A convex spherical mirror produces an image that is one-third the size of an object placed… — JAMB UTME

A convex spherical mirror produces an image that is one-third the size of an object placed in front of it. If the distance of the object from the mirror is doubled, what is the new linear magnification of the image?

$\frac{1}{5}$ Cevap
B
$\frac{1}{4}$
C
$\frac{1}{6}$
D
$\frac{1}{2}$

Cevap

The new linear magnification of the image is

\frac{1}{5}

For a convex mirror, the linear magnification

m

relates object distance

u

and focal magnitude

f

m = \frac{f}{f + u}

. Given

m = \frac{1}{3}

, solving

\frac{1}{3} = \frac{f}{f + u}

yields

u = 2f

. When the object distance is doubled to

u' = 4f

, the new magnification becomes

m' = \frac{f}{f + 4f} = \frac{1}{5}

Adım Adım Çözüm

Express linear magnification in terms of object distance and focal length for a convex mirror

m = \frac{f}{f + u}

For a convex mirror, the focal length is negative under the Cartesian sign convention, making the virtual image distance

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

, so magnification

m = -\frac{v}{u} = \frac{f}{f + u}

Substitute the initial magnification

m = \frac{1}{3}

to express initial object distance

u

in terms of focal length

f

\frac{1}{3} = \frac{f}{f + u} \implies f + u = 3f \implies u = 2f

This establishes that the object was originally located at a distance equal to twice the focal length of the mirror.

Calculate the new object distance

u'

when distance is doubled

u' = 2u = 2(2f) = 4f

The problem states the object distance from the mirror is doubled.

Compute the new linear magnification

m'

m' = \frac{f}{f + u'} = \frac{f}{f + 4f} = \frac{f}{5f} = \frac{1}{5}

Substituting

u' = 4f

into the magnification formula yields the final reduced magnification.

Anahtar Kavram

Linear magnification and sign convention for convex mirrors

Tahmini Süre:2m 0s

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