7.5} = \dfrac{1}{6} + \dfrac{1}{7.5} = \dfrac{1}{6} + \dfrac{2}{15}

\dfrac{1}{v} = \dfrac{5 + 4}{30} = \dfrac{9}{30} = \dfrac{3}{10}

v = \dfrac{10}{3} \text{ cm}

Since

v

is positive, the image is formed

\dfrac{10}{3} \text{ cm}

behind the mirror.

Magnification

m = -\dfrac{v}{u} = -\dfrac{\dfrac{10}{3}}{-7.5} = \dfrac{10}{3 \times 7.5} = \dfrac{10}{22.5} = \dfrac{4}{9}

Height of the image

h_i = m \times h_o = \dfrac{4}{9} \times 3 \text{ cm} = \dfrac{4}{3} \text{ cm} \approx 1.33 \text{ cm}

Since

m

is positive, the image is virtual and erect.

Practice Geometrical Optics on Quantrex Academy →

More from Geometrical Optics