An extended object is placed at a distance of 5.0 cm from a convex lens of focal length 8.0 cm . Find the position of the image from the lens formula, applying the standard Cartesian sign convention (with the object

Correct answer: B. -(40)/(3) cm Given the object distance u = -5.0 cm and the focal length of the convex lens f = +8.0 cm . Using the lens formula: (1)/(v) - (1)/(u) = (1)/(f) (1)/(v) - (1)/(-5.0) = (1)/(8.0) (1)/(v) + (1)/(5.0) = (1)/(8.0) (1)/(v) = (1)/(8.0) - (1)/(5.0) (1)/(v) = (5 - 8)/(40) = -(3)/(40) v = -(40)/(3) 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

An extended object is placed at a distance of 5.0 cm from a convex...

An extended object is placed at a distance of

5.0 \text{ cm}

from a convex lens of focal length

8.0 \text{ cm}

. Find the position of the image from the lens formula, applying the standard Cartesian sign convention (with the object placed at

u = -5.0 \text{ cm}

Options

A $+\dfrac{40}{3} \text{ cm}$
B $-\dfrac{40}{3} \text{ cm}$
C $+\dfrac{40}{13} \text{ cm}$
D $-\dfrac{40}{13} \text{ cm}$

Correct answer

B. $-\dfrac{40}{3} \text{ cm}$

Step-by-step solution

Given the object distance

u = -5.0 \text{ cm}

and the focal length of the convex lens

f = +8.0 \text{ cm}

.

Using the lens formula:

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

\dfrac{1}{v} - \dfrac{1}{-5.0} = \dfrac{1}{8.0}

\dfrac{1}{v} + \dfrac{1}{5.0} = \dfrac{1}{8.0}

\dfrac{1}{v} = \dfrac{1}{8.0} - \dfrac{1}{5.0}

\dfrac{1}{v} = \dfrac{5 - 8}{40} = -\dfrac{3}{40}

v = -\dfrac{40}{3} \text{ cm}

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