A concave mirror forms a real image that is twice the size of an object. When the object i… — JAMB UTME

A concave mirror forms a real image that is twice the size of an object. When the object is shifted $10\text{ cm}$ closer to the mirror, a virtual image of the same magnification is produced. What is the focal length of the mirror?

$10\text{ cm}$ Answer
B
$15\text{ cm}$
C
$20\text{ cm}$
D
$30\text{ cm}$

Answer

The focal length of the concave mirror is

10\text{ cm}

For a concave mirror forming a real image of magnification

2

v_1 = 2u_1

, yielding

u_1 = 1.5f

. When the object is moved

10\text{ cm}

closer, a virtual image of magnification

2

is formed, so

v_2 = -2u_2

, yielding

u_2 = 0.5f

. Subtracting the two object positions (

1.5f - 0.5f = 10\text{ cm}

) directly gives

f = 10\text{ cm}

Step-by-Step Solution

Set up the mirror equation for the first case (real image).

u_1 = \frac{3}{2}f

For a real inverted image with magnification

m = 2

v_1 = +2u_1

. Substituting into

\frac{1}{f} = \frac{1}{u_1} + \frac{1}{v_1}

gives

\frac{1}{f} = \frac{1}{u_1} + \frac{1}{2u_1} = \frac{3}{2u_1}

Set up the mirror equation for the second case (virtual image).

u_2 = \frac{1}{2}f

For a virtual erect image with magnification

m = 2

, sign convention dictates

v_2 = -2u_2

. Substituting into

\frac{1}{f} = \frac{1}{u_2} + \frac{1}{v_2}

gives

\frac{1}{f} = \frac{1}{u_2} - \frac{1}{2u_2} = \frac{1}{2u_2}

Use the given displacement between the two object positions to solve for

f

f = 10\text{ cm}

The object is moved

10\text{ cm}

closer, so

u_1 - u_2 = 10\text{ cm}

. Substituting the expressions yields

\frac{3}{2}f - \frac{1}{2}f = 10\text{ cm} \implies f = 10\text{ cm}

Key Concept

Mirror Formula and Sign Convention for Spherical Mirrors

Estimated Time:2m 0s

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