An object is placed 15.0\text{ cm} in front of a thin diverging lens with a focal length o… — JAMB UTME

An object is placed $15.0\text{ cm}$ in front of a thin diverging lens with a focal length of $10.0\text{ cm}$ . What is the image distance formed by the lens?

$-6.0\text{ cm}$ ( $6.0\text{ cm}$ on the same side as the object)Cevap
B
$+30.0\text{ cm}$ ( $30.0\text{ cm}$ on the opposite side of the object)
C
$+6.0\text{ cm}$ ( $6.0\text{ cm}$ on the opposite side of the object)
D
$-30.0\text{ cm}$ ( $30.0\text{ cm}$ on the same side as the object)

Cevap

The image distance is

-6.0\text{ cm}

, indicating a virtual image located

6.0\text{ cm}

in front of the lens on the same side as the object.

Using the thin lens equation

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

with

u = +15.0\text{ cm}

and

f = -10.0\text{ cm}

for the diverging lens gives

\frac{1}{v} = -\frac{1}{10} - \frac{1}{15} = -\frac{1}{6}

, resulting in

v = -6.0\text{ cm}

. The negative sign confirms the image is virtual and formed on the same side as the object.

Adım Adım Çözüm

Identify the given optical parameters and apply the proper sign convention.

Object distance

u = +15.0\text{ cm}

; Focal length for a diverging lens

f = -10.0\text{ cm}

By the real-is-positive sign convention, real object distance

u

is positive, while the focal length

f

of a concave/diverging lens is strictly negative.

Set up the thin lens formula

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

and solve for

\frac{1}{v}

\frac{1}{-10.0} = \frac{1}{15.0} + \frac{1}{v} \implies \frac{1}{v} = -\frac{1}{10.0} - \frac{1}{15.0}

Isolating the reciprocal image distance term requires subtracting

\frac{1}{u}

from both sides.

Calculate the common denominator and evaluate

v

\frac{1}{v} = \frac{-3 - 2}{30.0} = -\frac{5.0}{30.0} = -\frac{1}{6.0} \implies v = -6.0\text{ cm}

Inverting the reciprocal yields the final signed image distance.

Anahtar Kavram

Thin Lens Formula and Sign Conventions for Diverging Lenses

Tahmini Süre:1m 15s

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